Question

When a third degree polynomial f(x) is divided by (x-3), the quotient is Q(x) and the remainder is zero. Also when f(x) is divided by [Q(x) + x + 1], the quotient is (x-4) and remainder is R(x). Find
the remainder R(x)
(A) Q(x) + 3x + 4 - x²
(B) Q(x) + 4x + 4 - x²
(C) Q(x) + 3x + 4 + x²
(D) Q(x) - 3x + 4-x²​

Answer 1

When f(x) is divided by [Q(x)+x+1], the remainder is

R(x) = Q(x) + 3x + 4 - x²

Option (A) is correct.

Step-by-step explanation:

When polynomial f(x) is divided by (x-3), the quotient is Q(x) and remainder is zero

Therefore, by Euclid's division lemma we can write that

$f(x)=Q(x)(x-3)$

AGain When f(x) is divided by Q(x) + (x+1), the remainder is R(x) and quotient is (x-4)

Therefore,

$f(x)=[Q(x)+(x+1)](x-4)+R(x)$

$\implies f(x)=Q(x)(x-4)+(x+1)(x-4)+R(x)$

$\implies Q(x)(x-3)=Q(x)(x-4)+x^2-3x-4+R(x)$

$\implies R(x)=Q(x)+3x+4-x^2$

Therefore, Option (A) is correct.

Hope this answer was helpful.

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