Question

factorise (x^2-2x)^2-11(x^2-2x)+24 plz give solution

Answer 1

The factors of the equation are ( x - 4 ) ( x + 2 ) ( x - 3 ) ( x + 1 ).

Given: (x^2-2x)^2-11(x^2-2x)+24

To Find: Factorisation of the given equation

Solution: ( x² - 2x )² - 11 ( x² - 2x ) + 24

Let, y = ( x² - 2x ) ( Equation 1)

Now, we will substitute the value of ( x² - 2x ) with y:

y² - 11y + 24

-8 and -3 are the factors of the above equation.

Now, putting - 8y and - 3y in place of - 11y:

y² - 8y - 3y + 24

y(y - 8 ) -3(y - 8)

(y - 8)(y - 3)

Now, we will substitute the value of y from equation 1:

(x² - 2x - 8)(x² - 2x - 3)

Solving the above equation:

( x² - 4x + 2x - 8 ) ( x² - 3x + 1x - 3)

[ x ( x- 4 ) +2 ( x - 4 ) ] [x ( x - 3 ) +1 ( x - 3 )]

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

Therefore, the factors are ( x - 4 ) ( x + 2 ) ( x - 3 ) ( x + 1 ).

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Answer 2

Answer:

Given: (x^2-2x)^2-11(x^2-2x)+24

To find: factorize

Step-by-step explanation:

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 gives:

= 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 factorize 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 factorization.

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