Chapter y. 8. n cd (x y) 14. (2a b) 15. (a) 3(x 2y) = 3x 3(2y) = 3x 6y. 16. (a)


 Stuart Shepherd
 3 years ago
 Views:
Transcription
1 Chapter 6 Chapter 6 opener A. B. C. D. 6 E. 5 F. 8 G. H. I. J y 8. n 9. w z. 5cd.. xy z 5r s t. (x y). (a b) 5. (a) (x y) = x (y) = x 6y x 6y = x (y) = (x y) 6. (a) a (5 a+ b) = a (5 a) + a ( b) = 5a + a b 5a + a b= 5 a a+ a b= a (5 a+ b) 7. p + = p + = (p + 8. q 5 = q 5 = (q Section 6. Practice Exercises. (a) product prime (c) greatest common factor (d) prime (e) greatest common factor (GCF) (f) grouping.,,,, 6, 8,, c c+ 5= 5 c 5 c+ 5 = 5( c c+ 6d + d + d = (8 d)( d ) + (8 d)( d) + (8 d)( = 8 d(d + d + 5 x + x = x x + x = x ( x + y y = y y y = y ( y)
2 9 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual. t t+ 8t = tt t+ 8 tt = t( t + 8 t) 5 9y + y y = y( y + y r r + r = 7r r r + r r = r (7 r + r) ab + a b = ab + ab( a ) = ab(+ a ) t 9t = (8t + 9t+ ) 6x x 5 = ( (6x + x+ 5 p p = 5 p ( p+ ) 6. 5u v 5uv = 5 uv( u v) 5uv = 5 uv( u v. m m = m ( + m) x y 9x y = 9 x y() 9 x y( y ) = 9 x y( y ) 5 ab + 6ab= ab(5 ab) + ab( = ab(5ab +. pt+ pt + 6 pt = pt(6p pt t). mn + 6mn 9mn = mn( m n m + n) (a) x y 8xy z = 6 xy ( x y z) 7 7 5mp q + m q = mq (5p q + m ) 5+ 7y is prime because it is not factorable. w 5u v is prime because it is not factorable. pq + pq 7 pq = 7 pq (6p + p q ) 8mn mn + mn = mn(n 6 n+ m) 5 t + rt t + r t = t ( t + rt t + r ) u v+ 5u v u + 8uv = uuv ( + 5uv u+ 8 v) 8. (a) x x + 8x= x( x + x x x + 8x= x( x x+ 5 9y + y y = y(y y x 6y z = (7x + 6y + z) 6. a + 5b c = (a 5b + c) 7. (a + 6) b(a + 6) = (a + 6)( b) ( x + yx ( + = ( x + (7 y) 8( vw ) + ( w ) = 8 vw ( ) + ( w ) = ( w )(8v+ 5. tr ( + ) + ( r+ ) = tr ( + ) + ( r+ ) = ( r+ )( t x( x+ + 7 x ( x+ = 7 x( x+ ( + x) = 7 xx ( + 5 y( y ) 5 yy ( ) = 5 yy ( )( y 8a ab+ 6ac bc = a( a b) + c( a b) = ( a b)(a+ c) x + x y+ xy + y = x (x+ y) + y (x+ y) = (x+ y)( x + y )
3 Chapter 6 Factoring Polynomials q+ p+ qr+ pr = ( q+ p) + r( q+ p) = ( q+ p)( + r) 56. xy xz+ 7y 7 z = x( y z) + 7( y z) = ( y z)( x+ 7) x + x+ x+ = x(x+ + (x+ = (x+ (x+ ) y + 8y+ 7y+ = y( y+ ) + 7( y+ ) = ( y+ )(y+ 7) t + 6t t = t( t+ + ( ( t+ = (t ( t+ p p p+ = p( p + ( ( p = (p ( p 6y y 9y+ = y(y + ( (y = (y (y 5a + a a = 5 a( a+ 6) + ( )( a+ 6) = ( a+ 6)(5a ) b + b b = b ( b+ + ( ( b+ 5 = ( b+ ( b 8w + w w 5 = w (w + + ( (w + = (w + (w j k + 5k + j + 5= k( j + + ( j + ab 6ac + b c = ( ab c + ( b c = ( b c)(a wx + 7w x = ( j + (k + = 7 w (x + + ( (x + = (x + (7w p x x 9 p + p = x(9p ) + ( p)(9p ) = (9 p )( x p) 69. ay + bx + by + ax = ay + ax + by + bx = ay ( + x) + by ( + x) = ( a+ b)( y+ x) 7. c+ ay+ ac+ 6y = c+ ac+ 6y+ ay = c( + a) + y( + a) = ( c+ y)( + a) vw + w wv = vw vw + w = vw( w + ( w = ( vw + ( w x + 6m+ + x m= 6m+ x m+ + x = m(6 + x ) + (6 + x ) = ( m+ )(6 + x ) 5x + 5x y + x y+ xy = 5 x(x + xy + x y+ y ) = 5 x( x( x + y ) + y( x + y )) = 5 xx ( + y )(x+ y) ab ab+ ab 6b = ba ( a + 6a = ba ( ( a ) + 6( a )) = ba ( )( a + 6) abx b x ab + b = ( bax bx a+ b) = (( b x a b) ( a b)) = ( ba b)( x p q pq rp q + rpq = pq( p q rp+ rq) = pq(( p q) r( p q)) = pq( p q)( r)
4 96 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual st 8st 6t + 8t = 6( tst s t + t ) = 6( tst ( t ( t ) = 6( tt ( s t ) 5 j j k 5 j k + jk = 5 j(j jk jk + k ) = 5 j( j( j k) k ( j k)) = 5 j(j k)( j k ) 79. P = l + w P = (l + w) 8. P = a + b P = (a + b) 8. S = π r + πrh S = π r( r+ h) 8. A = P + Prt A = P( + rt) ( x + x = x + x (6 y y+ = y y (5 w + w+ = w + w+ 9) 5 ( p p+ = p p Answers may vary. For example: 6x + 9x 88. Answers may vary. For example: y y + 7 y 89. Answers may vary. For example: 6 p q + 8p q p q 9. Answers may vary. For example: 8ab ab Section 6. Practice Exercises. (a) positive. different (c) ( x+ ( x = ( x ( x+ = x + x x = x 7x Both are correct. (d) The factorization ( x 6)( x 6) factored further as ( x+ 6)( x+ ). ( x+ 6)( x+ ) is the complete factorization ab 7ab ab = ab(ab 9ab. tt ( 6( t = ( t ( t ). (x ) + 8 x(x ) = (x )( + x) 5. ax + bx 5a b = x( a+ b) 5( a+ b) = ( a+ b)( x m mx pm+ px = mm ( x) pm ( x) = ( m x)( m p) x + x+ 6 = ( x+ 8)( x+ ) y + 8y+ 8 = ( y+ ( y+ 8) z z+ 8 = ( z 9)( z ). w 7w+ = ( w ( w. z z 8 = ( z 6)( z+. w + w = ( w+ 6)( w ) + + can be
5 Chapter 6 Factoring Polynomials 97. p p = ( p 8)( p+. a a+ 9 = ( a 9)( a 5. t + 6t = ( t+ ( t 6. m m+ = ( m ( m 7. x x+ is prime 8. y + 6y+ 8 is prime w w = w 8w+ 65 = ( w ( w 7t+ t + 7 = t + 7t+ 7 = ( t+ 8)( t+ 9) t+ t + 7 = t + t+ 7 = ( t+ 8)( t+ q + q = q + q = ( q ( q+ 9. n + 8n+ 6 = ( n+ ( n+ = ( n+. v + v+ 5 = ( v+ ( v+ = ( v+. a. c. c. b 5. They are both correct because multiplication of polynomials is a commutative operation. 6. They are both correct because multiplication of polynomials is a commutative operation. 7. The expressions are equal and both are correct 8. The expressions are equal and both are correct 9. It should be written in descending order.. To factor a trinomial, write the trinomial in descending order such as x + bx + c. For all factoring problems, always factor out the GCF from all terms.. x+ x = x x = ( x ( x+ ) y 6 + y = y + y 6 = ( y+ )( y 8) 7. x x 7 = ( x x = ( x ( x+ ) 8. z + z 98 = ( z + z 99) = ( z+ ( z 9) p p + p= 8 p( p 5p+ = 8 p( p ( p 5w 5w + 5w = 5 w ( w 7w+ y z y z + 6y z = y z ( y y+ 6) = 5 w ( w )( w = y z ( y 6)( y 6) or y z ( y 6) t u + 6t u + 9t u = t u ( t + 6t+ 9) = t u ( t+ ( t+ or t u ( t+. x + x = ( x x+ = ( x ( x 6). 5. y y 5 = ( y + y+ = ( y+ ( y+ 7) 5a + 5ax+ x = 5( a ax 6 x ) = 5( a x)( a+ x)
6 98 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual 6. m + m+ n = ( m 5m 6 n ) = ( m+ n)( m 6 n) 6. t+ t = t t = ( t 8)( t c 6c= ( c + c+ ) = ( c+ )( c+ d d = ( d + d + = ( d + ( d + x y 9x y + 6xy = xy ( x 9x + 6) = xy ( x ( x y z + 7 yz + 6z 5 = z ( y + 7y+ 6) 5 = z ( y+ ( y+ p 96 p+ 8 = ( p 8 p+ 7) = ( p 7)( p 5w w 5= 5( w 8w 9) = 5( w 9)( w+ 6. x + x+ = ( x+ ( x+ = ( x + 6. z z+ = ( z ( z = ( z 65. t + 8t = ( t+ )( t ) 66. d + d 99 = ( d + ( d 9) 67. The student forgot to factor out the GCF before factoring the trinomial further. The polynomial is not factored completely, because (x has a common factor of. 68. The student forgot to factor out the GCF before factoring the trinomial further. The polynomial is not factored completely, because (5y has a common factor of ( x ( x+ = x + 9x m + m = ( m m+ = ( m ( m x 6x 8= ( x + x+ 7) = ( x+ 9)( x+ 55. c + 6cd + 5 d = ( c+ 5 d)( c+ d) 56. x + 8xy + y = ( x + 6 y)( x + y) 57. a 9ab+ b = ( a b)( a 7 b) 58. m 5mn+ n = ( m n)( m n) 59. a + a + 8 is Prime 6. b 6a + 5 is Prime 6. q+ q 6= q + q 6 = ( q 7)( q+ 9) (a) 7. (a) ( q 7)( q+ = q + q 7 5x+ x + x 5+ x = x + 9x x + 9x = ( x + x ( x )( x ) = + y + y + y+ 6 = y + y+ 6 y + y+ 6= ( y + 6y+ 8) 7. ( y )( y ) = + + x + x + 9 = ( x + ( x + 9) 7. y + y = ( y + 7)( y w + w 5 = ( w + ( w 75.
7 Chapter 6 Factoring Polynomials p p + = ( p 8)( p. w + 5w = (w ( w , 5, 7, 5 78., 7,, For example, c = 6 8. For example, c = 7 Section 6. Practice Exercises. (a) ( x+ ( x = ( x ( x+. = x 8x+ x = x 5x Both are correct. The factorization ( x ( x ) be factored further as ( x ( x ( x ( x+ is the complete factorization. + can +. 5uv u v + 5u v = 5 uv( v u + 5 uv). mn m n + = m( n ( n = ( n ( m ). 5x xy+ y = 5( x ) y( x ) = (5 y)( x ) a 6a a 8 = 6( a 5a = 6( a 7)( a+ ) b + b = ( b + b = ( b+ 6)( b. y y = (y+ ( y ). a + 7a+ 6 = (a+ ( a+ ) 5. 5x x = (5x+ ( x 6. 7y + 9y = (7y ( y+ ) 7. c 5c = (c+ (c ) 8. 6z + z = (z (z w + 7w= w + 7w = (w ( w+ + p + p= p + p = (p+ (5p ) 5q 6+ 6q = 6q 5q 6 = (q+ )(q 7a + a = a + 7a is prime 6b + b = b + 6b is prime 8+ 7x 8x= 7x 8x+ 8 = (7x ( x ) 8+ 5m m= 5m m 8 = (5m+ )(5m 6. 8q + q = (8q ( q+ 7. 6y + 9xy x = (6y 5 x)( y + x) 8. c 9. b. c. n + n+ = (n+ ( n y 7yz + 6 z = ( y z)( y 6 z) m m 8 = ( m 6m = ( m ( m+
8 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual. c c+ 7= ( c c+ = ( c 8)( c. 5 y + y + 6 y = y (y + y+ 6) = y (y+ ( y+ 6) w 5w = ( w+ ( w 8) 6t + 7t = (t+ (t p 9p+ = (p ( p ) u u + u = u (u u+ 6 = u (u ( u 9. m m+ 5 = (m (m = (m. a 5a+ = ( a + 5a = ( a+ 7)( a ). x 7x = ( x + 7x+ = ( x+ ( x+ ) r + r+ 9 = (r+ (r+ = (r+ 5c c+ is prime 7s + s+ 9 is prime 5. 8m + mp+ p = (8m mp p ) = ( m+ p)(m p) x 9xy + y = (x 5 y)(x y) 5 p + pq q = ( p q)(5 p+ q) 6. 6w 55wz+ 5z = (6w + 55wz 5 z ) = ( w+ z)(6w 5 z) m + mn 5 n = (m+ 5 n)( m n) a + 5ab 6 b = (a b)( a+ b) x + x + 9 = ( x + ( x + 9) 7. y + y = ( y ( y + 7) w + w 5 = ( w + ( w p p + = ( p 8)( p.. x 7x 5 = (x + ( x. 5y + y + = (5y + ( y + ) r + 5r = 5(6r + r ) = 5(r+ )(r 6x 8x = (8x 9x ) = (6x+ (x ) s 8 st+ t is prime 6u uv+ 5 v is prime t t 5 = (t (5t+.. z + z 8= ( z z+ 9) = ( z 9)( z 5t + 5t = 5( t t+ 6) = 5( t )( t n + n+ = (8n+ (n+ w + w = (7w (w+ x 6x+ 5 = (6x (x 5. q q+ = ( q 6)( q 7) 65. a a = ( a ( a+ )
9 Chapter 6 Factoring Polynomials b + 6b 7 = ( b+ 7)( b x + 9xy + y = ( x + 5 y)( x + y) p pq+ 6 q = ( p 9 q)( p q) a + ab+ b = ( a+ b)( a+ b) x 7xy 8 y = ( x 8 y)( x + y) t t+ = ( t 7)( t z 5z+ 6 = ( z ( z 6 5 5d + d d = d(5d + d y y + y= y(y y b b 8b= b( b b ) = bb ( + ( b w + w+ = ( w + w+ = ( w+ 7)( w+ x y xy + y = y ( x x+ = y ( x ( x pq pq + q = q( p p+ = q ( p ( p u u + u = u(6u + u = u(u )(u+ 8. (a) ft. 8. (a) 85. (a) h= 6t + t+ () () = = = 6 The height of the rock after sec is 6 6t + t+ = ( t t = ( t+ ( t ) ( )( ) ( )( ) ( )( ) h= t+ 5 t = + 5 = 9 = 6 Yes, the result is the same. h= 6t 8t+ ( ) ( ) = = = The height of the ball after sec is ft. 6t 8t+ = 8( t + t = 8 ( t ( t+ ( )( ) ( )( ) ( )( ) h= 8t 5 t+ = = 8 5 = Yes, the result is the same. x x = ( x ( x+ ) x x+ = ( x 6)( x 8. 8z + 5z + z = z (6z 5z 8. = z (z (z+ 8x + x + = (x + (x + 6y 5y = (y (y (a) 87. (a) 88. (a) x x = ( x ( x+ ) x x+ = ( x ( x x 5x 6 = ( x 6)( x+ x 5x+ 6 = ( x )( x x x+ 9 = ( x 9)( x
10 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual x + x+ 9 = ( x+ 9)( x+ 5. w 9w+ = w 8w w+ = ww ( ) ( w ) = ( w )(w Section 6. Practice Exercises. (a) ( 5x+ ( x = ( x ( 5x+ = x 5x+ x = x x Both are correct. The factorization ( x ( x 6) be factored further as ( x ( x ) ( x+ ( x ) is the complete + can +. factorization.. 5x(x ) (x ) = (x )(5x ). 8(y + + 9y(y + = (y + (8 + 9y). 6ab + b a 8 = 6( ab + b a 8) = 6( ba ( + ( a+ ) = 6( a+ ( b ) 5. and 6. and 7. 8 and 8. and 9. 5 and. and. 9 and. and 6.. x + x+ = x + x+ x+ = ( xx+ + ( x+ = ( x+ (x+ y + 7y+ 6= y + y+ y+ 6 = y(y+ + (y+ = (y+ ( y+ ) 6. p p = p p+ p = p( p ) + ( p ) = ( p )(p+ 7. x + 7x 8= x + 9x x 8 = xx ( + 9) ( x+ 9) = ( x+ 9)( x ) y y y y y 6 = + = yy ( + ( y = ( y ( y+ m + 5m = m + 6m m = mm ( + ( m+ = ( m+ (m 6n + 7n = 6n + 9n n = n(n+ (n+ = (n+ (n 8k 6k 9= 8k k + 6k 9 = k(k + (k = (k (k +. 9h h = 9h 6h+ h = ( h h ) + (h ) = (h )(h+.. k k + 5 = k k k + 5 = k(k 5(k = (k (k = (k 6h + h+ 9 = 6h + h+ h+ 9 = h(h+ + (h+ = (h+ (h+ = (h +
11 Chapter 6 Factoring Polynomials 5. Prime 6. Prime 8. 8t t t = t(t t = ( t t ( t+ ) 7. 9z z+ = 9z 5z 6z+ = ( z z (z = (z (z ) a + 5a + 6 = ( a + ( a + ) 9. y y 5 = ( y 7)( y x + x = x + 6x x = xx ( + ( x+ = ( x+ (x y + 8yz 5z = y + 8yz yz 5z = 6 y(y+ z) 5 z(y+ z) = (y+ z)(6y 5 z) a + ab 9 b = (5a ( b a+ b y + 5y+ = (7 y + 5y+ = (7 y y y = (7 yy ( + + ( y+ ) = ( y+ (7 y+ + w+ w = (w + 5w = (w w+ 8w = [ w(w + (w ] = (w ( w+ 5w + w+ 5 = (5w w = (w (5w+ 6z + z+ 5 = (6z z = (8z+ (z x + xy 8y = (x 5xy + y ) = (x y)( x y) 6x x 5 = (x (x +. 8t + t = (t + (t.. 8p + 7p 5 = (8p ( p + a + a + = (a + 7)( a + ) p 9 p+ = p 5p p+ = 5 p(p (p = (p (5 p p + 5pq 6q = p + 8pq pq 6q = p( p+ q) q( p+ q) = ( p+ q)(p q) 6u 9uv+ v = 6u 5uv uv+ v = u(u 5 v) v(u 5 v) = (u 5 v)(u v) 5m + mn n = 5m + 6mn 5mn n = m(5m+ n) n(5m+ n) = (5m+ n)( m n) a + ab 5b = a + 5ab ab 5b = a(a+ 5 b) b(a+ 5 b) = (a+ 5 b)( a b) 6. 6p pq 9q = (p + 7pq+ q ) = ( p+ q)( p+ q) 5. r rs s = r + 6rs 7rs s = ( rr+ ) s 7( sr+ ) s = ( r+ s)(r 7 s) 7. 8y + 6y + y= 6 y(y + y+ 7) = 6 y(y+ 7)( y+
12 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual 5. h + 9hk k = h + hk hk k = hh ( + 7 k) kh ( + 7 k) = ( h+ 7 k)(h k) u + uv 5v = u + 6uv 5uv 5v 5. = ( uu+ v 5( vu+ v = ( u+ v)(u 5 v) 5. Prime t t 6= t t+ t 6 = tt ( + ( t = ( t ( t+ ) m + m 6= m + m m 6 = mm ( + ( m+ = ( m+ ( m ) 6. v + v+ 5 prime 5. Prime z z+ = 6z 8z 6z+ = 8( z z (z = (z (8 z 6w + w+ = 6w + 8w+ w+ = 8 w(w+ + (w+ = (w+ (8w+ b 8b+ 6= b b b+ 6 = bb ( ( b = ( b ( b = ( b q q+ = q q q+ = qq ( ( q = ( q ( q = ( q 5x + 5x = 5( x 5x+ 6) = 5( x x x+ 6) = 5( xx ( ( x ) = 5( x ( x ) a + a 8= ( a a+ 9) = ( a a 9a+ 9) = ( aa ( 9( a ) = ( a ( a 9) x x prime 7x + 8x = (6x + 9x = (6x + x x = ( x(x+ (x+ ) = (x+ ( x y 78y 8 = (y 9y = (y y+ y = ( yy ( + ( y ) = ( y ( y+ p 6p 7 p= p( p 6p 7) 5 = p( p 9p+ p 7) = p( p( p 9) + ( p 9)) = p( p 9)( p+ w w + 8 w = w ( w w+ 8) = w ( w 7w w+ 8) = w ( w( w 7) ( w 7)) = w ( w 7)( w x + x + 7 x= x(x + x+ 7) = x(x + x+ 7x+ 7) = x( x( x+ + 7( x+ ) = x(x+ 7)( x+
13 Chapter 6 Factoring Polynomials 5 7. r + r r = r(r + r = r(r + 8r 5r = r( r( r+ ) 5( r+ )) = rr ( + )(r 77. n n+ 9 = ( n + n = ( n + 6n 5n = ( nn ( + 6) 5( n+ 6)) = ( n+ 6)( n p 8p + p = p( p 9 p+ 6) = p( p p 5p+ 6) = p( p( p 5( p ) = p( p ( p q q 8q= q( q q ) x y + x y+ x = x ( y + y+ = x ( y + y+ y+ = x ( y( y+ + ( y+ ) = x ( y+ ( y+ ab + ab + b = b ( a + a+ = b ( a + a+ a+ = b ( a( a+ + ( a+ ) = b ( a+ ( a+ = qq ( 5q+ q ) = qqq ( ( + ( q ) = qq ( ( q k 7k = ( k + 7k + = ( k + 5k + k + = ( kk ( + + ( k+ ) = ( k + ( k + ) m 5m+ = ( m + 5m = ( m m+ 7m = ( mm ( ) + 7( m )) = ( m )( m+ 7) h + 8h 9= ( h h+ = ( h 5h 9h+ = ( hh ( 9( h ) = ( h ( h 9) x 7x + = ( x ( x ) m + m + = ( m + ( m + 7) No. (x + contains a common factor of. 8. No. (5x contains a common factor of (a) c= x + x 7 = ( ) + ( ) 7 = + 7 = 8 There are 8 customers. x + x 7= ( x x+ 6) = ( x 8)( x ) ( )( ) ( )( ) ( )( ) c= x 8 x = 8 = 8 8 = 8 Yes, the result is the same. 8. (a) i= d + d + = ( ) + ( ) + = = 5 His income is $5.
14 6 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual 85. (a) 86. (a) d + d + = ( d 5d ( d )( d ) = + 5 i= ( d )( d + = ( )( + = ( 8)( 7) = 5 Yes, the result is the same. s = n + n = ( 6) + 6 = = n + n= n( n+ s= n( n+ = 66 ( + = 67 ( ) = Yes, the result is the same. n + n + n s = 6 ( ) + ( ) + = = 6 8 = 6 = ( + + ) n + n + n n n n = 6 6 n n = 6 n( n+ ( n+ s = 6 ( + ( + = 6 5 ( )( 9) = 6 8 = 6 = ( + ( n+ Yes, the result is the same. Section 6.5 Practice Exercises. (a) difference; ( a+ b)( a b).. sum (c) is not (d) square (e) ( a+ b) ; ( a b) x + x = (x ( x+ ) 6x 7x+ 5 = (x (x ab+ ab= ab( + a) 6 5x y xy = 5 xy (x y ) 5pq+ p pq p = p(5pq+ p q = p(5 p( q+ ( q+ ) = pq ( + (5p 7. ax + ab 6x 6 b = a( x + b) 6( x + b) = ( x+ b)( a 6)
15 Chapter 6 Factoring Polynomials x+ 5+ x = x 6x+ 5 = ( x ( x 8. y x = ( y x)( y + x) 9. 6y + y = y + 6y = ( y+ ( y 9. 5p q = (5 pq (5 pq +. a + 7a + is Prime.. x n 5 = x 5 = n 9.. 8st = (9st (9st+ c = () c = c+ c (p q)(p+ q) = p 9q (7x y)(7x+ y) = 9x 6y x 6 = ( x+ 6)( x 6) r 8 = ( r+ 9)( r 9) w = ( w = ( w+ ( w t 9 = ( t+ 7)( t 7) a b = ( a) ( b) = (a+ b)(a b) 9 x y = ( x) y = ( x+ y)( x y) 9m 6 n = (7 m) ( n) = (7m+ n)(7m n) a 9 b = ( a) (7 b) = (a+ 7 b)(a 7 b) 9q + 6 is prime 6 + s is prime y z = ( y+ z)( y z) b c = ( b+ c)( b c) a b = ( a+ b )( a b ) z = ( z) = z+ z 5 t = (5 6 t ) = 5 ( t) = (5 + t)(5 t) 6 7h = 7(9 h ) = 7( + h)( h) z 6 = ( z + ( z = ( z + ( z+ )( z ) 6. y 65 = ( y + ( y = ( y + ( y+ ( y 7. 8 a = ( 9 a )( 9 + a ) ( a)( a)( 9 a) = z = ( z )( + z ) = ( z)( + z)( + z ) 9. x + 5x 9x 5 = x ( x+ 9( x+ ( x ) = ( x+ 9 = ( x+ ( x+ ( x. y + 6y y = y ( y+ 6) ( y+ 6) = ( y+ 6)( y = ( y+ 6)( y+ )( y )
16 8 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual.. c c 5c+ 5 = c ( c 5( c ( c ) = ( c 5 = ( c ( c+ ( c t + t 6t = t ( t+ ) 6( t+ ) ( t ) = ( t+ ) 6 = ( t+ )( t+ ( t. x 8 + x y 9y = ( x 9) + y( x 9) = ( x 9)( + y) = ( x + ( x ( + y) a 5+ a b b= 5( a + b( a x y 9x y + 6 = x ( y 9) ( y 9) = ( a (5 + b) = ( a+ ( a (5 + b) = ( y 9)( x = ( y+ ( y ( x+ )( x ) w z w 5z + 5 w ( z 5( z ( z ( w = = = ( z+ ( z ( w+ ( w y 8y+ 6= y (( y) + = ( y 5z z+ = (5 z) (5 z)() + = (5z ) 6 p + 6 p+ 5 = (6 p) + (6 p)( + 5 9a + ab+ 9b = (6 p + = (7 a) + (7 a)( b) + ( b) = (7a+ b) 5m mn+ 9n = (5 m) (5 m)( n) + ( n) = (5m n) y+ y + = y y+ = ( y + w w= w w+ = ( w ) 8z + zw+ 5w = 5(6z + zw+ 9 w ) = 5(( z) + ( z)( w) + ( w) ) = 5(z+ w) 7. (x+ = 9x + x (y 7) = y 8y (a) x + x+ is a perfect square trinomial 6. 6 p pq+ q = (9 p 6 pq+ q ) = (( p) ( p)( q) + q ) = ( p q) x + x+ = ( x+ ) ; x + 5x+ = ( x+ ( x+ 5. (a) x + x+ 6 is a perfect square trinomial x + x+ 6 = ( x+ 9)( x+ ; x + x+ 6 = ( x+ 6) 5. x + 8x+ 8 = x + (9)( x) + 9 = ( x+ 9) 6. 9y + 78y+ 5 = (y+ (y+ 6. y + y+ 9 = (y+ 9)(y+ 6. a a+ 5= ( a a+ = ( a 6. t t t t = ( ) = ( t +
17 Chapter 6 Factoring Polynomials x + x+ 9 is prime 66. c c+ 6 is prime x + xy + y = ( x) + ( x) y + y = ( x+ y) y + yz + z = ( y) + ( y) z + z = ( y+ z) x 6x + x= x( x x+ = ( xx = ( xx ( x ) y + y + 5y = 5 y(y + y+ = 5 y(y+ (y+ = 5 y(y+ ( y 9 = (( y + (( y = yy ( 6) (a) (a b = + = (a 5+ b)(a 5 b) [(a b][ (a b] (k + 7) 9m = = (k m)(k m) 8. (a) [(k 7) 7 m][ (k 7) 7m] a (area of outer square) b (area of inner square) Total area = a b a b = ( a+ b)( a b) g (area of outer square) h (area of inner square) total area = g h g h = ( g + h)( g h) ( x ) = (( x ) + )(( x ) ) = xx ( ( p+ 6 = (( p+ + 6)(( p+ 6) = (p+ 7)(p (q+ 5 = ((q+ + ((q+ = (q+ 8)(q ) = ( q+ ) (q = 8( q+ )(q 6 ( t + ) = ( + ( t+ ))( ( t+ )) = ( + t+ )( t ) = ( t+ 6)( t+ ) or ( t+ 6)( t ) 8 ( a + = (9 + ( a+ )(9 ( a+ ) = (9 + a+ (9 a = ( a+ ( a+ or ( a+ ( a Section 6.6 Practice Exercises. (a) sum; cubes difference; cubes.. (c) ( a b)( a + ab+ b ) (d) ( a+ b)( a ab+ b ) 6 6x = 6( x ) = 6( x)( + x) 5t = 5( t ) = 5( t)( + t). ax + bx + 5a + 5 b = x( a + b) + 5( a + b) = ( a+ b)( x+ 5. t+ u+ st+ su = ( t+ u) + s( t+ u) = ( t+ u)( + s) 6. 5y + y 6 = (5y )( y+
18 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual v + 5v = (v ( v+ ab 6ab + ab = 8 ab(5ab b+ a) c c 5 = ( c + c+ = ( c+ z + 6z 9 = ( z 6z+ 9) = ( z x,8, y,7 q, w, r s 9 5 z,8, y,7 a,. t,, 7,., z, 6 ab, 5, w, 6 y 9 p,, 8 y 8= y = ( y )( y + y+ x + 7 = x + = ( x+ ( x x+ 9) p = ( p)( + p+ p ) q + = ( q+ ( q q+ w + 6 = w + = ( w+ ( w w+ 6) b 5 = (6 b) 5 = (6b (6b + b+ n = n = n n + n m = + ( m ) 7 = + m m + m 9 5x + 8 y = (5 x) + ( y) = (5x + y)(5x xy+ y ) 7t + 6 u = ( t) + ( u) = (t+ u)(9t tu+ 6 u ) x = ( x ) = ( x + )( x ) b 5 = ( b ) 5 = ( b + ( b a + 9 is prime. w + 6 is prime. t + 6 = t + = ( t+ ( t t+ 6). 8 t = t = ( t)(+ t+ t ) 6. u + 7 = u + = ( u+ ( u u+ 9) x y = x ( y) = ( x y)( x + xy+ y ) 8r 7 t = ( r) ( t) = (r t)(r + 6rt+ 9 t ) 6t + = ( t) + = (t+ (6t t+ 5r + = (5 r) + = (5r+ (5r 5r+ a + 7 = ( a) + = (a+ (a a+ 9) g is prime. h 5 is prime. b + 8 = ( b + 7) = ( b+ ( b b+ 9) c = ( c 8) = ( c )( c + c+ 5 p 5 = 5( p = 5( p+ ( p q 8 = ( q = ( q + )( q )
19 Chapter 6 Factoring Polynomials h = ( h) 6 = h + h+ h k = + ( k ) 5 5 = + k k + k x p 6 = ( x ) = ( x + ( x = ( x + ( x+ )( x ) 8 = ( p ) 9 = ( p + 9)( p 9) = ( p + 9)( p+ ( p x x w = w 9 x x = w + w 9 5w z = 5 w ( z ) 5. ( ) 6 ( 5w z )( 5w 5wz z ) = + + x + x x = x (x+ (x+ 5. = (x+ ( x = (x+ ( x+ ( x x + x x = x (x+ (x+ = (x+ ( x = (x+ ( x+ )( x ) 6 x y = ( x + y )( x y ) = ( x + y )( x+ y)( x y) t = ( + t )( t ) = ( + t )( + t)( t) 8y 6 = (9y + (9y = (9y + (y+ )(y ) u 56 u = u( u 6 ) = uu ( + 6)( u 6) = uu ( + 6)( u+ ( u y 6 y x = x 5 5 y y = x + x 5 5 q 6 6 = ( q ) 8 = ( q + 8)( q 8) = ( q+ )( q q+ ( q )( q + q+ a 6 = ( a ) = ( a + ( a = ( a+ ( a a+ ( a ( a + a x + 6 y = ( x ) + ( y) 6 = ( x + y)( x x y+ 6y ) a + b = a + ( b ) 6. 6 = ( a+ b )( a ab + b ) u v = ( u ) v = ( u v)( u + u v+ v ) 6. x y = ( x + y )( x y ) = ( x + y )( x+ y)( x y) 6. a b = ( a + b )( a b ) = ( a + b )( a+ b)( a b) 6. k + k 9k 6 = k ( k + 9( k + = ( k + ( k 9) = ( k + ( k + ( k
20 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual w w w+ 8 = w ( w ) ( w ) 6. = ( w )( w = ( w )( w+ )( w ) = ( w+ )( w ) 65. t t t+ = t ( t ( t = ( t ( t = ( t ( t+ ( t 7. (a) + x+ x x + x + x 8 x ( x x ) x + x (x x) x 8 (x 8) a + 7a a = 9 a ( a+ ( a+ = ( a+ (9a = ( a+ (a+ )(a ) 6 p q 5 8 = p q 5 6 = p q p + pq+ q r + s 7 = r + s = r+ s r rs+ s 5 9 a + b = ( a ) + ( b ) a 9 9 b 8 8 = ( a + b )( a a b + b ) = ( a ) ( b ) 6 6 = ( a b )( a + a b + b ) 6 6 = ( a b)( a + ab+ b )( a + a b + b ) 7. (a) x p 75. x y 5 x 8 = ( x )( x + x+ y y+ 9 y+ y + y + y+ 7 ( y + y ) y + y ( y 9 y) 9y + 7 (9y + 7) y + 7 = ( y+ ( y y+ 9) + x+ 5p+ 5 Problem Recognition Exercises. A prime factor cannot be factored further.. Factor out the GCF.. Look for the difference of squares: a b, a difference of cubes: a b, or a sum of cubes: a + b. Grouping 5. (a) Difference of squares
21 Chapter 6 Factoring Polynomials a 6 = ( a 8 = ( a+ 9)( a 9) 6. (a) Nonperfect square trinomial y + y+ = ( y+ ( y+ 7. (a) None of these 6w 6w= 6 w( w 8. (a) Difference of squares 6z 8 = (z + 9)(z+ (z 9. (a) Nonperfect square trinomial t + t+ = (t+ ( t+. (a) Sum of cubes 5r + 5 = 5( r + = 5( r+ ( r r+. (a) Four termsgrouping ac + ad bc bd = a( c+ d) b( c+ d) = ( c+ d)( a b). (a) Difference of cubes x + 5 = ( x+ ( x 5x+. (a) Sum of cubes y + 8 = ( y+ )( y y+. (a) Nonperfect square trinomial 7p 9p+ = (7p ( p 5. (a) Nonperfect square trinomial q 9q = ( q q = ( q ( q+ 6. (a) Perfect square trinomial x + 8x 8= ( x x+ 7. (a) None of these = ( x ) 8a + a= 6 a(a+ ) 8. (a) Difference of cubes 5 y = (7 y ) = ( y)(9 + y+ y ) 9. (a) Difference of squares t = ( t = ( t+ ( t. (a) Nonperfect square trinomial t t 8 = (t+ ( t 8). (a) Nonperfect square trinomial c + c+ = ( c + c+. (a) Four termsgrouping xw x+ yw 5y = xw ( + yw ( = ( w (x+ y). (a) Sum of cubes x +. = ( x+.( x.x+.. (a) Difference of squares q 9 = (q+ (q 5. (a) Perfect square trinomial 6 + 6k + k = 8 + (8)( k) + k = (8 + k) 6. (a) Four termsgrouping
22 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual st+ 5t+ 6s + = ts ( + + 6( s + = ( s + ( t+ 6) 7. (a) Nonperfect square trinomial 6x x 5= (x 7x = (x+ ( x 7. (a) Four termsgrouping x + x xy y = xx ( + yx ( + = ( x+ ( x y) 8. (a) Sum of cubes w + y = ( w+ y)( w wy+ y ) 9. (a) Difference of cubes a c = ( a c)( a + ac+ c ). (a) Nonperfect square trinomial y + y + is prime. (a) Nonperfect square trinomial c + 8c + 9 is Prime. (a) Perfect square trinomial a + a+ = ( a+. (a) Perfect square trinomial b + b+ 5 = ( b+. (a) Nonperfect square trinomial t t+ = ( t + t = ( t+ 8)( t 8. (a) Nonperfect square trinomial y y+ = (y 7 y+ = (5y ( y 9. (a) None of these 5a bc 7 abc = abc (5ac 7). (a) Difference of squares 8a 5= (a = (a (a+. (a) Nonperfect square trinomial t + t 6 = ( t 7)( t+ 9). (a) Nonperfect square trinomial b + b 8 = ( b 8)( b+. (a) Four termsgrouping ab + ay b by = ab ( + y) bb ( + y) = ( b+ y)( a b). (a) None of these 5 6x y + x y = x y ( x+ y) 5. (a) Nonperfect square trinomial 5. (a) Nonperfect square trinomial u uv+ v = (7u v)( u v) p 5p p= p( p + 5p+ = p( p+ ( p+ 6. (a) Difference of squares x y 9 = ( xy + 7)( xy 7) 6. (a) Nonperfect square trinomial 9 p 6 pq+ q is prime 7. (a) Nonperfect square trinomial q 8q 6= (q q
23 Chapter 6 Factoring Polynomials 5 8. (a) Nonperfect square trinomial 58. (a) Nonperfect square trinomial 9w + w 5= (w + w q + q 7 is prime 9. (a) Sum of squares 9m + 6 n is prime 5. (a) Perfect square trinomial 5b b+ 5= 5( b 6b+ 9) = 5( b 5. (a) Nonperfect square trinomial 6r + r+ = (r+ (r+ 5. (a) Nonperfect square trinomial s + s 5 = (s (s+ 5. (a) Difference of squares 6a = (a + (a 5. (a) Four termsgrouping = (a + (a+ (a p + p c 9p 9c = p ( p+ c) 9( p+ c) = ( p+ c)( p 9) = ( p+ c)( p+ ( p 55. (a) Perfect square trinomial 8u 9uv+ 5v = (9 u) (9 u)(5 v) + (5 v) = (9u 5 v) 56. (a) Sum of squares x + 6= ( x (a) Nonperfect square trinomial x 5x 6 = ( x 6)( x+ 59. (a) Four termsgrouping ax 6ay + bx by = ax ( y) + bx ( y) = ( x y)( a+ b) 6. (a) Nonperfect square trinomial 8m m m= m(8m m = m(m+ (m 6. (a) Nonperfect square trinomial x y+ x y+ x y = x y(x + x+ = x y(7x+ )(x+ = ( p+ c)( p+ ( p 6. (a) Difference of squares m 8 = ( m 6 = ( m + 8)( m 8) 6. (a) Four termsgrouping 8uv 6u + v 9 = u(v + (v = (v (u+ 6. (a) Four termsgrouping t t+ st 5s= t( t + s( t = ( t ( t+ s) 65. (a) Perfect square trinomial x x+ = (x x+ = ([ x] ( x)( + ) = (x 66. (a) Perfect square trinomial p + pq+ q = ( p+ q)
24 6 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual 67. (a) Nonperfect square trinomial 6n + 5n n = n(6n + 5n = n(6n n+ 8n = n[ n(n + (n ] = n[(n (n+ ] = n(n (n+ 68. (a) Nonperfect square trinomial k + k k = k(k + k = k(k k + 6k = k[ k(k + (k ] = k[(k (k + ] = k(k (k (a) Difference of squares 6 y = 8 y = (8 y)(8 + y) 7. (a) Difference of squares 6 b b = b(6 b ) = b(6 b ) = b(6 b)(6 + b) 7. (a) Nonperfect square trinomial b b+ is prime. 7. (a) Nonperfect square trinomial y + 6y+ 8 = ( y+ ( y+ ) 7. (a) Nonperfect square trinomial c c + = ( c ( c ) Section 6.7 Practice Exercises. (a) quadratic. 6a 8 ab+ b= (a b(a = (a ( b) b b+ = ( b b+ = ( b 6)( b 8u v uv = uv(uv x + x 8 = (x )( x+ h 75 = ( h = ( h+ ( h x + 6y = ( x + y ) 8. Linear 9. Neither. Quadratic. Quadratic. Neither. Linear. (x (x + = x 5= or x+ = x= 5 x= 5. (x + (x = x+ = or x = x= x= 6. (x )(x + ) = x = or x+ = x= x= x= x= 7. (x 7)(x + 7) = x 7= or x+ 7= x= 7 x= x= x= ;
25 Chapter 6 Factoring Polynomials 7 8. ( x 7)( x 7) = x 7= x = 7 9. ( x+ ( x+ = x + 5= x = 5. (x )(x = x = or x = x= x= x= x=. x(5x = x= or 5x = x= or 5x= x= or x= 5. x(x + 8) = x= or x+ 8= x= or x= 8 x= or 8 x=. The polynomial must be factored completely before applying the zero product rule.. The equation must have one side equal to zero and the other side factored completely p p 5= ( p+ ( p = p+ = or p 5= p= p= 5 y 7y 8= ( y 8)( y+ = y 8= or y+ = y = 8 y = z + z = ( z+ ( z ) = z+ = or z = z = z = w w+ 6 = ( w 8)( w ) = w 8= or w = w= 8 w= q 7q = (q+ ( q = q+ = or q = q= q= x x = (x+ ( x = x+ = or x = x= x= = 9x = (x+ )(x ) x+ = or x = x= x= a 9= (a+ 7)(a 7) = a+ 7= or a 7= 7 7 a= a= k 8k + 96= ( k k + 8) = ( k 6)( k 8) = k 6= or k 8= k = 6 k = 8 = t + t+ 5 ( t + t+ = ( t+ ( t+ = t + 5= t = 5
26 8 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual = m 5m m m(m 5m = m(m+ ( m = m= or m+ = or m = m= m= m= n + n + n= n(n + n+ = n(n+ ( n+ = n= or n+ = or n+ = n= n= n= 7. (p+ ( p ( p+ 6) = p+ = or p = or p+ 6= p= or p= or p= 6. xx ( + 7)(x = x= or x+ 7 = or x 5 = x = or x = 7 or x = 5 5 x = x= or x= 7 or.. x 6x= xx ( 6) = xx ( + ( x = x= or x+ = or x = x= or x= or x= t 6t = tt ( 6) = tt ( + 6)( t 6) = t = or t+ 6= or t 6= t = or t = 6 or t = 6 8. (x (x (x + 7) = x = or x = or x+ 7= x= or x= or x= 7 9. x(x (x + = x= or x = or x+ = x= or x= or x= x = x = or x = or. x(x+ ( x+ = x= or x+ = or x+ = x= or x= or x= x = x= or or x=. 5 x(x+ 9)( x = 5x= or x+ 9= or x = x = 5 or x = 9 or x = 9 x = x= or or x= x + 8x= xx ( + 6) = x= or x+ 6= x= or x= 6 y y= yy ( = y = or y = y = or y= 6m = 9 6m 9 = (m+ (m = m+ = or m = m= or m= 9n = 9n = (n+ (n = n+ = or n = n= or n=
27 Chapter 6 Factoring Polynomials 9 9. y + y = y y + y + y= yy ( + 7y+ = yy ( + ( y+ ) = y = or y+ 5= or y+ = y = or y= 5 or y= 55. b + b + b= bb ( + b+ = bb ( + )( b+ ) = b = b b b= or b+ = or b+ = b= or b= or b= 5. d 6d = d d 6d d = dd ( d 8) = dd ( ( d+ ) = d = or d = or d + = d = or d = or d = 56. x + 6x= x x x + 6x= xx ( x+ 6) = xx ( 6)( x 6) = x= or x 6= or x 6= x= or x= 6 or x= t ( t 7) = 5t t+ = t + = t = 5. 8h= 5( h 9) + 6 8h= 5h 5+ 6 h + 9= h = ( cc 8) = c 6c+ = ( c 8c+ = ( c ( c = c 5= or c = c= 5 or c= qq ( = q 9q = ( q q = ( q ( q+ = q = or q+ = q= or q= ( a + a) = a 9 a + 6a= a 9 a + 6a+ 9= ( a+ ( a+ = a + = a = 9( k = k 9k 9= k k + 9k 9= (k ( k + = k = or k + = k = k = nn ( + ) = 6 n + n 6= ( n + n = ( n+ ( n = n+ = or n = n= n=
28 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual p( p = 8 p p 8= ( p p 6) = ( p ( p+ ) = p = or p+ = p= p= x(x+ = x + x+ x + 5x = x + x+ x = x = ( zz ) z= z + z 6z z = z + 7z = z = 7 7q = 9q 7q 9q= 9 q(q = 9q= or q = q= q= w = w w w= 7 w(w ) = 7w= or w = w= w= ( c c) = ( cc ) = c= or c = c= c= ( d + d) = d(d + = d = or d + = d = d = y y y+ = y ( y ( y = ( y ( y+ )( y ) = y = or y+ = or y = y = y = y = t + t 6t = t ( t+ ) 6( t+ ) = ( t+ )( t+ ( t = t+ = or t+ = or t = t = t = t = ( x ( x+ ) = 8 x + x 8= x + x = ( x+ ( x = x+ 5= or x = x= 5 x= ( w+ ( w = w + w 5 = w + w 5= ( w+ 7)( w = w+ 7= or w 5= w= 7 w= 5 ( p+ )( p+ = p p + 5p+ 6= p p + 6p+ 5= ( p+ ( p+ = p+ 5= or p+ = p= 5 p=
29 Chapter 6 Factoring Polynomials 7. ( k 6)( k = k k 7k + 6= k k 6k + 8= ( k ( k ) = k = or k = k = k = 6. (a) 6a 7a = (a+ (a = a+ = or a = a= a= 6a 7a = (a+ (a Problem Recognition Exercises. (a) x + 6x 7 = ( x+ 7)( x ( x+ 7)( x =. (a) x+ 7= or x = x= 7 x= c + 8c+ = ( c+ 6)( c+ ) ( c+ 6)( c+ ) =. (a) c+ 6= or c+ = c= 6 c= y + 7y+ = (y+ ( y+ (y+ ( y+ = y+ = or y+ = y = y =. (a) x 8x+ 5 = (x ( x (x ( x = x 5= or x = 5 x= x= 5. (a) 5q + q = (5q ( q+ = 5q = or q+ = q= 5 q= 5q + q = (5q ( q+ 7. (a) 8. (a) 9. (a). (a). (a) a 6 = ( a+ 8)( a 8) = a+ 8= or a 8= a= 8 a= 8 a 6 = ( a+ 8)( a 8) v = ( v+ ( v = v+ = or v = v= v= v = ( v+ ( v b 8 = (b+ 9)(b 9) (b+ 9)(b 9) = b+ 9= or b 9= 9 9 b= b= 6t 9 = (6t+ 7)(6t 7) (6t+ 7)(6t 7) = 6t+ 7= or 6t 7= 7 7 t = t = 6 6 8x + 6x+ 6= (x + 8x+ = (x+ (x+ = x+ = or x+ = x= x= 8x + 6x+ 6= (x+ (x+
30 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual. (a) y + y+ = (y + y+ 8) = (y+ ( y+ ) = y+ = or y+ = y= y= y + y+ = (y+ ( y+ ). (a) x 8x x= x( x 8x ) = xx ( ( x+ ) xx ( ( x+ 8) = x= or x = or x+ = x= x= x=. (a) k + 5k k = k( k + 5k = kk ( + 7)( k ) kk ( + 7)( k ) = k = or k + 7= or k = k = k = 7 k = 5. (a) b + b 9b 9= b ( b+ 9( b+ = ( b+ ( b 9) = ( b+ ( b+ ( b = b+ = or b+ = or b = b= b= b= b + b 9b 9 = ( b+ ( b ( b+ 6. (a) x 8x x+ = x ( x 8) ( x 8) = ( x 8)( x = ( x 8)( x+ )( x ) = x 8 = or x+ = or x = x= 8 x= x= x 8x x+ = ( x 8)( x+ )( x ) 7. s 6s+ rs r = s( s + r( s = ( s ( s+ r) 8. 6t + t+ tu+ 5u = t(t+ + 5 u(t+ = (t+ (t+ 5 u) 9. 8x x= x(x = x(x (x+ = x= or x = or x+ = x= x= x =. b 5b= b( b = bb ( ( b+ = b= or b 5 = or b+ 5 = b= b= 5 x= 5 x x + x= x( x x+ = xx ( = x= or x = or x = x= x= x=. t + 8t + 7t = t( t + 6t+ 9) = ( tt+ = t = or t+ = or t+ = t = t = t =.. 7c c+ = 7( c + c) 7c c+ = 7c + 7c = 9c c = 9 c =. ( z z+ = 7+ 6z 6z + z = 7+ 6z z = 7 7 z =
31 Chapter 6 Factoring Polynomials 5. 8w + 7 = ( w) + ( = (w+ (w 6w+ 9) q = ( q) ( = (q ( q + q+ 5z + z = 7 5z + z 7= (5z+ 7)( z = 5z+ 7= or z = 7 z = 5 z = h + 5h= 6 h + 5h+ 6= (h+ ( h+ 6) = h+ = or h+ 6= h= h= 6 bb ( + 6) = b b + 8b= b b + 7b+ = (b+ )( b+ = b+ = or b+ 5= b= b= 5 y + = y( y y + = y y y + y+ = (y+ ( y+ = y+ = or y+ = y = y =. 5(x (x+ = x x 5 6x = x x 7= x 7x = x =. 6a= (a 6a= 8a+ 6a= 8a+ a = a = s = 6 s 6= ( s 6) = ( s+ ( s = s+ = or s = s= s= 8v = 6 8v 6 = 9(9v = 9(v+ )(v ) = v+ = or v = v= v= ( x ( x = 6 x 7x+ = 6 x 7x+ 6= ( x 6)( x = x 6= or x = x= 6 x= ( x+ ( x+ 9) = x + x+ 5 = x + x+ = ( x+ ( x+ ) = x+ = or x+ = x= x= Section 6.8 Practice Exercises. (a) x + x + (c) x +
32 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual (d) LW (e) (f) bh a + b = c. (6x+ ( x+ = 6x+ = or x+ = x= 6 x=. 9 x(x+ ) = 9x= or x+ = x= x= x = (x+ (x = x+ = or x = x= x= x 5x= 6 x 5x 6= ( x 6)( x+ = x 6= or x+ = x= 6 x= xx ( ) = x x+ = ( x = x = or x = x= x= 6x 7x = (6x+ ( x ) = 6x+ 5= or x = 5 x= 6 x= 8. A factored expression must be set equal to zero to use the zero product rule. 9. Let x = the number. Then, x + = 6 x 9 = ( x+ 7)( x 7) = x+ 7= or x 7= x= 7 x= 7 The numbers are 7 and 7.. Let x = the number. Then, x+ x = 6 x + x 6= (x+ 9)( x = x+ 9= or x = 9 x= x= 9 The numbers are and.. Let x = a number. Then, + 6x= x 8 x 6x = ( x+ ( x = x+ = or x = x= x= The numbers are and.. Let x = a number. Then, x = x+ x x = ( x+ ( x = x+ = or x 5= x= x= 5 The numbers are and 5.. Let x = first integer. Then the next consecutive odd integer is (x + ). xx ( + ) = 6 x + x 6= ( x+ 9)( x 7) = x+ 9= or x 7= x= 9 x= 7 The numbers are 9 and 7, or 7 and 9.
33 Chapter 6 Factoring Polynomials 5. Let x = first integer. Then the next consecutive even integer is (x + ). xx ( + ) = 8 x + x 8= ( x+ 8)( x 6) = x+ 8= or x 6= x= 8 x= 6 The numbers are 8 and 6, or 6 and Let x = first integer. Then the next consecutive integer is (x +. x + ( x+ = 6 x + x + x+ = 6 x + x 6= ( x+ 6)( x = x+ 6= or x 5= x= 6 x= 5 The numbers are 6 and 5, or 5 and Let x = first integer. Then the next consecutive integer is (x + ). x + ( x+ ) = 5 x + x + x+ = 5 x + x 8= ( x + x = ( x+ 6)( x = x+ 6= or x = x= 6 x= The numbers are 6 and , or and Let x = width of the painting and x + is the length of the painting. Then, A = (length)(width) 99 = xx ( + ) 99 = x + x x + x 99= ( x+ ( x ) = x = or x = 9 The height of the painting is ft and the width is 9 ft. 8. Let x = length of painting and x be the length of the painting. Then, A = (length)(width) = xx ( ) = x x x x = ( x+ ( x = x = or x = The painting has length in. and width in. 9. Let x = length and x be the width. (a) A= lw 8 = xx ( x x 8= ( x+ ( x 7) = x = or x = 7 The dimensions are 7 m by m. P = w + l P = ( + (7) P = 8 + = m. Let x = length and x 7 be the width. (a) A= lw 78 = xx ( 7) x 7x 78= ( x+ 6)( x = x = 6 or x = The dimensions are in. by 6 in. P = w + l P = (6) + ( P = + 6 = 8 in.. Let x = height and x + = the base. A= bh = ( x)( x+ 8 = x + x x + x 8= ( x+ 7)( x = x+ 7= or x = x = 7 x = The height is ft and the base is 7 ft.
34 6 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual. Let x = base and x + 5 = the height. A= bh 5 = ( x)( x+ 5 = x + 5x x + 5x 5 = ( x+ ( x = x+ 5 = or x = x = 5 x = The base is cm and the height is 5 cm.. Let x = base and x 7 = the height. A= bh = ( x)(x 7) = x 7x x 7x = (x+ 8)( x = x+ 8= or x 5= 8 x = x = 5 x 7= ( 7 = 8 The base is 5 cm and the height is 8 cm.. Let x = height and x = the base. A= bh 6 = ( x)(x ) = x x x x = (x x 6) = (x+ ( x ) = x+ = or x = x = x = x = () = 6 The base is 6 ft and the height is ft. 5. If you let h =, then, = 6t + = 6( t 9) 6( t+ ( t = t+ = or t = t = t = It will take seconds to hit the ground. 6. If you let h =, then = 6t + 56 = 6( t 6) 6( t+ ( t = t+ = or t = t = or t = It will take seconds to hit the ground. 7. Ground level is when h =. Then, = 6t + t = 8 t(t 8t = or t = t = t = =.5 The times are seconds and.5 seconds. 8. Ground level is when h =. Then, = 6t + 6t = 6 tt ( 6t = or t = t = t = The times are seconds and seconds. 9. Pictures may vary.. Given a right triangle with legs a and b and hypotenuse c, then a + b = c.
35 Chapter 6 Factoring Polynomials c = a + b c c = + 7 = 65 c 65 = ( c+ ( c = c = 5 or c = 5 c = 5 cm c = a + b c c = + = 5 c 5 = ( c+ ( c = c = 5 or c = 5 c = 5 m c = a + b 7 = a = a + 6 a 5 = ( a+ ( a = a = 5 or a = 5 a = 5 in. c = a + b 5 = 9 + b 5 = 8+ b b = ( b+ ( b = b = or b = b = yd 5. Let x = c. c = a + b x x = 6 + = x = ( x+ )( x ) = x = or x = The brace is in. long. 6. Let h = a. c = a + b 5 = h + h = 8 h 8 = ( h+ 9)( h 9) = h = 9 or h = 9 The height is 9 km 7. Let h = height of the kite. Then, a = h. c = a + b = ( h + 9 = h h h h = ( h+ 7)( h 9) = h = 7 or h = 9 The kit is 9 yd high. 8. Let c = distance between the two cars. Then, a = 8 and b = 6. c = a + b c = c = 6 c 6 = ( c+ 8)( c 8) = c = 8 or c = 8 They were 8 mi apart. 9. Let x = distance between the base of the ladder and the bottom of the house. Then, the distance between the top of the ladder and the ground is x + 7 ft. c = a + b 7 = x + ( x+ 7) 89 = x + x + x+ 9 x + x = ( x+ ( x 8) = x = 5 or x = 8 The bottom of the ladder is 8 ft from the house. The distance from the top of the ladder to the ground is 5 ft.
36 8 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual. Let x = distance traveled by the first boat. Then, the distance traveled by the second boat is x + mi. c = a + b 5 = x + ( x+ 5 = x + x + x+ x + x = ( x+ ( x = x = or x = The first boat traveled mi; the second boat traveled mi.. Let x = hypotenuse. Then, one leg is x m and the other leg is x m. x = ( x + ( x ) x = x 8x+ 6+ x x+ x = x x+ x x+ = ( x ( x ) = x= or x= The hypotenuse is m.. Let x = length of the shorter leg. Then, the longer leg is x cm and the hypotenuse is x + cm. (x+ = x + (x x + x+ = x + x x+ x + x+ = 5x x+ x 8x= xx ( 8) = x= 8 or x= The length of the shorter leg is 8 cm. Group Activity. Answers will vary. For example: x x + x= x( x 5x+. Answers will vary. For example: xy + 5y x = (x + (5y 7). Answers will vary. For example: x 5x = ( x+ )( x 7). Answers will vary. For example: 6x 7x 5 = (x+ (x 5. Answers will vary. For example: 7x x x= 7 x( x ( x+ 6. Answers will vary. For example: 6x 8 y = (x + 9 y)(x 9 y) 7. Answers will vary. For example: 6 5 b = (6 5 b)(6 + b+ 5 b ) 8. Answers will vary. For example: u + 7 v = ( u+ v)( u uv+ 9 v ) 9. Answers will vary. For example: ( x ( x+ 7) = x + x 8=. Answers will vary. For example: x x+ = x + 6x= Chapter 6 Review Exercises Section 6.. GCF: a b. GCF: x + 5. GCF: c(c. GCF: yz or yz x + x 8x= x(x+ x 5 w y w y = w y ( w y ) t + 5 t = t( t or t( t + 6 u u = u(6u+ or u( 6u 9. b(b + ) 7(b + ) = (b + )(b 7)
37 Chapter 6 Factoring Polynomials 9. (5x + 9) + 8x(5x + 9) = (5x + 9)( + x).... 7w + w+ wb+ b= 7 w( w+ ) + b( w+ ) = ( w+ )(7 w+ b) b b+ yb y = b( b ) + y( b ) = ( b )( b+ y) Section y 5y y+ 9 = (y 5y y+ = (5 y( y ( y ) = (y (5 y 6a a ab+ a b= a(6 a b+ ab) = a(( a) b( a)) = a( a)( b) x x+ = x 7x x+ = xx ( 7) ( x 7) = ( x ( x 7) 6. y 9y+ 88 = y y 8y+ 88 = yy ( 8( y = ( y 8)( y 7. 6z+ z 7= z 6z 7 = z + 6z z 7 = zz ( + 6) ( z+ 6) = ( z ( z+ 6). m + 6m + 8m = m ( m + m+ = m ( m + 5m+ 8m+ = m ( m( m+ + 8( m+ ) = m ( m+ 8)( m+.. t + t 6 = ( t t+ 6) = ( t t 8t+ 6) = (( tt ) 8( t )) = ( t 8)( t ) w w+ = ( w + w ) = ( w + 5w w ) = ( ww ( + ( w+ ) = ( w ( w+ a + ab+ b = a + ab+ ab+ b = aa ( + b) + ba ( + b) = ( a+ b)( a+ b). c cd 8d = c + cd 6cd 8d = cc ( + d) 6 dc ( + d) = ( c 6 d)( c+ d). Section Different 6. Both negative 7. Both positive q q= q q+ 9 p w+ 6pw+ 6w = wp ( + p+ ) = q + q q+ 9 = qq ( + ( q+ = ( q ( q+ = wp ( + p+ p+ ) = wpp ( ( + ) + ( p+ ) = wp ( + ( p+ ) 8. Different 9... y 5y = y 8y+ y = yy ( + ( y = (y+ ( y w 5w 6= w 8w+ w 6 = ww ( ) + ( w ) = (w+ ( w ) z + 9z+ = z + z+ 5z+ = (5 z z+ ) + 5(5z+ ) = (z+ (5z+ )
38 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual z + 6z 9= 8z + z 6z 9 = z(z+ (z+ = (z (z+ p 5p+ is prime. 5r r+ 7 is prime. w 6w 7 = ( w 6w 7) = ( w 9w+ w 7) = ( ww ( 9) + ( w 9) = ( w+ ( w 9) Section 6.. 5,., c 5c = c 6c+ c = cc ( ) + ( c ) = ( c )(c+ y + y+ = y + y+ y+ = yy ( + + ( y+ = ( y+ (y y 8y 8 = ( y 6y 6) 9c cd + 5d = ( y 8y+ y 6) = ( yy ( 8) + ( y 8)) = ( y+ )( y 8) = 9c 5cd 5cd + 5d = ( c c 5 d) 5 d(c 5 d) = (c 5 d)(c 5 d) = (c 5 d) x + x+ 6 = x + 6x+ 6x+ 6 = xx ( + 6) + 6( x+ 6) = ( x+ 6)( x+ 6) = ( x + 6) t + tw+ w = t + tw+ tw+ w = tt ( + w) + ( t+ w) = ( t+ w)( t+ w) x + 7x 5= x + x x = x ( x + ( x + = ( x + (x w + 7w + = w + 5w + w + = w ( w + + ( w + ) = ( w + ( w + ) p 8pq+ 5q = p pq 5pq+ 5q = p( p q) 5 q( p q) = ( p q)( p 5 q) g + 7gh + h = (g + h)( g + h) m mn+ 5 n = (6 m n)(m 5 n) v v = ( v ) v = ( v ) + v v = v ( v + ( v + = ( v ( v v + v 6 = (v + v = (v + 5v v = (5 v(v+ (v+ ) = (v+ (5v s + s = (s + s = (s + 8s 5s = ( ss ( + ) 5( s+ )) = ( s+ )(s. x + 7x + = ( x ) + 7x + = ( x ) + x + 5x + = x ( x + ) + 5( x + ) = ( x + ( x + )
39 Chapter 6 Factoring Polynomials ab ab + ab = ab( a ab + b ) = ab( a 6ab ab + b ) = ab( a( a 6 b) b( a 6 b)) = ab( a 6 b)( a b) 6 5 z + 8z z = z ( z + z m+ 9m prime + 6p + 9p prime = z ( z + 7z z = z ( z( z+ 7) ( z+ 7)) = z ( z+ 7)( z y + ; this is a sum of squares, not a difference. y + y+ 6 = ( y+ 6) t + 6t+ 6 = ( t+ 8) 9a a+ = (a ) 5x x+ 6 = (5x v v = ( v + v+ = ( v + ) x + x 5= ( x x+ = ( x 57. 9x + x+ = (7 x) + (7 x)( + = (7x c 8= ( c 9) = ( c + ( c 7x y = (6 x y ) = (6 x + y)(6 x y) 58. Section w 6wz+ z = ( w) + ( w)( z) + ( z) = ( w z) a b = ( a b)( a+ b) 6. a + b is prime p + p 6p 8 = p + p 6( p+ = p ( p+ 6( p+ = ( p 6)( p+ = ( p ( p+ ( p+ k 8 k + k = ( k )( k)( + k) or = ( k ) ( k + ) a 9 = a 7 = ( a+ 7)( a 7) d 6 = d 8 = ( d + 8)( d 8) 8t = (9 t) = ( + 9 t)( 9 t) Section a + b = ( a+ b)( a ab+ b ) 78. a b = ( a b)( a + ab+ b ) k = (5 k) = (+ 5 k)( 5 k) x + 6; this is a sum of squares, not a difference a = + a = ( + a)( a+ a ) = ( + a)(6 a+ a )
40 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual 8. 5 b = 5 b p q = (5 b)(5 + 5 b+ b ) = (5 b)(5 + 5 b+ b ) + 8 = ( p ) + = ( p + )(( p ) p + ) = ( p + )( p p + = ( q ) 7 = q ( q ) + q + = q q + q x 8= 6( x 8) = 6( x ) = 6( x )( x + x+ 8. 7y + 7= 7( y = 7( y+ ( y y+ x 6 x= xx ( 6) = xx ( + 6)( x 6) q 6 q= q( q ) = qq ( ( q + q+ 6) 8h + = (h + m 8 m= m( m 8) 9. Section 6.7 m = ( m = ( m ( m + m+ 9. The equation (x (x + = can be solved directly by the zero product rule because it is a product of factors set equal to zero. 9. (x (x + ) = x = or x+ = x= x= 95. (a 9)(a = a 9= or a = a= 9 a= 96. w(w + (5w + ) = w= or w+ = or 5w+ = w= w= w= u(u 7)(u 9) = 6u = or u 7 = or u 9 = u = u = 7 9 u = 98. 7k 9k = (7k + ( k ) = 7k + 5= or k = 5 k = 7 k = x + x x = x ( x+ ( x+ = ( x+ ( x = ( x+ ( x+ ( x 5pq q = 5 qp ( q) = 5 q( p + q)( p q) 8 n+ n = n(8 + n ) = n( + n)( n+ n ) 99.. h h 6= (h+ ( h 6) = h+ = or h 6= h= h= 6 q = ( q+ ( q = q+ = or q = q= q=
41 Chapter 6 Factoring Polynomials r = 5 r 5 = ( r+ ( r = r+ 5= or r 5= r = 5 r = 5 5v v= v(5v = v= or 5v = v= v= 5 xx ( 6) = 8 x 6x+ 8= ( x ( x ) = x = or x = x= x= 6t + 6t = 5 6t + 6t+ 5 = (6t+ (6t+ = 6t + 5= 5 t = 6 9s + s= 9s + s+ = (s+ )(s+ ) = s + = s = ( y + = y y + = y y y+ = (y )( y 6) = y = or y 6= y = y = Section 6.8 ( p 66) = p p + p = ( p ( p+ = p = or p+ = p= p= y 8y = 8y y 8y + 8y = yy ( 9y+ = yy ( 7)( y ) = y = or y 7 = or y = y = or y = 7 or y = x x= xx ( = xx ( + )( x ) = x= or x+ = or x = x= or x= or x=. Let x = height and x + be the base. A = (height)(base) 78 = ( x)(x+ 78 = x + x x + x 78= (x+ ( x 6) = x= or x= 6 The height is 6 ft and the base is ft.. At h =, the ball is ground level. = 6x + 6x = 6 xx ( x = or x = The ball is at ground level at and second.
42 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual. Let x represent the length of the ramp. + 5 = x + 5 = x 69 = x = x The ramp is ft long.. Let x = shorter leg. Then, the other leg is x + and the hypotenuse is x. x + ( x+ ) = (x ) x + x + x+ = x 8x+ x x= xx ( 6) = x = or x = 6 The legs are 6 ft and 8 ft; the hypotenuse is ft. Chapter 6 Test x x+ 6x = x(5x + x ) 7a 5 a + 5a= 7( a a( a = ( a (7 a) 6w w+ 7 = (6w ( w 7) 69 p = p = ( + p)( p) q 6q+ 6 = ( q 8) 8 + t = ( + t)( t+ t ). Let x = a number. Then, 6 x = x 6 = ( x+ 8)( x 8) = x = 8 or x = 8 The numbers are 8 and x = first integer and x + is the next integer. xx ( + = ( x+ x+ + x + x= 8x+ + x 7x 58 = ( x+ )( x 9) = x = and x = 9 The numbers are and or 9 and a + a+ = ( a+ ( a+ 8) x + x = ( x+ 7)( x 6) y 7y+ 8 = (y ( y 8) 6z + 9z+ 8 = (z+ (z+ 8) 9t = (t+ (t v 8 = ( v+ 9)( v 9) a + 7ab+ 5b = ( a + 9ab+ 8 b ) = ( a+ 6 b)( a+ b) 6. Let x = height, x + be the base. A= bh 8 = x(x+ 6 = x + x x + x 6= (x+ 9)( x = 9 x= or x= The height is m and the base is 9 m.. c = ( c + ( c = ( c + ( c+ ( c 5. xy 7x + y = x( y 7) + ( y 7) = ( y 7)( x p is prime. u + u = ( u u+ ) = ( u )( u t 75 = (t = (t+ (t
43 Chapter 6 Factoring Polynomials y 5y+ 5 = 5( y y+ = 5( y ( y = 5( y q + q= 7 q(q+ ) x + x 8x = x (x+ (x+ = (x+ ( x = (x+ ( x+ )( x ) y 5 = ( y ( y + 5y+ mn 8 = ( mn+ 9)( mn 9).. x(5x+ = 5x + x = (5x ( x+ = 5x = or x+ = x= 5 x= y + y 9y 9 = y ( y+ 9( y+ = ( y 9)( y+ = ( y+ ( y ( y+ = y+ = or y = or y+ = y = or y = or y = a 6b = 6( a b ) = 6( a+ b)( a b) 6 6x 7y = (x y )(6 x + xy + 9 y ) x y 6xy y= y( x ( x+ ) 7. (x ( x+ = x = or x+ 5= x= x= x 7x= xx ( 7) = x= or x 7= x= x= 7 x 6x= 6 x 6x 6= ( x+ )( x 8) = x+ = or x 8= x= x= 8. Let x = width and x + is the length. A= lw = (x + )( x) = x + x x + x = ( x + x 56) = ( x+ ( x = x = or x = The tennis court is yd by 6 yd.. Let x represent the first odd integer. Then x + represents the second odd integer. xx ( + ) = 5 x + x= 5 x + x 5= ( x+ 7)( x = x+ 7= or x 5= x= 7 or x= 5 x+ = 5 or x+ = 7 The two integers are 5 and 7 or 5 and 7.. Let x represent the length of the base. Then x 5 represents the height of the triangle.
44 6 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual A = (base)(height) = ( x)( x 8 = ( x)( x 8 = x 5x x 5x 8= ( x ( x+ 7) = x = or x+ 7 = x= or x= 7 x 5= 7 The base is in. and the height is 7 in. 5. Let x = shorter leg, x the longer leg, and x the length of the hypotenuse. ( x) + (x = (x ) x + 9x 8x+ 9= 9x x+ x 6x+ 5= ( x ( x = x = or x = 5 The shorter leg is 5 ft. 6. h= 6t + 6 = 6t + 6 = 6( t = ( t+ )( t ) t+ = or t = t = t = The stone hits the ground in sec. Cumulative Review Exercises Chapters 6. 5 ( ( = ( = = 7 = 5. 5 ( t+ = t+ 5 t 8= t+ t = t+ 5t = 5 5 t = = 5. x y = 8 y = 8 x y 8 x = x 8 8 x y = or y =. Let x = number of quarters, (x + ) the number of nickels, and (x the number of dimes. Value Value Value of + of + of = $.8 quarters nickels dimes.5x+.5( x+ ) +.( x =.8.5x+.5x+. +.x. =.8 5..x = x = = There are quarters, nickels, and 7 dimes. 5 5 x 5 5 x 5 5 x [, ) 6. (a) Yes m = (c) (, (d) = x + x = (, ).
45 Chapter 6 Factoring Polynomials 7 (e).. (p 5p (p = 8p p p p + 5 p+ = 8p p + p+ (w 7) = ( w) ( w)(7) + 7 = w 8w (a) Vertical line Undefined (c) (5, ) (d) Does not exist y y = m( x x ) y 5= [ x ( ] y 5= ( x+ y 5= x+ 9 y = x x y = 5x 6y = Multiply the first equation by to obtain opposite coefficients on y. Then add the equations and solve the resulting equation for x. x+ 6y = 8 5x 6y = x = 5 ( y = y = y = 6 y = Solution: (5, ). y y 7 y + y + 5y = y y y y 5y 7 = y 5y r + 5r + 5r+ r r + r + r 5r + ( r r ) 5r + r (5r 5 r ) 5r 5r (5r 5 r) r + (r r + 5r + 5r+ + r 5 7 c c c = = c = c c c 7 6ab 5 9 = a b = a b = 5 9 ab 8 8 = 6 = w 6 = ( w ) ( 6) = ( w + ( w = ( w + ( w+ )( w ) 8. ax + bx ya 5yb = ( x a+ b ( y a+ b = ( a+ 5 b)(x y) 9. x 8x 5 = (x+ (x ab
46 8 Miller // O Neill // Hyde Beginning Algebra Instructor s Solution Manual. x(x (x + = x= or x = or x+ 5= x= x= x= 5
Algebra I. Book 2. Powered by...
Algebra I Book 2 Powered by... ALGEBRA I Units 47 by The Algebra I Development Team ALGEBRA I UNIT 4 POWERS AND POLYNOMIALS......... 1 4.0 Review................ 2 4.1 Properties of Exponents..........
More informationMultiplication of Polynomials
Summary 391 Chapter 5 SUMMARY Section 5.1 A polynomial in x is defined by a finite sum of terms of the form ax n, where a is a real number and n is a whole number. a is the coefficient of the term. n is
More informationMath 10  Unit 5 Final Review  Polynomials
Class: Date: Math 10  Unit 5 Final Review  Polynomials Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Factor the binomial 44a + 99a 2. a. a(44 + 99a)
More information5.1, 5.2, 5.3 Properites of Exponents last revised 6/7/2014. c = Properites of Exponents. *Simplify each of the following:
48 5.1, 5.2, 5.3 Properites of Exponents last revised 6/7/2014 Properites of Exponents 1. x a x b = x a+b *Simplify each of the following: a. x 4 x 8 = b. x 5 x 7 x = 2. xa xb = xa b c. 5 6 5 11 = d. x14
More information( ) Chapter 7 ( ) ( ) ( ) ( ) Exercise Set The greatest common factor is x + 3.
Chapter 7 Exercise Set 7.1 1. A prime number is an integer greater than 1 that has exactly two factors, itself and 1. 3. To factor an expression means to write the expression as the product of factors.
More informationLESSON 7.1 FACTORING POLYNOMIALS I
LESSON 7.1 FACTORING POLYNOMIALS I LESSON 7.1 FACTORING POLYNOMIALS I 293 OVERVIEW Here s what you ll learn in this lesson: Greatest Common Factor a. Finding the greatest common factor (GCF) of a set of
More informationLecture Guide. Math 90  Intermediate Algebra. Stephen Toner. Intermediate Algebra, 3rd edition. Miller, O'Neill, & Hyde. Victor Valley College
Lecture Guide Math 90  Intermediate Algebra to accompany Intermediate Algebra, 3rd edition Miller, O'Neill, & Hyde Prepared by Stephen Toner Victor Valley College Last updated: 4/17/16 5.1 Exponents &
More informationPolynomial Functions
Polynomial Functions NOTE: Some problems in this file are used with permission from the engageny.org website of the New York State Department of Education. Various files. Internet. Available from https://www.engageny.org/ccsslibrary.
More information( ) Chapter 6 ( ) ( ) ( ) ( ) Exercise Set The greatest common factor is x + 3.
Chapter 6 Exercise Set 6.1 1. A prime number is an integer greater than 1 that has exactly two factors, itself and 1. 3. To factor an expression means to write the expression as the product of factors.
More informationAdding and Subtracting Polynomials
Adding and Subtracting Polynomials Polynomial A monomial or sum of monomials. Binomials and Trinomial are also polynomials. Binomials are sum of two monomials Trinomials are sum of three monomials Degree
More informationreview To find the coefficient of all the terms in 15ab + 60bc 17ca: Coefficient of ab = 15 Coefficient of bc = 60 Coefficient of ca = 17
1. Revision Recall basic terms of algebraic expressions like Variable, Constant, Term, Coefficient, Polynomial etc. The coefficients of the terms in 4x 2 5xy + 6y 2 are Coefficient of 4x 2 is 4 Coefficient
More informationMath 101 Study Session Spring 2016 Test 4 Chapter 10, Chapter 11 Chapter 12 Section 1, and Chapter 12 Section 2
Math 101 Study Session Spring 2016 Test 4 Chapter 10, Chapter 11 Chapter 12 Section 1, and Chapter 12 Section 2 April 11, 2016 Chapter 10 Section 1: Addition and Subtraction of Polynomials A monomial is
More informationTest 4 also includes review problems from earlier sections so study test reviews 1, 2, and 3 also.
MATD 0370 ELEMENTARY ALGEBRA REVIEW FOR TEST 4 (1.110.1, not including 8.2) Test 4 also includes review problems from earlier sections so study test reviews 1, 2, and 3 also. 1. Factor completely: a 2
More informationGet Ready. 6. Expand using the distributive property. a) 6m(2m 4) b) 8xy(2x y) c) 6a 2 ( 3a + 4ab) d) 2a(b 2 6ab + 7)
Get Ready BLM 5 1... Classify Polynomials 1. Classify each polynomial by the number of terms. 2y x 2 + 3x + 2 c) 6x 2 y + 2xy + 4 d) x 2 + y 2 e) 3x 2 + 2x + y 4 6. Expand using the distributive property.
More informationTHE RING OF POLYNOMIALS. Special Products and Factoring
THE RING OF POLYNOMIALS Special Products and Factoring Special Products and Factoring Upon completion, you should be able to Find special products Factor a polynomial completely Special Products  rules
More information= The algebraic expression is 3x 2 x The algebraic expression is x 2 + x. 3. The algebraic expression is x 2 2x.
Chapter 7 Maintaining Mathematical Proficiency (p. 335) 1. 3x 7 x = 3x x 7 = (3 )x 7 = 5x 7. 4r 6 9r 1 = 4r 9r 6 1 = (4 9)r 6 1 = 5r 5 3. 5t 3 t 4 8t = 5t t 8t 3 4 = ( 5 1 8)t 3 4 = ()t ( 1) = t 1 4. 3(s
More informationAlgebra I Polynomials
Slide 1 / 217 Slide 2 / 217 Algebra I Polynomials 20140424 www.njctl.org Slide 3 / 217 Table of Contents Definitions of Monomials, Polynomials and Degrees Adding and Subtracting Polynomials Multiplying
More informationSYMBOL NAME DESCRIPTION EXAMPLES. called positive integers) negatives, and 0. represented as a b, where
EXERCISE A1 Things to remember: 1. THE SET OF REAL NUMBERS SYMBOL NAME DESCRIPTION EXAMPLES N Natural numbers Counting numbers (also 1, 2, 3,... called positive integers) Z Integers Natural numbers, their
More informationTo Find the Product of Monomials. ax m bx n abx m n. Let s look at an example in which we multiply two monomials. (3x 2 y)(2x 3 y 5 )
5.4 E x a m p l e 1 362SECTION 5.4 OBJECTIVES 1. Find the product of a monomial and a polynomial 2. Find the product of two polynomials 3. Square a polynomial 4. Find the product of two binomials that
More informationAlgebra I. Polynomials.
1 Algebra I Polynomials 2015 11 02 www.njctl.org 2 Table of Contents Definitions of Monomials, Polynomials and Degrees Adding and Subtracting Polynomials Multiplying a Polynomial by a Monomial Multiplying
More informationWhen factoring, we ALWAYS start with the (unless it s 1).
Math 100 Elementary Algebra Sec 5.1: The Greatest Common Factor and Factor By Grouping (FBG) Recall: In the product XY, X and Y are factors. Defn In an expression, any factor that is common to each term
More informationAlgebraic Expressions
Algebraic Expressions 1. Expressions are formed from variables and constants. 2. Terms are added to form expressions. Terms themselves are formed as product of factors. 3. Expressions that contain exactly
More informationCollecting Like Terms
MPM1D Unit 2: Algebra Lesson 5 Learning goal: how to simplify algebraic expressions by collecting like terms. Date: Collecting Like Terms WARMUP Example 1: Simplify each expression using exponent laws.
More informationFactoring Polynomials. Review and extend factoring skills. LEARN ABOUT the Math. Mai claims that, for any natural number n, the function
Factoring Polynomials GOAL Review and extend factoring skills. LEARN ABOUT the Math Mai claims that, for any natural number n, the function f (n) 5 n 3 1 3n 2 1 2n 1 6 always generates values that are
More information254 CHAPTER 9 RADICAL EXPRESSIONS AND EQUATIONS
Chapter 9 Radical Expressions and Equations Section 9.. Introduction to Radical Expressions Practice 9... Since negative so this is not defined using real numbers.. +  +. 8   9 . 88   8   08.
More informationAnswers. Chapter 9 A92. Angles Theorem (Thm. 5.6) then XZY. Base Angles Theorem (Thm. 5.6) 5, 2. then WV WZ;
9 9. M, 0. M ( 9, 4) 7. If WZ XZ, then ZWX ZXW ; Base Angles Theorem (Thm..6). M 9,. M ( 4, ) 74. If XZ XY, then XZY Y; Base Angles Theorem (Thm..6). M, 4. M ( 9, ) 7. If V WZV, then WV WZ; Converse of
More informationHow could you express algebraically, the total amount of money he earned for the three days?
UNIT 4 POLYNOMIALS Math 11 Unit 4 Introduction p. 1 of 1 A. Algebraic Skills Unit 4 Polynomials Introduction Problem: Derrek has a part time job changing tires. He gets paid the same amount for each tire
More informationANSWERS. CLASS: VIII TERM  1 SUBJECT: Mathematics. Exercise: 1(A) Exercise: 1(B)
ANSWERS CLASS: VIII TERM  1 SUBJECT: Mathematics TOPIC: 1. Rational Numbers Exercise: 1(A) 1. Fill in the blanks: (i) 21/24 (ii) 4/7 < 4/11 (iii)16/19 (iv)11/13 and 11/13 (v) 0 2. Answer True or False:
More informationInstructor: Richard Getso Course: Math 200.P10 TR 1:00 PM Spring 2016 (Getso)
1/8/016 Practice Test 1 (Chapter 11) Richard Getso Student: Richard Getso Date: 1/8/16 Instructor: Richard Getso Course: Math 00.P10 TR 1:00 PM Spring 016 (Getso) Assignment: Practice Test 1 (Chapter 11)
More informationLESSON 6.2 POLYNOMIAL OPERATIONS I
LESSON 6.2 POLYNOMIAL OPERATIONS I Overview In business, people use algebra everyday to find unknown quantities. For example, a manufacturer may use algebra to determine a product s selling price in order
More informationHomework 1/Solutions. Graded Exercises
MTH 3103 Abstract Algebra I and Number Theory S18 Homework 1/Solutions Graded Exercises Exercise 1. Below are parts of the addition table and parts of the multiplication table of a ring. Complete both
More informationF.3 Special Factoring and a General Strategy of Factoring
F.3 Special Factoring and a General Strategy of Factoring Difference of Squares section F4 233 Recall that in Section P2, we considered formulas that provide a shortcut for finding special products, such
More informationSolutions Key Exponents and Polynomials
CHAPTER 7 Solutions Key Exponents and Polynomials ARE YOU READY? PAGE 57. F. B. C. D 5. E 6. 7 7. 5 8. (0 9. x 0. k 5. 9..  (  5. 5. 5 6. 7. ( 6 (((((( 8 5 5 5 5 6 8. 0.06 9.,55 0. 5.6. 6 +
More informationRATIO AND PROPORTION, INDICES, LOGARITHMS
CHAPTER RATIO AND PROPORTION, INDICES, LOGARITHMS UNIT I: RATIO LEARNING OBJECTIVES After reading this unit a student will learn How to compute and compare two ratios; Effect of increase or decrease of
More informationSummer Prep Packet for students entering Algebra 2
Summer Prep Packet for students entering Algebra The following skills and concepts included in this packet are vital for your success in Algebra. The Mt. Hebron Math Department encourages all students
More informationMath 0312 EXAM 2 Review Questions
Name Decide whether the ordered pair is a solution of the given system. 1. 4x + y = 2 2x + 4y = 20 ; (2, 6) Solve the system by graphing. 2. x  y = 6 x + y = 16 Solve the system by substitution. If
More informationLATE AND ABSENT HOMEWORK IS ACCEPTED UP TO THE TIME OF THE CHAPTER TEST ON
Trig/Math Anal Name No LATE AND ABSENT HOMEWORK IS ACCEPTED UP TO THE TIME OF THE CHAPTER TEST ON HW NO. SECTIONS ASSIGNMENT DUE FS1 44 Practice Set A #157 eoo 45 Practice Set B #145 eoo, 57, 59 FS
More informationThe P/Q Mathematics Study Guide
The P/Q Mathematics Study Guide Copyright 007 by Lawrence Perez and Patrick Quigley All Rights Reserved Table of Contents Ch. Numerical Operations  Integers...  Fractions...  Proportion and Percent...
More informationExpanding brackets and factorising
CHAPTER 8 Epanding brackets and factorising 8 CHAPTER Epanding brackets and factorising 8.1 Epanding brackets There are three rows. Each row has n students. The number of students is 3 n 3n. Two students
More informationMaintaining Mathematical Proficiency
Chapter 7 Maintaining Mathematical Proficiency Simplify the expression. 1. 5x 6 + 3x. 3t + 7 3t 4 3. 8s 4 + 4s 6 5s 4. 9m + 3 + m 3 + 5m 5. 4 3p 7 3p 4 1 z 1 + 4 6. ( ) 7. 6( x + ) 4 8. 3( h + 4) 3( h
More information= 9 = x + 8 = = 5x 19. For today: 2.5 (Review) and. 4.4a (also review) Objectives:
Math 65 / Notes & Practice #1 / 20 points / Due. / Name: Home Work Practice: Simplify the following expressions by reducing the fractions: 16 = 4 = 8xy =? = 9 40 32 38x 64 16 Solve the following equations
More informationAlgebra I Vocabulary Cards
Algebra I Vocabulary Cards Table of Contents Expressions and Operations Natural Numbers Whole Numbers Integers Rational Numbers Irrational Numbers Real Numbers Order of Operations Expression Variable Coefficient
More informationNAME DATE PERIOD. Study Guide and Intervention. Solving Polynomial Equations. For any number of terms, check for: greatest common factor
55 Factor Polynomials Study Guide and Intervention For any number of terms, check for: greatest common factor Techniques for Factoring Polynomials For two terms, check for: Difference of two squares a
More informationMiniLecture 5.1 Exponents and Scientific Notation
MiniLecture.1 Eponents and Scientific Notation Learning Objectives: 1. Use the product rule for eponents.. Evaluate epressions raised to the zero power.. Use the quotient rule for eponents.. Evaluate
More information43 Solving Quadratic Equations by Factoring. Write a quadratic equation in standard form with the given root(s). 1. 8, 5 ANSWER: ANSWER: ANSWER:
Write a quadratic equation in standard form with the given root(s). 1. 8, 5 8. 9. (2x 5)(x + 6) 2. (4x 3)(4x 1) Solve each equation. 3. 10. 6, 6 4. Factor each polynomial. 5x(7x 3) 11. 12. 5. 6. (6x 1)(3x
More informationAdditional Practice Lessons 2.02 and 2.03
Additional Practice Lessons 2.02 and 2.03 1. There are two numbers n that satisfy the following equations. Find both numbers. a. n(n 1) 306 b. n(n 1) 462 c. (n 1)(n) 182 2. The following function is defined
More informationLesson 6. Diana Pell. Monday, March 17. Section 4.1: Solve Linear Inequalities Using Properties of Inequality
Lesson 6 Diana Pell Monday, March 17 Section 4.1: Solve Linear Inequalities Using Properties of Inequality Example 1. Solve each inequality. Graph the solution set and write it using interval notation.
More informationChapter 6. WorkedOut Solutions AB 3.61 AC 5.10 BC = 5
27. onstruct a line ( DF ) with midpoint P parallel to and twice the length of QR. onstruct a line ( EF ) with midpoint R parallel to and twice the length of QP. onstruct a line ( DE ) with midpoint Q
More informationUNC Charlotte 2005 Comprehensive March 7, 2005
March 7, 2005 1. The numbers x and y satisfy 2 x = 15 and 15 y = 32. What is the value xy? (A) 3 (B) 4 (C) 5 (D) 6 (E) none of A, B, C or D Solution: C. Note that (2 x ) y = 15 y = 32 so 2 xy = 2 5 and
More informationLos Angeles Southwest College. Mathematics Department. Math 115 Common Final Exam. Study Guide (solutions) Fall 2015
Los Angeles Southwest College Mathematics Department Math 5 Common Final Exam Study Guide (solutions) Fall 05 Prepared by: Instr. B.Nash Dr. L.Saakian Chapter. The Real Number System Definitions Placevalue
More informationAlgebra II First Semester Assignment #5 (Review of Sections 1.1 through 1.8)
Algebra II First Semester Assignment #5 (Review of Sections 1.1 through 1.8) Do not rely solely on this review to prepare for the test. These problems are meant only as a means to remind you of the types
More informationPolynomials 370 UNIT 10 WORKING WITH POLYNOMIALS. The railcars are linked together.
UNIT 10 Working with Polynomials The railcars are linked together. 370 UNIT 10 WORKING WITH POLYNOMIALS Just as a train is built from linking railcars together, a polynomial is built by bringing terms
More informationEuler s Multiple Solutions to a Diophantine Problem
Euler s Multiple Solutions to a Diophantine Problem Christopher Goff University of the Pacific 18 April 2015 CMC 3 Tahoe 2015 Euler 1/ 28 Leonhard Euler (17071783) Swiss Had 13 kids Worked in St. Petersburg
More informationAdding and Subtracting Polynomials
Adding and Subtracting Polynomials When you add polynomials, simply combine all like terms. When subtracting polynomials, do not forget to use parentheses when needed! Recall the distributive property:
More informationCONTENTS COLLEGE ALGEBRA: DR.YOU
1 CONTENTS CONTENTS Textbook UNIT 1 LECTURE 11 REVIEW A. p. LECTURE 1 RADICALS A.10 p.9 LECTURE 1 COMPLEX NUMBERS A.7 p.17 LECTURE 14 BASIC FACTORS A. p.4 LECTURE 15. SOLVING THE EQUATIONS A.6 p.
More information5.1 Modelling Polynomials
5.1 Modelling Polynomials FOCUS Model, write, and classify polynomials. In arithmetic, we use Base Ten Blocks to model whole numbers. How would you model the number 234? In algebra, we use algebra tiles
More informationRemember, you may not use a calculator when you take the assessment test.
Elementary Algebra problems you can use for practice. Remember, you may not use a calculator when you take the assessment test. Use these problems to help you get up to speed. Perform the indicated operation.
More informationUnit # 4 : Polynomials
Name: Block: Teacher: Miss Zukowski Date Submitted: / / 2018 Unit # 4 : Polynomials Submission Checklist: (make sure you have included all components for full marks) Cover page & Assignment Log Class Notes
More informationLESSON 7.2 FACTORING POLYNOMIALS II
LESSON 7.2 FACTORING POLYNOMIALS II LESSON 7.2 FACTORING POLYNOMIALS II 305 OVERVIEW Here s what you ll learn in this lesson: Trinomials I a. Factoring trinomials of the form x 2 + bx + c; x 2 + bxy +
More informationLESSON 6.2 POLYNOMIAL OPERATIONS I
LESSON 6. POLYNOMIAL OPERATIONS I LESSON 6. POLYNOMIALS OPERATIONS I 63 OVERVIEW Here's what you'll learn in this lesson: Adding and Subtracting a. Definition of polynomial, term, and coefficient b. Evaluating
More informationTopic 7: Polynomials. Introduction to Polynomials. Table of Contents. Vocab. Degree of a Polynomial. Vocab. A. 11x 7 + 3x 3
Topic 7: Polynomials Table of Contents 1. Introduction to Polynomials. Adding & Subtracting Polynomials 3. Multiplying Polynomials 4. Special Products of Binomials 5. Factoring Polynomials 6. Factoring
More informationAlgebra I. Slide 1 / 216. Slide 2 / 216. Slide 3 / 216. Polynomials
Slide 1 / 216 Slide 2 / 216 lgebra I Polynomials 20151102 www.njctl.org Table of ontents efinitions of Monomials, Polynomials and egrees dding and Subtracting Polynomials Multiplying a Polynomial by
More informationmn 3 17x 2 81y 4 z Algebra I Definitions of Monomials, Polynomials and Degrees 32,457 Slide 1 / 216 Slide 2 / 216 Slide 3 / 216 Slide 4 / 216
Slide 1 / 216 Slide 2 / 216 lgebra I Polynomials 20151102 www.njctl.org Slide 3 / 216 Table of ontents efinitions of Monomials, Polynomials and egrees dding and Subtracting Polynomials Multiplying a
More informationA2. Polynomials and Factoring. Section A2 1
A Polynomials and Factoring Section A 1 What you ll learn about Adding, Subtracting, and Multiplying Polynomials Special Products Factoring Polynomials Using Special Products Factoring Trinomials Factoring
More informationStudy Guide for Math 095
Study Guide for Math 095 David G. Radcliffe November 7, 1994 1 The Real Number System Writing a fraction in lowest terms. 1. Find the largest number that will divide into both the numerator and the denominator.
More informationNever leave a NEGATIVE EXPONENT or a ZERO EXPONENT in an answer in simplest form!!!!!
1 ICM Unit 0 Algebra Rules Lesson 1 Rules of Exponents RULE EXAMPLE EXPLANANTION a m a n = a m+n A) x x 6 = B) x 4 y 8 x 3 yz = When multiplying with like bases, keep the base and add the exponents. a
More informationWorksheets for GCSE Mathematics. Quadratics. mrmathematics.com Maths Resources for Teachers. Algebra
Worksheets for GCSE Mathematics Quadratics mrmathematics.com Maths Resources for Teachers Algebra Quadratics Worksheets Contents Differentiated Independent Learning Worksheets Solving x + bx + c by factorisation
More informationChapter 3: Section 3.1: Factors & Multiples of Whole Numbers
Chapter 3: Section 3.1: Factors & Multiples of Whole Numbers Prime Factor: a prime number that is a factor of a number. The first 15 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,
More informationCourse 2BA1: Hilary Term 2007 Section 8: Quaternions and Rotations
Course BA1: Hilary Term 007 Section 8: Quaternions and Rotations David R. Wilkins Copyright c David R. Wilkins 005 Contents 8 Quaternions and Rotations 1 8.1 Quaternions............................ 1 8.
More informationPRECALCULUS By: Salah Abed, Sonia Farag, Stephen Lane, Tyler Wallace, and Barbara Whitney
PRECALCULUS By: Salah Abed, Sonia Farag, Stephen Lane, Tyler Wallace, and Barbara Whitney MATH 141/14 1 PreCalculus by Abed, Farag, Lane, Wallace, and Whitney is licensed under the creative commons attribution,
More informationArithmetic Operations. The real numbers have the following properties: In particular, putting a 1 in the Distributive Law, we get
MCA AP Calculus AB Summer Assignment The following packet is a review of many of the skills needed as we begin the study of Calculus. There two major sections to this review. Pages 29 are review examples
More informationx 9 or x > 10 Name: Class: Date: 1 How many natural numbers are between 1.5 and 4.5 on the number line?
1 How many natural numbers are between 1.5 and 4.5 on the number line? 2 How many composite numbers are between 7 and 13 on the number line? 3 How many prime numbers are between 7 and 20 on the number
More informationAlgebra I Vocabulary Cards
Algebra I Vocabulary Cards Table of Contents Expressions and Operations Natural Numbers Whole Numbers Integers Rational Numbers Irrational Numbers Real Numbers Absolute Value Order of Operations Expression
More informationMathwithsheppard.weebly.com
Unit #: Powers and Polynomials Unit Outline: Date Lesson Title Assignment Completed.1 Introduction to Algebra. Discovering the Exponent Laws Part 1. Discovering the Exponent Laws Part. Multiplying and
More informationF.4 Solving Polynomial Equations and Applications of Factoring
section F4 243 F.4 ZeroProduct Property Many application problems involve solving polynomial equations. In Chapter L, we studied methods for solving linear, or firstdegree, equations. Solving higher
More informationPREFACE. Synergy for Success in Mathematics 8 is designed for Grade 8 students. The textbook contains
Synergy for Success in Mathematics 8 is designed for Grade 8 students. The textbook contains all the required learning competencies and is supplemented with some additional topics for enrichment. Lessons
More informationWeb Solutions for How to Read and Do Proofs
Web Solutions for How to Read and Do Proofs An Introduction to Mathematical Thought Processes Sixth Edition Daniel Solow Department of Operations Weatherhead School of Management Case Western Reserve University
More informationUNC Charlotte 2005 Comprehensive March 7, 2005
March 7, 2005 1 The numbers x and y satisfy 2 x = 15 and 15 y = 32 What is the value xy? (A) 3 (B) 4 (C) 5 (D) 6 (E) none of A, B, C or D 2 Suppose x, y, z, and w are real numbers satisfying x/y = 4/7,
More informationCBSE QUESTION PAPER CLASSX MATHS
CBSE QUESTION PAPER CLASSX MATHS SECTION  A Question 1: In figure, AB = 5 3 cm, DC = 4cm, BD = 3cm, then tan θ is (a) (b) (c) (d) 1 3 2 3 4 3 5 3 Question 2: In figure, what values of x will make DE
More informationChapter 8 Class Notes 8A1 (Lessons 81&82) Monomials and Factoring p Prime Factorization: a whole number expressed as the of factors.
Chapter 8 Class Notes Alg. 1H 8A1 (Lessons 81&8) Monomials and Factoring p. 404 Prime Factorization: a whole number epressed as the of factors. Tree Method: Ladder Method: Factored Form of a Monomial:
More information1.3 Algebraic Expressions. Copyright Cengage Learning. All rights reserved.
1.3 Algebraic Expressions Copyright Cengage Learning. All rights reserved. Objectives Adding and Subtracting Polynomials Multiplying Algebraic Expressions Special Product Formulas Factoring Common Factors
More information5.3. Polynomials and Polynomial Functions
5.3 Polynomials and Polynomial Functions Polynomial Vocabulary Term a number or a product of a number and variables raised to powers Coefficient numerical factor of a term Constant term which is only a
More informationLESSON 6.3 POLYNOMIAL OPERATIONS II
LESSON 6.3 POLYNOMIAL OPERATIONS II LESSON 6.3 POLYNOMIALS OPERATIONS II 277 OVERVIEW Here's what you'll learn in this lesson: Multiplying Binomials a. Multiplying binomials by the FOIL method b. Perfect
More informationUnit 3 Factors & Products
1 Unit 3 Factors & Products General Outcome: Develop algebraic reasoning and number sense. Specific Outcomes: 3.1 Demonstrate an understanding of factors of whole number by determining the: o prime factors
More informationMATH98 Intermediate Algebra Practice Test Form A
MATH98 Intermediate Algebra Practice Test Form A MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Solve the equation. 1) (y  4)  (y + ) = 3y 1) A)
More informationFor Your Notebook E XAMPLE 1. Factor when b and c are positive KEY CONCEPT. CHECK (x 1 9)(x 1 2) 5 x 2 1 2x 1 9x Factoring x 2 1 bx 1 c
9.5 Factor x2 1 bx 1 c Before You factored out the greatest common monomial factor. Now You will factor trinomials of the form x 2 1 bx 1 c. Why So you can find the dimensions of figures, as in Ex. 61.
More informationMSLC Math 1075 Final Exam Review. 1. Factor completely Solve the absolute value equation algebraically. g. 8x b. 4x 2 5x. f.
MSLC Math 07 Final Exam Review Disclaimer: This should NOT be used as your only guide for what to study.. Factor completely. a. x y xy xy mn n 7x x x x 0xy x y e. xy y x y f. z z 7 g. mn m n h. c d i.
More informationREAL WORLD SCENARIOS: PART IV {mostly for those wanting 114 or higher} 1. If 4x + y = 110 where 10 < x < 20, what is the least possible value of y?
REAL WORLD SCENARIOS: PART IV {mostly for those wanting 114 or higher} REAL WORLD SCENARIOS 1. If 4x + y = 110 where 10 < x < 0, what is the least possible value of y? WORK AND ANSWER SECTION. Evaluate
More informationSECTION 1.4 PolyNomiAls feet. Figure 1. A = s 2 = (2x) 2 = 4x 2 A = 2 (2x) 3 _ 2 = 1 _ = 3 _. A = lw = x 1. = x
SECTION 1.4 PolyNomiAls 4 1 learning ObjeCTIveS In this section, you will: Identify the degree and leading coefficient of polynomials. Add and subtract polynomials. Multiply polynomials. Use FOIL to multiply
More information91 Skills Practice Factors and Greatest Common Factors Find the factors of each number. Then classify each number as prime or composite
91 Skills Practice Factors and Greatest Common Factors Find the factors of each number. Then classify each number as prime or composite. 1. 10 2. 31 3. 16 4. 52 5. 38 6. 105 Find the prime factorization
More informationFind two positive factors of 24 whose sum is 10. Make an organized list.
9.5 Study Guide For use with pages 582 589 GOAL Factor trinomials of the form x 2 1 bx 1 c. EXAMPLE 1 Factor when b and c are positive Factor x 2 1 10x 1 24. Find two positive factors of 24 whose sum is
More informationJUST THE MATHS UNIT NUMBER 1.5. ALGEBRA 5 (Manipulation of algebraic expressions) A.J.Hobson
JUST THE MATHS UNIT NUMBER 1.5 ALGEBRA 5 (Manipulation of algebraic expressions) by A.J.Hobson 1.5.1 Simplification of expressions 1.5.2 Factorisation 1.5.3 Completing the square in a quadratic expression
More informationFinal Exam Review for DMAT 0310
Final Exam Review for DMAT 010 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Factor the polynomial completely. What is one of the factors? 1) x
More informationSolutions to Chapter Review Questions, Chapter 0
Instructor s Solutions Manual, Chapter 0 Review Question 1 Solutions to Chapter Review Questions, Chapter 0 1. Explain how the points on the real line correspond to the set of real numbers. solution Start
More informationMath 2 Variable Manipulation Part 3 Polynomials A
Math 2 Variable Manipulation Part 3 Polynomials A 1 MATH 1 REVIEW: VOCABULARY Constant: A term that does not have a variable is called a constant. Example: the number 5 is a constant because it does not
More informationBeginning Algebra MAT0024C. Professor Sikora. Professor M. J. Sikora ~ Valencia Community College
Beginning Algebra Professor Sikora MAT002C POLYNOMIALS 6.1 Positive Integer Exponents x n = x x x x x [n of these x factors] base exponent Numerical: Ex:  = where as Ex: () = Ex:  = and Ex: () = Rule:
More informationAlgebra I Lesson 6 Monomials and Polynomials (Grades 912) Instruction 61 Multiplying Polynomials
In algebra, we deal with different types of expressions. Grouping them helps us to learn rules and concepts easily. One group of expressions is called polynomials. In a polynomial, the powers are whole
More informationMA094 Part 2  Beginning Algebra Summary
MA094 Part  Beginning Algebra Summary Page of 8/8/0 Big Picture Algebra is Solving Equations with Variables* Variable Variables Linear Equations x 0 MA090 Solution: Point 0 Linear Inequalities x < 0 page
More informationQuiz # 6 Date: Name. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Quiz # 6 Date: Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find all numbers not in the domain of the function. ) f(x) = x + None  0 ) )
More informationChapter 5 Simplifying Formulas and Solving Equations
Chapter 5 Simplifying Formulas and Solving Equations Look at the geometry formula for Perimeter of a rectangle P = L W L W. Can this formula be written in a simpler way? If it is true, that we can simplify
More information