Factorize:
$16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z$

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Solution

We know that, the algebraic identity:
$a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = (a + b + c)^{2}$
Now, given expansion is,
$16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z$
Here, in $- 16 x y, - 12 y z$ common variable is $y$ and these are negative terms.
So, the variable $y$ is negative.
$\Rightarrow 16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z$
$= (4 x)^{2} + (- 2 y)^{2} + (3 z)^{2} + 2 (4 x) (- 2 y) + 2 (- 2 y) (3 z) + 2 (4 x) (3 z)$
$= (4 x - 2 y + 3 z)^{2}$
$∴ 16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z = (4 x - 2 y + 3 z)^{2}$

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Question 29
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Factorise the following:

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