(x2 – x) (4x2 – 4x – 5) – 6
(x2 – x) [(4x2 – 4x) – 5] – 6
(x2 – x) [4(x2 – x) – 5] – 6
Let us assume x2 – x = q
So, q[4q – 5] – 6
4q2 – 5q – 6
4q2 – 8q + 3q – 6
4q(q – 2) + 3(q – 2)
(q – 2) (4q + 3)
Now, substitute the value of q
(x2 – x – 2) (4(x2 – x) + 3)
(x2 – x – 2) (4x2 – 4x + 3)
(x2 – 2x + x – 2) (4x2 – 4x + 3)
[x(x – 2) + 1(x – 2)] (4x2 – 4x + 3)
(x – 2) (x + 1) (4x2 – 4x + 3)