Factorise: $12 {(z + 1)}^{2} + 25 (z + 1) (x + 2) + 12 {(x + 2)}^{2}$

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Solution

Let $(z + 1) = p$ and $(x + 2) = q$ , then the equation $12 (z + 1)^{2} - 25 (z + 1) (x + 2) + 12 (x + 2)^{2}$ becomes $12 p^{2} - 25 p q + 12 q^{2}$

Consider the expression $12 p^{2} - 25 p q + 12 q^{2}$ we can factorise it as follows:

$12 p^{2} - 25 p q + 12 q^{2} = 12 p^{2} - 16 p q - 9 p q + 12 q^{2} = 4 p (3 p - 4 q) - 3 q (3 p - 4 q) = (4 p - 3 q) (3 p - 4 q)$

Now, substitute the value of $p$ as $p = (z + 1)$ and $q = (x + 2)$ :

$(4 p - 3 q) (3 p - 4 q) = [4 (z + 1) - 3 (x + 2)] [3 (z + 1) - 4 (x + 2)] = (4 z + 4 - 3 x - 6) (3 z + 3 - 4 x - 8)$
$= (4 z - 3 x - 2) (3 z - 4 x - 5)$

Hence, $12 (z + 1)^{2} - 25 (z + 1) (x + 2) + 12 (x + 2)^{2} = (4 z - 3 x - 2) (3 z - 4 x - 5)$

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Factorise:
$4 x^{2} + 9 y^{2} + 25 z^{2} + 12 x y + 30 y z + 20 z x$

4 x^{2} + 9 y^{2} + 25 z^{2} + 12 x y - 30 y z - 20 z x

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