SSC CHSL 19 Mar 2020 Shift 3 Paper

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Question : 63 of 100

Marks: +1, -0

If

x^{3}-6 x^{2}+a x+b

is divisible by

(x^{2}-3 x+2)

, then the values of

a

and

b

are

$a=-11$ and $b=6$
$a=-6$ and $b=-11$
$a=11$ and $b=-6$
$a=6$ and $b=11$

Solution:

Let the quotient, when

x^{3}-6 x^{2}+a x+b

is divisible by

(x^{2}-3 x+2)

be

x-p

.
⇒

(x^{2}-3 x+2)(x-p)=x^{3}-6 x^{2}+a x+b

⇒

x^{3}-3 x^{2}+2 x-p x^{2}+3 p x-2 p

=x^{3}-6 x^{2}+a x+b

⇒

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

=x^{3}-6 x^{2}+a x+b

On comparing both sides,

-(3+p)=-6

⇒

p=6-3=3

2+3 p=a \⇒2+3 \×3=a \⇒a=11

and

-2 p=b \⇒-2 \×3=b \⇒b=-6

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