Factorise: $9 {(2 x - y)}^{2} - 4 (2 x - y) - 13$

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Solution

Let $(2 x - y) = p$ , then the equation $9 (2 x - y)^{2} - 4 (2 x - y) - 13$ becomes $9 p^{2} - 4 p - 13$

Consider the expression $9 p^{2} - 4 p - 13$ we can factorise it as follows:

$9 p^{2} - 4 p - 13 = 9 p^{2} + 9 p - 13 p - 13 = 9 p (p + 1) - 13 (p + 1) = (9 p - 13) (p + 1)$

Now, substitute the value of $p$ as $p = (2 x - y)$ :

$(9 p - 13) (p + 1) = (9 (2 x - y) - 13) (2 x - y + 1) = (18 x - 9 y - 13) (2 x - y + 1)$

Hence, $9 (2 x - y)^{2} - 4 (2 x - y) - 13 = (18 x - 9 y - 13) (2 x - y + 1)$

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