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Question

If the remainders of the polynomial f(x) when divided by x+1 and x-1 are 3, 7 then the remainder of f(x) when divided by (x21) is

A
x + 4
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B
2x + 3
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C
2x + 4
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D
2x + 5
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Solution

The correct option is D 2x + 5
According to remainder theorem

f(x)=Q1(x)(x21)+r(x)
r(x)=ax+b
So, f(x)=Q1(x)(x21)+(ax+b)
f(1)=a+b=3 (1)
f(1)=a+b=7 (2)
Add (1) and (2)

a+b+a+b=102b=10b=5

substitute it in (1) then

a=5b=53=2

b=5;a=2
So r(x)=2x+5

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