Question

8. Factorise : (x² - 2x)2 - 11(x?- 2x) + 24.​

Answer 1

Step-by-step explanation:

see this attachment .

hope it helps

Answer 2

Let, x² - 2x = p

∴ (x² - 2x)² - 11 (x² - 24) + 24

Substituting x² - 2x = p, we get:

= p² - 11p + 24

1 * 24 = 24 = 8 * 3 and 8 + 3 = 11:

= p² - (8 + 3) p + 24

Since (a + b) c = ac + bc, we get:

= p² - 8p - 3p + 24

Grouping of two consecutive terms give:

= p (p - 8) - 3 (p - 8)

Taking (p - 8) common, we have:

= (p - 8) (p - 3)

Putting p = x² - 2x, we get:

= (x² - 2x - 8) (x² - 2x - 3) ..... (1)

Now we factorise the in-bracket terms:

x² - 2x - 8 = x² - (4 - 2) x - 8

= x² - 4x + 2x - 8

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

= (x - 4) (x + 2)

& x² - 2x - 3 = x² - (3 - 1) x - 3

= x² - 3x + x - 3

= x (x - 3) + 1 (x - 3)

= (x - 3) (x + 1)

Putting the factors in (1), we write:

(x² - 2x)² - 11 (x² - 2x) + 24

= (x² - 2x - 8) (x² - 2x - 3)

= (x - 4) (x + 2) (x - 3) (x + 1)

This is the required factorisation.