Factorise
$1 - {(2 x - 3 y)}^{2}$

(1 - 2 x + y) (2 x + 3 y)

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(2 + 2 x - 3 y) (1 + 2 x + 3 y)

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(2 + x - 3 y) (2 x + 3 y)

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(1 + 2 x - 3 y) (1 - 2 x + 3 y)

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Solution

The correct option is D $(1 + 2 x - 3 y) (1 - 2 x + 3 y)$
$1 - {(2 x - 3 y)}^{2}$

Using,
$(a^{2} - b^{2}) = (a + b) (a - b)$

$1^{2} - {(2 x - 3 y)}^{2}$

$= (1 + 2 x - 3 y) (1 - 2 x + 3 y)$

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Similar questions

Factorise $1 - (2 x - 3 y)^{2}$

{(\frac{1}{2} x - \frac{1}{3} y)}^{3} + {(\frac{1}{3} y - \frac{1}{4} z)}^{3} + {(\frac{1}{4} z - \frac{1}{2} x)}^{3} = m (\frac{1}{2} x - \frac{1}{3} y) (\frac{1}{3} y - \frac{1}{4} z) (\frac{1}{4} z - \frac{1}{2} x)

m

(2 x + 3 y)^{2}

(2 x - 3 y)^{2}

(2 x + 3 y) (2 x - 3 y)

Factorise : $1 - (2 x + 3 y) - 6 (2 x + 3 y)^{2}$

(2 x - 3 y) (2 x + 3 y) = (2 x)^{2} - (3 y)^{2}

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