Question

x

4 4x3 – x2 – 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)

Answer 1

HEY!!

The given expression is p(x) = x4 + 4x3 – x2 – 16x – 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 x4 + 4x3 – x2 – 16x – 12 by x + 1 by long division, you can get the quotient as, x3 + 3x2 – 4x – 12
[Note: Here, remainder will be 0 as (x + 1) is a factor of p(x)]

p(x) = (x + 1)(x3 + 3x2 – 4x – 12) = (x + 1). q(x) where q(x) = (x3 + 3x2 – 4x – 12).
Now, you can observe that q(2) = 23 + 3(2)2 – 4(2) – 12 = 0.
So, by factor theorem, (x – 2) is a factor of q(x).

On dividing x3 + 3x2 – 4x – 12 by (x – 2) by long division, you can get the quotient as, x2 + 5x + 6
[Note: Here, remainder will be 0 as (x – 2) is a factor of p(x)]
∴ x3 + 3x2 – 4x – 12 = (x – 2) (x2 + 5x + 6) so that p(x) = (x + 1) (x – 2) (x2 + 5x + 6)

Lastly, you can factorise (x2 + 5x + 6) as,
x2 + 5x + 6
= x2 + 2x + 3x + 6 [2x + 3x = 5x and 2x × 3x = 6x2]
= x(x + 2) + 3(x + 2)
= (x + 2)( x+ 3)

So, you can now write as,
p(x) = (x + 1) (x – 2) (x2 + 5x + 6) = (x + 1) (x – 2) (x + 2)( x+ 3)

Thus, the correct answer is D.