Question

Factorise(a^2-a)(4a^2-4a-5)-6

Answer 1

Hi ☺

For the answer refer above attachment❤

Hope it helps!✔

Answer 2

Answer:

$factors \: ( {a}^{2} - a - 2)(4{a}^{2} - 4a + 3)$

Step-by-step explanation:

To factorise

$({a}^{2} - a)(4 {a}^{2} - 4a - 5) - 6$

by inspection it is clear it can't be factorised without substitution

preparation for substitution

$({a}^{2} - a)(4 ({a}^{2} - a )- 5) - 6 \\ let \: {a}^{2} - a = z \\ \\ so \\ \\ = > z(4z - 5) - 6 \\ \\ = > 4 {z}^{2} - 5z - 6$

now it is the Quadratic equation in z,it can easily be factorised by splitting the middle term

$= > 4 {z}^{2} - 8z + 3z - 6 \\ \\ = > 4z(z - 2) + 3(z - 2) \\ \\ = > (z - 2)(4z + 3)$

Now undo substitution

$( {a}^{2} - a - 2)(4( {a}^{2} - a) + 3) \\ \\ ( {a}^{2} - a - 2)(4{a}^{2} - 4a + 3)$

Hope it helps you.