Question
Factorise:$ 9(2a - b)^2 - 4(2a - b) - 13$
✓
Answer
Let $2a - b = c$
Then, $9(2a - b)^2 - 4(2a - b) - 13$
$= 9c^2 - 4c - 13 = 9c^2 - 13c + 9c - 13$
$= c(9c - 13) + 1(9c - 13)$
$= (c + 1)(9c - 13)$
Now, replacing $c$ by $(2a - b),$
we get $9(2a - b)^2 - 4(2a - b) - 13$
$= (2a - b + 1)[9(2a - b) - 13]$
$= (2a - b + 1)(18a - 9b - 13)$
Then, $9(2a - b)^2 - 4(2a - b) - 13$
$= 9c^2 - 4c - 13 = 9c^2 - 13c + 9c - 13$
$= c(9c - 13) + 1(9c - 13)$
$= (c + 1)(9c - 13)$
Now, replacing $c$ by $(2a - b),$
we get $9(2a - b)^2 - 4(2a - b) - 13$
$= (2a - b + 1)[9(2a - b) - 13]$
$= (2a - b + 1)(18a - 9b - 13)$
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