AP ICET 2017 Shift 2 Paper

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Question : 123 of 200

Marks: +1, -0

If

x^{2}+x-2

is a factor of

x^{4}+3x^{3}-8x^{2}+p(x)

,where

p(x)

is a linear polynomial then

p(x)

is a linear polynomial then

p(x)=

$-12x+16$
$12x-16$
$x+16$
$x-16$

Solution:

(x^2+x–2)=x^2+2x–x–2=x(x+2)–(x+2)

=(x+2)(x–1)

Since

(x^2+x–2)

is a factor of

x^4+3x^3–8x^2+p(x)

, the

(x+2)

and

(x-1)

are also its factors.
We have,
For

x=–2:(-2)^4+3(-2)^3-8(-2)^2+p(-2)=0

⇒16-24-32+p(-2)=0⇒p(-2)=40

For

x=1:(1)^4+3(1)^3-8(1)^2+p(1)=0

⇒1+3–8+p(1)=0⇒p(1)=4

p(x)=-12x+16

gives

p(-2)=24+16=40

, and

p(1)=–12+16=4

Hence, the correct answer is

p(x)=–12x+16

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