AP ICET 2017 Shift 1 Paper

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Question : 125 of 200

Marks: +1, -0

If

x^{2}+x-2

is a factor of the polynomial

x^{4}+3x^{3}+ax^{2}+bx+16

then the ordered pair

(a,b)=

$(8,12)$
$(8,-12)$
$(-8,12)$
$(-8,-12)$

Solution:

x^2+x-2=(x–1)(x+2)

p(x)=x^4+3x^3+ax^2+bx+16

Since

(x^2+x-2)

is a factor of p(x), and

(x-1)

and

(x+2)

are factors of

(x^2+x-2)

, then p(x) is divisible by

(x–1)

and

(x+2)

.

p(1)=1+3+a+b+16=0⇒a+b=-20

....(i)

p(-2)=16-24+4a-2b+16=0

⇒4a-2b=-8⇒2a-b=-4

...(ii)
(i) + (ii) gives

3a=–24⇒a=–8

Substituting it in (i) we get

b=-12

∴(a,b)=(-8,-12)

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