POLYNOMIALS. x + 1 x x 4 + x 3. x x 3 x 2. x x 2 + x. x + 1 x 1

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1 POLYNOMIALS A polynomial in x is an expression of the form p(x) = a 0 + a 1 x + a x +. + a n x n Where a 0, a 1, a. a n are real numbers and n is a non-negative integer and a n 0. A polynomial having only one term is called monomial. A polynomial having two terms is called binominal. A polynomial having three terms is called trinomial. Degree of polynomial: The highest power of x in a polynomial is called degree of a polynomial. A polynomial is said to be linear if it is of degree 1. A polynomial is said to be quadratic if it is of degree. A polynomial is said to be cubic if it is of degree. A polynomial is said to be biquadratic if it is of degree 4. Sum and difference of two polynomial Let p(x) and q(x) be two polynomials p(x) = x 4 + x x + 6x + 9 q(x) = x + x + 6x + The sum of two polynomial gives a polynomial Here p(x) + q(x)=x x + x + 1x + 1 Similarly we can find difference of two polynomials p(x) q(x)= x 4 + x 5x + 6 Multiplication of two polynomial p(x)q(x)= (x 4 + x x + 6x + 9)( x + x + 6x + ) Division of a polynomial by a polynomial- if we divide p(x) by g(x) and we get quotient q(x) and remainder r(x) then we can write p(x) = g(x) q(x) + r(x) degree r(x) < degree g(x) Illustration 1: Divide p(x) by g(x), where p(x) = x and g(x) = x + 1 Solution: x x + x 1 x + 1 x x 4 + x x + 1 x x x + 1 x + x x + 1 x 1 Here, the quotient q(x) = x x + x 1 and the remainder r(x) = We write x = (x x + x 1) (x + 1) +. Note: Notice that the degree of g(x) is less than the degree of p(x). Therefore it is always possible to divide a polynomial of higher degree by a polynomial of lower degree (in the same variable). The process above stops as soon as the remainder is zero or the degree of the remainder becomes smaller than that of the divisor.

2 Illustration : Let p(x) = x 4 + x x + x 1. Find the remainder when p(x) is divided by x. Solution: x + 4x + 5x + 11 x x 4 + x x + x 1 x 4 x 4x x + x 1 4x 8x 5x + x 1 5x 10x 11x 1 11x 1 Remainder is 1. In the above example let us evaluate p(). We get p() = 4 + ( ) x + 1 = = 1. Thus we find that p() is equal to the remainder when p(x), is divided by x. Remainder Theorem: Let p(x) be any polynomial of degree 1, and a any real number. If p(x) is divided by x a. then the remainder is p(a). Proof: Let us suppose that, when p(x) is divided by x a, the quotient is q(x) and remainder is r(x). So we have p(x) = (x a) q(x) + r(x), where r(x) = 0 or degree r(x) < degree (x a) Since degree of (x a) is 1, either r(x) = 0 or degree of r(x) = 0 (<1). So r(x) is a constant, say r. Thus for all values of x, p(x) = (x a) q (x) + r where r is a constant. In particular, for x = a, p(a) = 0. q(a) + r = 0 + r = r. This proves the theorem. Illustration : What would be the remainder when (4x + 7x 5x + ) is divided by (x + ). Solution: Let f(x) = 4x + 7x 5x +. We may write, (x + ) = [x ( )] Thus, when f(x) is divided by [x ( )], then by remainder theorem, remainder = f( ). Now, f( )= [4x ( ) + 7 x ( ) 5x( ) + ] = [4x ( 8) + 7 x ] = ( ) = 9. Required remainder = 9 Illustration 4: Using remainder theorem, find the remainder when (4x 1x + 15x ) is divided by (x 1). Solution: Let f(x) = 4x 1x + 15x. 1 We may write, (x 1) = x Now f 4 x 1x 15 x 8 4 Required remainder =. Synthetic Division The method of synthetic division can be illustrated from the following example: Find the quotient and remainder when x 4 + 5x + x + 7x is divided by (x )

3 x )x 4 + 5x x + 7x (x + 9x + 15x + 7 x 4 4x 9x x 9x 18x 15x + 7x 15x 0x 7x 7x Quotient is x + 9x + 15x + 7. Remainder is 71. The divided is x 4 + 5x x + 7x Divisor is x. The quotient will be of third degree. = 1st coefficient in quotient = = 1st coefficient in divided. 9 = nd coefficient in quotient = () + 5 = (1st coefficient in quotient) + nd coefficient in dividend. 15 = rd coefficient in quotient = (9) + ( ) = (nd coefficient in quotient) + rd term in dividend 7 = 4th coefficient in quotient = (15) + 7 = (rd coefficient in quotient) + 4th term in dividend. 71 = Remainder = (7) + ( ) = (4th term in quotient) + 5th term in dividend. The above process can be illustrated as follows: Write down the coefficient only of the dividend. 5 7 () (9) (15) Hence quotient is x + 9x + 15x + 7. Remainder is 71. Synthetic division to find quotient and remainder is also called the method of find quotient and remainder by the method of detached coefficients. Illustration 5: Divide x 4 + 5x x + 7x by (x + ) the method of detached coefficients. 5 7 Solution: ( ) 1( ) ( 5)( ) 17( ) Quotient is x + x 5x Remainder is 7. If p(x) is a polynomial of degree n > 0, then it follows the remainder theorem that p(x) = (x a). q(x) + p(a), where q(x) is a polynomial of degree n 1. If p(a) = 0, then p(x) = (x a), q(x) + p(a), where q(x) is a polynomial of degree n 1. If p(a) = 0, then p(x) = (x a). q(x). we say that (x a) is factor of p(x) Factor Theorem : Let p(x) be polynomial of degree n > 0. If p(a) = 0 for a real number a, then (x a) is a factor of p(x). Conversely, if (x a) is a factor of p(x), the remainder p (a) must be zero. Illustration 6: Show that (x + ) as well as (x + ) is a factor of the polynomial (x + 8x 5x 6). Solution : Let f(x) = x + 8x 5x 6. By factor theorem, (x + ) will be a factor of f(x), if f( ) = 0. Now, f( ) = ( ) + 8 ( ) 5( ) 6 = [ x ( 7) + 8 x ] = ( ) = 0. (x + ) is a factor of f(x) Again, (x + ) = x x. By factor theorem, (x + ) will be factor of f(x), if f 0. Now f x 8 x 5 x = x 8 x

4 Illustration 7: Given that (x 1) and (x + ) are factors of the polynomial (x + ax + bx 8). Find the values of a and b. With these values of a and b. Factorize the given polynomial. Solution : Let f(x) = (x + ax + bx 8). Then (x 1) is a factor of f(x). f(1) = 0 [ By factor theorem ] (1 + a x 1 + b x 1 8) = 0 [ f(1) = (1 + a x 1 + b x 1 8)] 1 + a + b 8 = 0 a + b = 7..(i) Again, (x + ) is a factor of f(x) f( ) = 0 [ By factor theorem ] ( ) + a x ( ) + b x ( ) 8 = 0 [ f( ) = ( ) + a x ( ) + b x ( ) 8 ] 8 + 4a b 8 = 0 4a b = 16 a b = 8..(ii) Adding (i) and (ii), we get a = 15 or a = 5. Substituting a = 5 in (i), we get 5 + b = 7, i.e., b = With these values of a and b, we have f(x) = x + x + x 8. Now, (x 1) and (x + ) are factors of f(x) (x + x ) is a factor of f(x) x + 4 x + x ) x + 5x + x 8 ( x + x x 4x + 4x 8 4x + 4x 8 0 On dividing f(x) = x + 5x + x 8 by (x + x ), we get Quotient = (x + 4) f (x) = x + 5x + x 8 = (x + x ) (x + 4) = (x 1) (x + ) (x + 4) Hence, (x + 5x + x 8) = (x 1) (x + ) (x + 4). Illustration 8: A Polynomial f(x) with rational coefficients leaves remainder 15 when divided by (x ) and remainder when divided by (x 1). Find the remainder when the polynomial is divided by (x 1) and (x ) Solution: f(x) = ax + bx + c f(x) = Q(x) (x ) + 15 f(x) = Q(x) (x 1) +. f(x) = Q(x) (x )(x 1) + (ax + b) f() = 0 + a + b = 15 f(1) = 0 + a + b = a = 1 a = 6 b = 6 = Remainder is 6x. Exercise 1: If f(x) is a polynomial in x, 4 and 10 are the remainders when f(x) is divided by (x 1) and (x ). Then find the remainder when f(x) is divided by (x 1)(x ). A.S.Rao, 006 SQUARE ROOT: If y x or y = x, then y is called the square root of x. As per the above definition 4 is the square root of 16 since 4 = 16. But also ( 4) = =4.

5 The positive square root of 16 is called the principal square root of 16. In general root we mean only the principal square root. Hence x = x when x > 0 x = x when x < 0 x = x. By square x = 0 when x = 0. In short x = x In what follows we consider only the non negative square roots. Modulus sign is to be put whenever it is needed. For instance Illustration 9: Find the square root of a 8 b 1 c 6 Solution: a ab b ( a b) ( a b) ( a b) if a b ) a b c (a b c ) a b c Illustration 10: Find the square root of Solution: 5a 4 b 4 = 5a 5 a = b 4b 5 4 a b Illustration11: Find 11a 9ab 5 6 Solution: 11a 9ab Square root by inspection: a + ab + b = (a + b) a ab + b = (a b) 5 6 = a ab b = (a + b) 11 a b = 11a b a ab b = (a b) If the given expression is reduced to one of the above forms, the square root can be written down by inspection. Extraction of square root by division method: Principle behind finding the square root is given below. Let us find the square root of 5x 0x + 9 Let 5x 0x = (5x + a) 5x 0x + 9 = 5x + 10ax + a 0x + 9 = 10ax + a Equating coefficient of x on both sides 10a = 0 or a = We can show this briefly in the division method 5x 5x 5x 0x + 9 5x 5(x) 10x 0x + 9 (10x ) ( ) 0x + 9 0

6 Procedure: (1) The square root of 5x is 5x. This is the first term of the square root. Multiply (5x) by (5x) which gives 5x. Substract 5x from the dividend. We get 0x + 9 as the next two terms. 0x () Multiply 5x by which gives 10x. gives. This becomes the second term of the square root. 10x Adding to 10x, we get 10x. Multiplying this by. This becomes the second term of the square root. Additing. This becomes the second term of the square root. Adding to 10x, we get 10x. Multiplying this by we get 0x + 9. The remainder becomes zero. Square root is (5x ) Illustration 1: Find the square root of 4x 4 + 1x y + 1x y + 6xy + y 4 Solution: x + xy + y 4x 4 +1x y + 1 x y + 6xy + y 4 x 4x 4 (x ) = 4x 4 4x + xy 1x y + 1 x y (4x + x) xy 1x y + 9x y 1x y since xy 4x 4x + 6xy + y 4x y + 6xy + y 4 (4x + 6xy + y ) y 4x y + 6xy + y 4 4x y since 4x 0 y Hence the square root is x + xy + y Square root by the method of Indeterminate coefficients Illustration 1: Find the square root of x 4 + 6x + 17x + 4x + 16 Solution: The expression is of the fourth degree in x. Hence the square root will be an expression of second degree in x. x 4 + 6x + 17x + 4x + 16 = (x + ax + b) Equating coefficients of the like powers of x on both sides, we get a = 6 a = a + b = 17 b = 17 a = 17 9 = 8 b = 8 or b = 4. Hence the required square root is x + x + 4. Illustration 14: If ax + bx + c is a perfect square, then prove that b = 4ac. Solution: Let ax + bx + c = (px + q) ax + bx + c = (px + q) Equating coefficients of like powers x p = a pq = b 4p q = b and q = c 4ac = b Hence b = 4ac Illustration 15: Show that (x + 1) (x + 4) (x + )(x + ) + 1 is a perfect square Solution: The given expression = (x + 1) (x + 4) (x + ) (x + ) + 1 = (x + 5x + 4) (x + 5x + 6) + 1 Put y = x + 5x The expression reduces to (y + 4)(y + 6) + 1 = y + 10y + 5 = (y + 5) = (x + 5x + 5) Which is a perfect square. Exercise : Find the square root of 1 4x + 10x 0x + 5x 4 4x x 6

7 1 1 Exercise : Find the square root of x + - x + + x x Exercise 4 : Find the square root of x y x y y 4x y x 4 Exercise 5 : Find the square root of x x x x Exercise 6 : Find the square root using undetermined coefficient method X 4 + 6x + 17x + 4x + 16 Exercise 7 : Find the square root using undetermined coefficient method X 6 4x 4 6x + 4x + 1x + 9 Exercise 8 : If x 4 + ax + bx + cx + d = (x + px + q), then show that q = b Homogeneous function a c = 4 a. An integral function is said to be homogeneous, if each of its term is of the same degree with respect to any set of variables. eg. x + 4y 5xy is a homogeneous expression of the second degree in x and y. The product of two homogeneous functions of degrees m and n respectively is a homogeneous function of degree m + n. eg. ( + 5y) (4x + 5y ) = 8x + 10xy + 0x y + 5y x + 5y is of first degree in x and y. 4x + 5y is of second degree in x and y and both are homogeneous in x and y. Their product is homogenous and of third degree in x and y. General forms of homogeneous integral functions of different degrees Degree Variables Forms of homogeneous functions 1 x, y ax + by x, y ax + bxy + cy x, y ax + bx y + cxy + dy 1 x, y, z ax + by + cz x, y, z ax + by + cz + fyz + gzx + hxy Symmetric functions A function is said to be symmetric with respect to any set of variables if the interchange of any pairs of the set of variables does not alter the value of the function. eg. 5 + x + y does not change in value if the variables x and y are interchanged. The function 5x + y + z is symmetric with respect to y and z, but it is not symmetric with respect to x and y. The expression xy + yz + zx + x + y + z is symmetric with respect to x, y, z.

8 Similarly p q r and qr rp pq p q r p q q r r p are symmetric with respect to p, q, r. Hence the necessary and sufficient condition that an integral function to be symmetrical is that all the terms of any one type shall have the same coefficient. If the variables x and y are interchanged ax + bxy + cy then the resulting expression is ay + byx + cx then the resulting expression is ay + byx + cx In order that ax + bxy + cy is symmetric, we must have a = c. Forms of symmetric integral functions: Degree Variables Functions 1 x, y a(x + y) + b x, y a (x + y ) + bxy + c (x +y) + d 1 x,y,z a (x + y + z) + b x,y,z a (x + y + z ) + b(xy + yz + zx) + c(x + y + z) + d Homogeneous symmetric integral function Degree Variables Functions 1 x, y a(x + y) x, y a (x + y ) + bxy 1 x,y,z a (x + y + z) x,y,z a (x + y + z ) + b(xy + yz + zx) Cyclic symmetry In the expression y z + z x +x y which consists of three terms (i) z x can be obtained from y z if we replace y by z and z by x. (ii) x y can be obtained from z x if we replace z by x and x by y. (iii) y z can be obtained from x y if we replace x by y and y by z. The selected term from which the other forms are obtained may be called the typical term. The function can evidently be determined when a typical term is given. Hence we can say that the function y z+ z x + x y possesses cyclic symmetry. Again the expression a (b c) + b (c a) + c (a b) possesses cyclic symmetry. We can get the second term the first by replacing a by b, b by c and c by a. We can get the third term from the second if we replace b by c, c by a and a by b. y z x It can also be seen that z x y Standard Formulae: 1. (a + b) = a + ab + b. (a b) = a ab + b. (a + b) (a b) = a b 4. (a + b) + (a b) = (a + b ) 5. (a + b) (a b) = 4ab 6. (a + b + c) = a + b + c + bc + ca + ab = a + bc 7. (a + b) = a + b + ab ( a + b) = a + a b + ab + b 8. (a b) = a b ab (a b) = a a b + ab b 9. (x + a) (x + b) = x + (a + b) x + ab 10. (x + a) (x + b) (x + c) = x + x (a + b + c) + x (ab + bc + ca) + abc 11. a + b = (a + b) (a ab + b ) 1. a b = (a b) (a + ab + b ) 1. a + b + c abc = (a + b + c) (a + b + c ab bc ca) 14. If a + b + c = 0, then a + b + c abc = 0 or a + b + c = abc possesses cyclic symmetry. The typical term can be taken as 15. a + b + c ab bc ca = 1 [(a b) + (b c) + (c a) ] y z.

9 Illustration 16: Factorise x + y + z xyz and hence prove that x + y + z = xyz if x + y + z = 0 Solution: x + y + z xyz = [(x + y) + z ] xy(x + y) xy (x + y + z) = (x + y + z) [(x + y) (x + y)z +z ] xy (x + y + z) = (x + y + z) [(x + y) (x + y)z + z xy = (x + y + z) (x + y + z xy yz zx) Illustration 17: Show that if x + y + z = xy + yz + zx then x = y = z Solution: x + y + z = xy + yz + zx x + y + z xy yz zx 1 [x + y + z xy yz zx] = 0 i.e 1 [(x y) + (y z) + (z x) ] =0 The sum of perfect squares cannot be equal to zero unless each quantity of equal to zero. (x y) = 0 x = y (y z) = 0 y = z x = y = z. Exercise 9 : Resolve into factors a (b + c) + b (c + a) + c (a + b) + abc Exercise 10 : Resolve into factors a - a. abc abc Exercise 11 : Resolve into factors (a + b) (a - b). abc Exercise 1 : Resolve into factors a 4 (b c) + b 4 (c a) + c 4 (a b) Exercise 1 : Resolve into factors (a + b + c) 5 a 5 b 5 c 5 Answers to exercises: 1. 6a. 4x x + x 1. 1 x 1 4. x y 1 x y x x x 6. x + x x x 9. (a + b) (b + c) (c + a) 10. (a + b) (b + c) (c + a) 11. (a b) (b c) (c a) 1. (a b) (b c) (c a) (a + b + c + ab + bc + ca) (a + b) (b + c) (c + a) [ 8(a + b + c ) + 145(ab + bc + ca)]

10 ASSIGNMENTS SUBJECTIVE LEVEL I 1. On dividing (kx + 9x + 4x 10) by (x ), we get 5 as remainder. Find the value of k.. The polynomials f(x) = ax + x and g(x) = x 5x + a, when divided by (x 4), leave the same remainder in each case. Find the value of a.. Using factor theorem, factorize the polynomial x + x 17x 0 and hence solve the equation x + x 17x 0=0. 4. If (x + a) is a common factor of the polynomials f(x) = x + mx + n and g(x) = x + px + q, show that a = q n. p m 5. Show that (1+x) is a factor of the polynomial f(x) = (x + 7x 4x 15) and hence factorize f(x) 6. Prove (a + b ) (x + y ) = (ax by) + (bx + ay) 7. Factorise (x + y + z) + (x y z) 8x 8. A polynomial leaves remainders 0, 1 and when divided by x, (x 1), (x ) and respectively. Find the remainder when polynomial is divided by x (x 1) (x ). 9. Find the remainder when (a b) x + (b c) x + (c a) is divided by x If a + b + c = abc then the value of a b b c c a + abc is.. 1ab 1bc 1 ca

11 LEVEL II 1. A polynomial f(x) leaves remainders and when divided by x +1 and (x ) respectively. Find the remainder when f(x) is divided by (x + 1) (x ).. The remainder when x 0 x 0 + x 01 x x 1 is divided by x x is.. Given that x + x 1 = (x + a) (x + b) (x + c). Find the value of (a) a + b + c (b) a + b + c 4. Find the remainder and quotient when x 5x x + is divided by (1 x). 5. Factorise (i) 1 a + a + 6a (ii) b c + c a + a b a 4 b 4 c 4

12 OBJECTIVE LEVEL I Select the correct alternative (A), (B), (C),(D)from each of the following,indicate your choice by writing the appropriate letter only. 1. If (x a) be a factor of x + px + q then (A) p = a ; q = a (B) p = a ; q =a (C) p =a ; q=a (D) None of these. The homogeneous function of the second degree in x and y having x y as a factor. A) x + xy y (B) x + xy + y (C) x + xy + y (D) None of these. The factors of the polynomial expression 15 x 6x are (A) (x + 5) and (x + ) (B) (5 x) and (x + ) (C) ( x) and (x + 5) (D) None of these 4. If x 1 x = 7 then the value of 1 x is x (A) (B) 4 (C) 64 (D) None of these 5. If x x + is a factor of the expression x 4 + ax + b, then the values of a and b are given by (A) a = 5; b = 4 (B) a = 4; b = 5 (C) a = 5; b = 4 (D) None of these 6. If a + b + c=6; bc + ca + ab = 11; abc = 6, then the value of (1 a)(1 b)(1 c) is (A) 1 (B) 1 (C) 0 (D) None of these 7. If x = a b+c, y = b c +a, z = c, then the value of xy + yz +zx+ xyz is a+b (A) 1 (B) (C) (D) None of these Determine whether each of the following statements is true or false. 8. If a + b + c = 0, then a 4 + b 4 + c 4 = (b c + c a + a b ). 9 The expression (ab + cd) (ad + bc) is divisible by (a c) (b d). 10. A certain algebraic expression is exactly divisible by x, the quotient being x x 6. Then the expression is also divisible by x +.

13 LEVEL II 1. If (a b) = 5 then 8a 7b 90ab is (A) 15 (B) 91 (C) 111 (D) None of these. The H.C.F. of the functions x + (a + b)x + (ab + 1) x + a and bx + (ab + 1)x + (a + b) x + 1 is (A) x + ax + 1 (B) x + bx + 1 (C) x + x + a (D) None of these. If a + b = 9, x = 5 and a b x = then value of (a b) [x ax + a x (a+b)b ] is (A) 445 (B) 5 (C) 76 (D) None of these 4. x 4 x x 5 = (x + 1) (...) (A) x 4x + 4x 5 (B) x + 4x 4x + 5 (C) x x 5 (D) None of these 5. If abx = (a b) (x + 1) then the value of is x x ab ab a (A) (B) (C) ab (D) None of these ab ab 6. For what value of x will 4x 4 + 1x 11x 15x 5 be a perfect square? (A) (B) (C) (D) None of these 7. The square root of 9a 4ab + 16b is (A) 4b a (B) a + 4b (C) a 4b (D) None of these 8. The L.C.M. of 6x 19x + 10 and x + x is (A) (x 5) (x + 1) (x + ) (B) (x + 1) (x + 5) (x ) (C) (x ) (x 5) (x + 1) (D) None of the above 9. The value of p for which the function 4x 4 1x + 17x 1x + p is a perfect square is (A) 4 (B) 4 (C) (D) None of these 10. What must be added to (x + 7x + 4) (x + 7x ) to make it a perfect square? (A) 9 (B) 9 (C) 1 (D) None of these

14 ANSWERS SUBJECTIVE LEVEL I 8 1. k = 9. a = 1 5.,, 5. (x ) (x + 5) (x + 1) 7. 6x(x + y + z) (y + z x) 8. x LEVEL II (x + 9). 151x + 15x 1. 4, 4. Rem: 59 7, x 1 x (i) (1 + a) (1 5a + 1a ) (ii) (a + b + c) (a + b c) (b + c a) (c + a b) OBJECTIVE LEVEL I 1. B. A. C 4. C 5. A 6. C 7. A 8. T 9. T 10. T LEVEL II 1. A. A. B 4. A 5. B 6. C 7. A 8. C 9. A 10. B

Factorisation CHAPTER Introduction

Factorisation CHAPTER Introduction FACTORISATION 217 Factorisation CHAPTER 14 14.1 Introduction 14.1.1 Factors of natural numbers You will remember what you learnt about factors in Class VI. Let us take a natural number, say 30, and write