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Question

Let f(x) be a polynomial function. If f(x) is divided by x−1,x+1 and x+2, then remainders are 5,3 and 2 respectively. When f(x) is divided by x3+2x2−x−2, then remainder is

A
x4
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B
x+4
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C
x2
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D
x+2
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Solution

The correct option is A x+4
We know that when a polynomial f(x) is divided by (xa), the remainder is f(a)

Let f(x) be the polynomial, q(x) be the quotient and r(x) be the remainder.

Degree of remainder is always less than divisor. So, let r(x)=ax2+bx+c

Given that f(1)=5; f(1)=3; f(2)=2

x3+2x2x2=(x1)(x+1)(x+2)

f(x)=q(x)×(x1)(x+1)(x+2)+r(x)

f(1)=q(1)×0+r(1)

5=r(1) ..(1)

f(1)=q(1)×0+r(1)

3=r(1) ..(2)

f(2)=q(2)×0+r(2)

2=r(2) ..(3)

Substituting these values from (1),(2),(3) in r(x) we get

r(x)=ax2+bx+c

r(1)=a+b+c=5

r(1)=ab+c=3

r(2)=4a2b+c=2

Solving for a,b,c we get

a=0;b=1;c=4

Therefore, r(x)=x+4

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