CHAPTER Points to Remember : POLYNOMIALS 7. A symbol having a fied numerical value is called a constant. For e.g. 9,,, etc.. A symbol which may take different numerical values is known as a variable. We usually denotes variable by, y, z etc.. A combination of constants and variables which are connected by basic mathematical operations, is known as Algebraic Epression. For e.g. 7 +, y etc. 4. An algebraic epression in which variable have only whole numbers as a power is called a Polynomial. 5. Highest power of the variable is called the degree of the polynomial. For e.g. 7 9 + 7 is a polynomial in of degree. 6. A polynomial of degree is called a linear polynomial. For e.g. 7 + is a linear polynomial in. 7. A polynomial of degree is called a Quadratic Polynomial. For e.g. y 7y + is a Quadratic polynomial in y. 8. A polynomial of degree is called a Cubic Polynomial. For e.g. t 7t + t is a cubic polynomial in t. 9. According to number of terms, a polynomial having one non-zero term is a monomial, a polynomial having two non-zero terms is a bionomial and a polynomial have three non-zero terms is a trinomial. 0. Remainder Theorem : Let f() be a polynomial of degree n and let a be any real number. If f() is divided by linear polynomial ( a), then the remainder is f(a).. Factor Theorem : Let f() be a polynomial of degree n > and a be any real number. (i) If f(a) = 0, then a is a factor of f(). (ii) If ( a) is a factor of f() then f(a) = 0.. Algebraic Identities : (i) ( + y) = + y + y (ii) ( y) = y + y (iii) y = ( y) ( + y) (iv) ( + y + z) = + y + z + y + yz + z (v) ( + y) = + y + y ( + y) (vi) ( y) = y y ( y) (vii) y = ( y) ( + y + y ) (viii) + y = ( + y) ( y + y ) (i) + y + z yz = ( + y + z) ( + y + z y yz z). If + y + z = 0 then, + y + z = yz. Eample. ILLUSTRATIVE EXAMPLES Which of the following epressions are Polynomials? In case of a polynomial, give degree of polynomial. (i) 4 + 7 + (ii) y 7 y 6 (iii) (iv) (v) 7 (vi) / 4 POLYNOMIALS MATHEMATICS IX
Solution. (i) is a polynomial of degree 4. (ii) is a polynomial of degree. (v) is a polynomial of degree 0. Eample. Verify whether the following are zeros of the polynomial, indicated against them. 4 (i) p ( ), (ii) p( ) 5, 5 (iii) p() =, =, (iv) p() = ( + ) ( ), =, m (v) p() =, = 0 (vi) p() = l + m, l (vii) p() =,, (viii) p() = +, Solution. (i) We have, p() = + At, p 0 is a zero of p(). (ii) We have, p() = 5 4 4 4 At, p 5 4 0 5 5 5 4 is not a zero of p(). 5 (iii) We have, p( ) At =, p() = () = = 0 also, at =, p( ) = ( ) = = 0, both are zeros of p(). (iv) We have, p() = ( + ) ( ) At =, p( ) = ( + ) ( ) = 0 ( ) = 0 also, at =, p() = ( + ) ( ) = 0 = 0, both are zeros of p(). (v) We have, p() = At = 0, p(0) = 0 = 0 0 is a zero of p(). (vi) We have, p() = l + m m m m At,, p l m m m 0 l l l m is a zero of p(). l (vii) We have, p() = NCERT MATHEMATICS IX POLYNOMIALS 5
6 POLYNOMIALS MATHEMATICS IX At, 0 p, and at, 0 4 4 p, is a zero of p(), but is not a zero of p(). (viii) We have p() = + At 0 p, is not a zero of p(). Eample. Find the zero of the polynomial in each of the following cases : (i) p() = + 5 (ii) p() = 5 (iii) p() = + 5 (iv) p() = (v) p() = (vi) p() = a ; a 0 (vii) p() = c + d, c 0, c, d are real numbers. NCERT Solution. We know, finding a zero of p(), is the same as solving the equation p() = 0. (i) p() = 0 + 5 = 0 = 5 5 is a zero of p(). (ii) p() = 0 5 = 0 = 5 5 is a zero of p(). (iii) p() = 0 + 5 = 0 = 5 5 5 is a zero of p(). (iv) p() = 0 = 0 = is a zero of p(). (v) p() = 0 = 0 = 0 0 is a zero of p(). (vi) p() = 0 a = 0 = 0 ( Given a 0) 0 is a zero of p(). (vii) p() = 0 c + d = 0 c = d c d c d is a zero of p().
Eample 4. Using remainder theorem, find the remainder when p() = 5 + 9 8 is divided by ( ). Solution. Remainder = p() = () 5() + 9 () 8 = 54 45 + 7 8 = 8 5 = 8 Ans. Eample 5. Find the remainder when is divided by : (i) + (ii) (iii) (iv) + (v) 5 + NCERT Solution. (i) + = 0 = by remainder theorem, the required remainder is p ( ). Now, p() = + + + p( ) = ( ) + ( ) + ( ) + = + + = 0 Remainder = 0 (ii) 0. by remainder theorem, the required remainder is p. Now, p 6 8 8 4 8 7 8 7 Remainder 8 (iii) = 0 by remainder theorem, the required remainder is p(0). Now, p(0) = (0) + (0) + (0) + = Remainder = (iv) + = 0 = by remainder theorem, the required remainder is p( ). Now, p( ) = ( ) + ( ) + ( ) + = + + Remainder = + +. 5 (v) 5 + = 0 = 5 = 5 By remainder theorem, the required remainder is p. 5 5 5 5 Now, p 5 75 5 8 4 MATHEMATICS IX POLYNOMIALS 7
5 50 60 8 7 8 8 7 Remainder. 8 Eample 6. Using Factor theorem, show that ( + ) is a factor of 4. Solution. Let p() = 4. Eample 7. Solution. Eample 8. Now, + = 0 =. ( + ) is a factor of p() p( ) = 0 Now, p( ) = ( ) 4 ( ) = 6 4 = 0, which shows that ( + ) is a factor of p(). Use the factor theorem to determine whether g() is a factor of p() is each of the following cases: (i) p() = +, g() = + (ii) p() = + + +, g() = + (iii) p() = 4 + + 6, g() = NCERT (i) In order to prove that g() = + is a factor of p() = +, it is sufficient to show that p( ) = 0. Now, p( ) = ( ) + ( ) ( ) = + + = 0 g() is a factor of p(). (ii) In order to prove that g() = + is a factor of p() = + + +, it is sufficient to show that p( ) = 0. Now, p ( ) = ( ) + ( ) + ( ) + = 8 + 6 + = 0. g() is not a factor of p(). (iii) In order to prove that g() = is a factor of p() = 4 + 6, it is sufficient to show that p() = 0 Now, p() = () 4() + 6 = 7 6 + 6 = 0. g() is not a factor of p(). If + a + b + 6 has as a factor and leaves a remainder, when divided by, find the value of a and b. Solution. Let p() = + a + b + 6 Since, is a factor of p() p() = 0 ( factor theorem) () + a() + b() + 6 = 0 a + b = 7...() Also, p() leaves remainder, when divided by. p() = ( Remainder theorem) () + a() + b() + 6 = a + b = 0...() Solving () and (), we get a =, b = Ans. 8 POLYNOMIALS MATHEMATICS IX
Eample 9. Factorize the following: (i) 6 8y (ii) + 7 7 (iii) 6a 8b (iv) + 5 4 (v) 9 + 8 Solution. (i) 6 ( y) (ii) ( + 7) ( + 7) = ( ) ( + 7) = ( ) ( + ) ( + 7) (iii) (4a) (9b) = (4a + 9b) (4a 9b) (iv) 8 4 ( 8) ( 8) = ( ) ( + 8) (v) 9 8 4 + 8 = 9 ( ) 4( ) = ( ) ( 9 4) Eample 0. Factorise : (i) 7 + (ii) + 7 + (iii) 6 + 5 6 (iv) 4 NCERT Solution. (i) 7 + here, p + q = coefficient of = 7 pq = coefficient of constant term = = p + q = 7 = 4, pq = = ( 4) ( ) 7 + = 4 + = 4 ( ) ( ) = (4 ) ( ) Ans. (ii) + 7 + = + + 6 + = ( + ) + ( + ) = ( + ) ( + ) (iii) 6 + 5 6 = 6 + 9 4 6 = ( + ) ( + ) = ( ) ( + ) (iv) 4 = + 4 4 = ( + ) 4( + ) = ( 4) ( + ) Eample. Factorize 9 5 using factor theorem. Solution. Let p() = 9 5. factors of constant term are ± and ±5. Now, p() = () 9() 5 0 p( ) = ( ) ( ) 9( ) 5 0. ( + ) is a factor of p(). p() = ( + ) ( 4 5) = ( + ) [ 5 + 5] = ( + ) [( 5) + ( 5)] = ( + ) ( + ) ( 5) Ans. Eample. Factorise : + + + 0 NCERT Solution. Let p() = + + + 0 Now, factors of constant term 0 are ±, ±, ± 5, ± 0 and ± 0. Now, p() = () + () + () + 0 = + + + 0 = 66 0 p( ) = ( ) + ( ) + ( ) + 0 = + + 0 = + = 0 ( + ) is a factor of + + + 0. Now, divide p() by +. MATHEMATICS IX POLYNOMIALS 9
p() = ( +) ( + + 0) = ( + ) [ + + 0 + 0] = ( + ) [( + ) + 0 ( + )] = ( + ) ( + ) ( + 0) Ans. Eample. Epand the following : (i) (a b + c) (ii) ( + y) (iii) ( y) Solution. (i) (a) + ( b) + (c) + (a) ( b) + ( b) (c) + (a) (c) = a + 4b + 9c 4ab bc + 6ac (ii) () + (y) + ()(y) [ + y] = 8 + y + 6y ( + y) = 8 + y + y + 6y (iii) () (y) () (y) [ y] = 7 8y 8y ( y) = 7 8y 54 y + 6y Eample 4. Factorize the following : (i) + 7y (ii) 7a 64b (iii) a 8b + 64c + 4abc Solution. (i) () + (y) = ( + y) ( y + 9y ) [ a + b = (a + b) (a ab + b )] (ii) (a) (4b) = (a 4b) (9a + ab + 6b ) [ a b = (a b) (a + ab + b )] (iii) (a) + ( b) + (4c) a ( b) (4c) = (a b + 4c) (a + 4b + 6c + ab + 8bc 4ac) [ a + b + c abc = (a + b + c) (a + b + c ab bc ca)] Eample 5. Verify that : + y + z yz ( y z).[( y) ( y z) ( z ) ] NCERT Solution. Consider, LHS = + y + z yz = ( + y + z) ( + y + z y yz z) ( + y + z). ( + y + z y yz z) ( + y + z) ( + y + z y yz z) ( + y + z) ( + + y + y + z + z y yz z) ( + y + z) [( + y y) + (y + z yz) + (z + z)] ( + y + z) [( y) + (y z) + (z ) ] = RHS. Hence verify. Eample 6. If + y + z = 0, show that + y + z = yz. NCERT Solution. We have + y + z = 0 + y = z Cubing both sides, we get ( + y) = ( z) + y + y ( + y) = z + y + y ( z) = z ( + y = z) 0 POLYNOMIALS MATHEMATICS IX
+ y + z yz = 0 + y + z = yz. Hence shown. Eample 7. Give possible epressions for the length and breadth of each of the following rectangles, in which their areas are given : (i) Area : 5a 5a + (ii) 5y + y NCERT Solution. (i) Given, Area of rectangle = 5a 5a + = 5a 5a 0a + = 5a (5a ) 4 (5a ) = (5a ) (5a 4) Possible length and breadth are (5a ) and (5a 4) units. (ii) Area of given rectangle = 5y + y = 5y + 8y 5y = 7y (5y + 4) (5y + 4) = (5y + 4) (7y ). Possible length and breadth are (5y + 4) and (7y ) units. Eample 8. What are the possible epressions for the dimensions of the cuboid whose volumes are given below: (i) Volume : (ii) Volume : ky + 8 ky 0 k NCERT Solution. (i) Volume : = ( 4) = ( 4) Possible dimensions of cuboid are, and ( 4) units. (ii) Volume = ky + 8ky 0k = 4 k (y + y 5) = 4k (y y + 5y 5) = 4k [y (y ) + 5 (y )] = 4k (y ) (y + 5) Possible dimensions of a cuboid are 4k, (y ) and (y + 5) units. PRACTICE EXERCISE. Which of the following epressions are polynomials in one variable? (i) + 7 + (ii) 4 (iii) y 7y (iv) y / y (v) 4 7 +. Write degree of each the following polynomial: (i) 7 4 9 + + 4 (ii) 7 y + y (iii) 5t 7t (iv) 0 (v) 7 + 4. Classify each of the following as linear, quadratic and cubic polynomial: (i) 7 (ii) 4 y + y (iii) 7r (iv) (v) y y 4. Find the value of p() = 5 + + 4 at (i) = 0 (ii) = (iii) = 5. Find the value of p(0), p() and p( ) where p() = +. 6. Verify whether the following are zeros of the polynomial, indicated against them: (i) p ( ) ; (ii) p() = ( + ) ( 4) ; =, 4 (iii) p() = + 6; =, (iv) p() = ; = 0,, MATHEMATICS IX POLYNOMIALS
7. Find the zero of the polynomial in each of the following : (i) p() = 4 (ii) p() = + 4 (iii) p() = 7 (iv) p() = r + s; r 0, r, s are real numbers. 8. Using Remainder Theorem, find the remainder when : (i) 4 7 + is divided by (ii) 7 + 6 + 4 is divided by (iii) + + is divided by + (iv) 4 4 + is divided by + (v) a + 5 + a is divided by a (vi) + a 6 + a is divided by + a 9. If 5 is the remainder when + a + 7 is divided by 4, find the value of a. 0. Without actual division, prove that + + 70 is eactly divisible by.. Show that is a factor of 5 + 6.. Find the value of p for which the polynomial 4 + p + + 8 is eactly divisible by +.. Find values of a and b so that the polynomial 4 + 7 + 4 + a + b is eactly divisible by and +. 4. The polynomials a + 7 and + 7 a are divided by +. If the remainder in each case is same, find the value of a. 5. The polynomials a + 4 and 4 + 4 a when divided by, leaves the remainder R and R respectively. Find the value of a if R = R. 6. Find the integral zeros of +. 7. Find the integral zeros of + 4 4. 8. Show that is a factor of p() = 6 + 5 4. 9. Show that + 4 is a factor of 4 + 4 + + 4. 0. Find value of a for which + a is a factor of f() = + a + + a + 4.. Without actual division, prove that 4 + + 5 6 is eactly divisible by +.. For what value of a is the polynomial a + 8 + a + 4 is eactly divisible by +.. If + a + b has as a factor and leaves a remainder 0 when divided by +, find a and b. 4. Factorise the following, using factor theorem : (i) 6 + 6 (ii) 6 + + 0 (iii) 5 + 6 (iv) + (v) + 4 0 (vi) 4 + 9 9 0 5. Factorize 6 + 5 7 6, given + 6 is one of its factor. 6. Factorize + 5 4 6, given + 9 is one of its factor. 7. Use suitable identity to find the following products : (i) ( + 4) ( + 6) (ii) ( ) ( + 8) 5 (iii) ( + ) ( ) (iv) t 5 t 8. Evaluate the following products without multiplying directly: (i) 87 9 (ii) 06 94 POLYNOMIALS MATHEMATICS IX
9. Factorize the following using appropriate identities : y (i) 9 44 (ii) 9 6 4 4 (iii) y (iv) 9 0y 5y 0. Factorize by splitting the middle term : (i) + 08 (ii) y y 6 (iii) t t 4 (iv) 6z + 5z 6 (v) 8 5 (vi) 0 + 84. Epand each of the following using suitable identity : (i) ( y z) a (ii) b c (iii) ( y) (iv) y. Evaluate the following, using suitable identity: (i) (0) (ii) (99). Factorize the following: (i) 4 + y + 9z + 4y 6yz z (ii) 64a 44a b + 08ab 7b (iii) + 6 y + y + 8y (iv) 8 + 5 (v) 7a 64b (vi) 7a + 8b + c 8abc 4. Without actually calculating the cubes, find the value of each of the following : (i) ( 4) + (8) + (6) (ii) (9) + ( ) + ( 8) 5. If + y + z = 0, prove that + y + z = yz. 6. Prove that: + y + z yz ( y z) [( y) ( y z) ( z ) ] 7. Using factor theorem, show that a b, b c and c a are the factors of a(b c ) + b (c a ) + c (a b ). 8. Factorize the following : (i) ab a b (ii) 4 + 5 + 9 (iii) 5 5 0 8 5 (iv) 9( y) 4( y ) 6( y) (v) 6 y 6 (vi) 6 + y 6 9. Factorize the following : (i) (y z) + y (z ) + z ( y) (ii) (a 4b) + (4b c) + (c a) 40. Prove that : (a + b) + (b + c) + (c + a) (a + b) (b + c) (c + a) = (a + b + c abc) MM : 0 General Instructions : PRACTICE TEST Q. -4 carry marks, Q. 5-8 carry marks and Q. 9-0 carry 5 marks each.. Find the remainder when 8 + 9 0 is divided by.. Find the value of a for which + a + 4 is eactly divisible by +.. Factorize : 64a 7b 44a b + 08ab 4. Evaluate : (0), using suitable identity. MATHEMATICS IX POLYNOMIALS Time : hour
5. Factorize : + 0 6. Epand the following : (i) ( y z) (ii) y 7. Without actually calculating the cubes, evaluate the following : (0) + ( 5) + ( 5) 8. Factorize : 64y 8z 4yz. 9. Factorize 6 + 5 + 0 using factor theorem. 0. Prove that : a + b + c abc = ( a b c) ( a b) ( b c ) ( c a) ANSWERS OF PRACTICE EXERCISE. (i), (iii) and (v). (i) 4 (ii) (iii) (iv) 0 (v). (i) Binomial (ii) Trinomial (iii) Monomial (iv) Bionomial (v) Bionomial 4. 4, 46, 5 5., 5, 40 6. (i) yes (ii) yes (iii) no (iv) yes 4 5 7. (i) 4 (ii) (iii) 0 (iv) r 8. (i) (ii) 75 (iii) (iv) 4 (v) 6a (vi) 8a 09 9. a =. p = 9. a =, b = 9 4. a 5. a 5 0 6.,, 7.,, 4 0. a =. a. a =, b = 8 4. (i) ( ) ( ) ( ) (ii) ( + ) ( ) ( 5) (iii) ( + ) ( ) ( ) (iv) ( ) ( ) ( ) (v) ( + ) ( ) ( + 5) (vi) ( ) ( + ) (4 + 5) 5. ( + 6) ( + ) ( ) 6. ( + 9) ( 7) ( ) 5 7. (i) + 0 + 4 (ii) + 5 4 (iii) 6 + 5 6 (iv) t 4 4 y y 8. (i) 809 (ii) 9964 9. (i) (ii) ( ) ( ) (iii) ( y)( y)( y ) (iv) ( + 5y) ( + 5y) 0. (i) ( 9) ( ) (ii) (y ) (4y + ) (iii) (t + ) (t 4) (iv) (z + ) (z ) (iv) ( ) (4 + 5) (vi) (4 7) (5 + ) a. (i) + 4y + 9z 4y yz + 6z (ii) b 9c ab 6bc ac 9 (iii) 7 8y 54 y + 6y y (iv) 8 6 y y 8. (i) 000 (ii) 97099 4 POLYNOMIALS MATHEMATICS IX
. (i) ( + y z) ( + y z) (ii) (4a b) (4a b) (4a b) (iii) ( + y) ( + y) ( + y) (iv) ( + 5) (4 0 + 5) (v) (a 4b) (9a + ab + 6b ) (vi) (a + b + c) (9a + 4b + c 6ab bc ac) 4. (i) 06 (ii) 506 8. (i) ( + a + b) ( a b) (ii) ( + + ) ( + ) (iii) ( 5 ) (5 4 5) (iv) ( + 7y) ( + 7y) (v) ( y) ( + y) ( y + y ) ( + y + y ) (vi) ( + y ) ( 4 y + y 4 ) 9. (i) yz (y z) (z ) ( y) (ii) (a 4b) (4b c) (c a) ANSWERS OF PRACTICE TEST. 6. a =. (4a b) (4a b) (4a b) 4. 0977 5. ( ) ( + 5) 6. (i) + 4y + 9z 4y + yz 6z (ii) y 7 y y 7. 4500 8. ( 4y z) ( + 6y + 4z + 4y 8yz + z) 9. ( + ) ( + 5) ( ) MATHEMATICS IX POLYNOMIALS 5