NAVODAYA VIDYALA SAMITI

NOIDA, UP

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E- CONTENT

MATHEMATICS

CLASS X

CHAPTER 2

POLYNOMIALS

Division Algorithm for Polynomials

Division algorithm of polynomials

4x³- 7x²+ 10x +10

4x + 1

4x³ +x²

-8x²+ 10x + 10

-2x

-8x² - 2x

12x+ 10

+ 3

12x+ 3

7

Eg. Divide p(x) =4x³- 7x²+ 10x +10 by g(x)= 4x +1 and hence

obtain the quotient and remainder.

x²

Dividend, p(x) = 4x³- 7x²+ 10x +10

Divisor, g(x) = 4x + 1

Quotient, q(x) = x²-2x + 3

Remainder, r(x) = 7

g(x) X q(x) + r(x) =( 4x + 1) (x²-2x + 3) + 7

=4x³-8x²+12x + x²-2x+3+7

=4x³- 7x²+ 10x +10

= p(x)

ie, Dividend = Divisor X Quotient + Remainder.

This relation is called division algorithm of polynomials.

Note that remainder , r(x) =0 or degree of r(x) < degree of g(x).

Note :- When r(x) =0, then the divisor polynomial

g(x) is a factor of p(x).

This fact is used in the process of finding the zeroes of higher order polynomials.

For example, to find all zeroes of the polynomial 2x³ -3x²-4x +2 , given that x=2 is one zero of it, we proceed in the following way.

Solution : -

Given that x=2 is a zero of p(x)= 2x³ -3x²-3x +2, we get, x-2 is a factor of p(x).

Now, Divide p(x) by x-2.

2x³ -3x²-3x +2

x-2

2x²

2x³-4x²

X²- 3x+2

+x

X²- 2x

- 1

-x +2

-x +2

0

Now, by division algorithm, we get

2x³ -3x²-3x +2 = (x-2)( 2x²+x – 1)

Therefore, Zeroes are given by

(x-2)( 2x²+x – 1)=0

ie, x-2 =0, or 2x²+x – 1=0

ie, x = 2 (which was given)

and 2x² + x – 1=0

Factorising 2x²+x – 1, by splitting the middle term, we get 2x²+2x –x – 1=0

ie, 2x(x + 1) -1(x + 1) =0

ie, (x + 1 ) (2x-1) =0

ie, x + 1=0 or 2x-1=0

ie, x = -1 or x = ½

Therefore the zeroes of p(x) are x= -1, ½ and 2

Question :- If the zeroes of the polynomial x³- 3x²+ x + 1 are a - b, a and a + b, find a and b.

Let α = a – b , β = a and γ = a + b be the zeroes of the polynomial x³- 3x²+ x + 1

By the relation, we have α + β + γ = -b/a,

Therefore, a –b + a + a + b = 3 ( since a=1, b=-3)

ie, 3a= 3 , so that a = 1

Again α β + β γ + α γ = c/a

Therefore, a(a-b) + a(a+b) +(a-b)(a+b) =1

ie, 1(1-b) + 1(1 +b) +(1-b)(1+b)=1

ie, 1-b+1+b +1- b²=1

ie, 3 - b²=1

ie, - b² =-2

ie, b² =2

Therefore, b =±√2

ie, a = 1 and b=±√2

PREPARED BY

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JOAN A LUKE

JNV KOLLAM

KERALA HYDERABAD REGION