Answer to Question #11094 in Algebra for n lakshmi

Question

factorize

(x^2-4x)(x^2-4x-1)-20

Answer 1

For simplicity denote t = x^2-4x

Then
A = (x^2-4x)(x^2-4x-1)-20
= t (t-1) -20
= t^2 - t - 20

It is easy to see that the equation
t^2 - t - 20 = 0
has two roots:
t1=5
t2=-4

Hence

A = t^2 - t - 20
= (t-5)(t+4)
= ( x^2 - 4x - 5 ) ( x^2 - 4x + 4 )
= ( x^2 - 4x - 5 ) ( x^2 - 4x + 4 )


The equation
x^2 - 4x - 5 = 0
has two roots:
x1=-1
x2=5
Hence
x^2 - 4x - 5 = (x+1)(x-5).

Moreover,
x^2 - 4x + 4 = (x-2)^2.

Therefore

A = ( x^2 - 4x - 5 ) ( x^2 - 4x + 4 )
= (x+1)(x-5)(x-2)^2.