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Remainder Theorem

Polynomial Px...

Question

Polynomial $P (x)$ contains only terms of odd degree. When $P (x)$ is divided by $(x - 3)$ , the remainder is $6$ . If $P (x)$ is divided by $(x^{2} - 9)$ , then the remainder is $g (x)$ . Then the value of $g (2)$ is

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Solution

$P (x)$ contains only terms of odd degree.
So, $P (x)$ is an odd function.
$\Rightarrow P (- 3) = - P (3) = - 6$

Let $P (x) = Q (x) (x^{2} - 9) + a x + b$
where $Q (x)$ is the quotient and $a x + b = g (x)$ is the remainder.

Now, $P (3) = 3 a + b = 6 \dots (1)$
$P (- 3) = - 3 a + b = - 6 \dots (2)$
Solving $(1)$ and $(2)$ , we get
$a = 2, b = 0$
$∴ g (x) = 2 x$
$\Rightarrow g (2) = 4$

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Polynomial P(x) contains only terms of odd degree. When P(x) is divided by (x−3), the remainder is 6. If P(x) is divided by (x2−9), then the remainder is g(x). Then the value of g(2) is

P(x) contains only terms of odd degree. So, P(x) is an odd function. ⇒P(−3)=−P(3)=−6 Let P(x)=Q(x)(x2−9)+ax+b where Q(x) is the quotient and ax+b=g(x) is the remainder. Now, P(3)=3a+b=6 ⋯(1) P(−3)=−3a+b=−6 ⋯(2) Solving (1) and (2), we get a=2,b=0 ∴g(x)=2x ⇒g(2)=4

Polynomial $P (x)$ contains only terms of odd degree. When $P (x)$ is divided by $(x - 3)$ , the remainder is $6$ . If $P (x)$ is divided by $(x^{2} - 9)$ , then the remainder is $g (x)$ . Then the value of $g (2)$ is