Let $p (x) = 9 x^{2} + y^{2} + 4 z^{2} + 6 x y - 4 y z - 12 x z$ factorize $P (x)$ for $x = 1, y = 1, z = 1$

1

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4

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9

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16

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Solution

The correct option is A $4$
We know that one of the formula is $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$ .
In the given polynomial $9 x^{2} + y^{2} + 4 z^{2} + 6 x y - 4 y z - 12 x z$ , we observe that $a = 3 x, b = y$ and $c = - 2 z$ , therefore, comparing with the above formula, we get:
$9 x^{2} + y^{2} + 4 z^{2} + 6 x y - 4 y z - 12 x z = (3 x)^{2} + (y)^{2} + (- 2 z)^{2} + (2 \times 3 x \times y) + (2 \times y \times (- 2 z)) + (2 \times (- 2 z) \times 3 x)$
$= (3 x + y - 2 z)^{2}$
Let us substitute $x = 1, y = 1$ and $z = 1$ in $(3 x + y - 2 z)^{2}$ as shown below:
$[3 \times 1) + 1 - (2 \times 1)]^{2} = (3 + 1 - 2)^{2} = 2^{2} = 4$
Hence, $9 x^{2} + y^{2} + 4 z^{2} + 6 x y - 4 y z - 12 x z = 4$ if $x = 1, y = 1$ and $z = 1$ .

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