Question

15 (2x-3y)^2 -4 (4x-6y)-16

Answer 1

The factorisation of $15 (2x-3y)^2 -4 (4x-6y)-16$ is (10x - 15y + 4)(6x  - 9y - 4).

Step-by-step explanation:

We have,

$15 (2x-3y)^2 -4 (4x-6y)-16$

To find, the factorisation of $15 (2x-3y)^2 -4 (4x-6y)-16$ = ?

∴ $15 (2x-3y)^2 -4 (4x-6y)-16$

$=15(2x-3y)^2 -8(2x-3y)-16$

Let a = 2x - 3y, we get

$=15a^2 -8a-16$

$=15a^2 -20a+12a-16$

= 5a(3a - 4) + 4(3a - 4)

= (5a + 4)(3a - 4)

Put a = 2x - 3y, we get

= [5(2x - 3y) + 4][3(2x - 3y) - 4]

= (10x - 15y + 4)(6x  - 9y - 4)

∴ The factorisation of $15 (2x-3y)^2 -4 (4x-6y)-16$

= (10x - 15y + 4)(6x  - 9y - 4)

Thus, the factorisation of $15 (2x-3y)^2 -4 (4x-6y)-16$ is (10x - 15y + 4)(6x  - 9y - 4).