# Answer to Question #11094 in Algebra for n lakshmi

Question #11094

factorize

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

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

Expert's answer

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.

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.

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