Question

Factorise :-
(x² - 4x) (x² - 4x - 1) - 20​

Answer 1

Let (x^2–4x)=p

= p(p-1)-20

= p^2 -p -20

= p^2 -5p + 4p -20

= p (p-5 ) + 4 (p - 5 )

= (p - 5 ) (p + 4 )

put p = x^2 -4x

= (x^2 - 4 x - 5 ) (x ^2 -4 x + 4 )

=[ x^2–5x + x -5 ] ( x - 2 )^2

=[ x (x- 5) +1( x-5 )](x-2)^2

= (x-5) (x + 1) (x - 2 )^2

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Answer 2

Solution -

We have,

➝ (x² - 4x) (x² - 4x - 1) - 20

➝ (x² - 4x) (x² - 4x) - (x² - 4x) - 20

➝ (x² - 4x)² - (x² - 4x) - 29

⠀

Let,

• y = x² - 4x

➝ y² - y - 20

➝ y² - 5y + 4y - 20

➝ y(y - 5) + 4(y - 5)

➝ (y - 5) (y + 4)

⠀

Replacing y by x² - 4x

➝ (x² - 4x - 5) (x² - 4x + 4)

➝ (x² - 5x + x - 5) (x² - 2 × x × 2 + 2²)

➝ {x(x - 5) + 1(x - 5)} (x - 2)²

➝ (x - 5) (x + 1) (x - 2)²

⠀

Identity used here

• (a - b)² = a² + b² - 2ab

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