x

^{4} + 4x^{3} – x^{2} – 16x – 12, on factorisation, gives as

(A) (x –1) (x – 2) (x + 2) (x + 3)

(B) (x + 1) (x – 2) (x + 2) (x – 3)

(C) (x – 1) (x – 2) (x + 2) (x – 3)

(D) (x + 1) (x – 2) (x + 2) (x + 3)

Hi Martin!

Here is the answer to your question.

The given expression is

*p*(*x*) =*x*^{4}+ 4*x*^{3}–*x*^{2}– 16*x*– 12.It can be observed that,

*p*(–1) = (–1)^{4}+ 4(–1)^{3}– (–1)^{2}– 16(–1) – 12 = 1 – 4 – 1 + 16 – 12

= 17 – 17

= 0

So, by factor theorem, {

*x*– (–1)} or (*x*+ 1) is a factor of*p*(*x*).Now, on dividing

*x*^{4}+ 4*x*^{3}–*x*^{2}– 16*x*– 12 by*x*+ 1 by long division, you can get the quotient as,*x*^{3}+ 3*x*^{2}– 4*x*– 12[

**Note:**Here, remainder will be 0 as (*x*+ 1) is a factor of*p*(*x*)]*p*(

*x*) = (

*x*+ 1)(

*x*

^{3}+ 3

*x*

^{2}– 4

*x*– 12) = (

*x*+ 1).

*q*(

*x*) where

*q*(

*x*) = (

*x*

^{3}+ 3

*x*

^{2}– 4

*x*– 12).

Now, you can observe that

*q*(2) = 2^{3}+ 3(2)^{2}– 4(2) – 12 = 0.So, by factor theorem, (

*x*– 2) is a factor of*q*(*x*).On dividing

*x*^{3}+ 3*x*^{2}– 4*x*– 12 by (*x*– 2) by long division, you can get the quotient as,*x*^{2}+ 5*x*+ 6[Note: Here, remainder will be 0 as (

*x*– 2) is a factor of*p*(*x*)]∴

*x*^{3}+ 3*x*^{2}– 4*x*– 12 = (*x*– 2) (*x*^{2}+ 5*x*+ 6) so that*p*(*x*) = (*x*+ 1) (*x*– 2) (*x*^{2}+ 5*x*+ 6)Lastly, you can factorise (

*x*^{2}+ 5*x*+ 6) as,*x*

^{2}+ 5

*x*+ 6

=

*x*^{2}+ 2*x*+ 3*x*+ 6 [2*x*+ 3*x*= 5*x*and 2*x*× 3*x*= 6*x*^{2}]=

*x*(*x*+ 2) + 3(*x*+ 2)= (

*x*+ 2)(*x*+ 3)So, you can now write as,

*p*(

*x*) = (

*x*+ 1) (

*x*– 2) (

*x*

^{2}+ 5

*x*+ 6) = (

*x*+ 1) (

*x*– 2) (

*x*+ 2)(

*x*+ 3)

Thus, the correct answer is D.

Hope that this explanation will help you.

Cheers! Go ahead to solve more problems on maths.

**
**