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Difference of Two Squares

Factorization...

Question

Factorization of $x^{2} - 6 y + 9 - 4 z^{2}$ is equal to

(x - 3 - 2 z) (x - 3 + 2 z)

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(x - 3 + 2 z) (x - 3 + 2 z)

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(x - 3 - 2 z) (x - 3 - 2 z)

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(x + 3 - 2 z) (x + 3 + 2 z)

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Solution

The correct option is A $(x - 3 - 2 z) (x - 3 + 2 z)$
$x^{2} - 6 x + 9 - 4 z^{2}$ can be written as
= $x^{2} - 6 x + 9 - 4 z^{2}$
= $x^{2} - 2 \times 3 x + 9 - (2 z)^{2}$
= $(x - 3)^{2} - (2 z)^{2}$ [ $(a + b)^{2} = a^{2} + b^{2} + 2 a b$ ]
= $(x - 3 - 2 z) (x - 3 + 2 z)$ [ $a^{2} - b^{2} = (a + b) (a - b)$ ]
Hence, the factorize expression is

$(x - 3 - 2 z) (x - 3 + 2 z)$

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