Factorise :

$8 {(3 x - 2 y)}^{2} - 6 x + 4 y - 1$

(2 x - 4 y - 1) (2 x - 8 y + 1)

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(6 x - 4 y - 1) (12 x - 8 y + 1)

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(2 x - 4 y + 1) (12 x - y + 1)

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(6 x - 4 y + 1) (x - 8 y + 1)

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Solution

The correct option is A $(6 x - 4 y - 1) (12 x - 8 y + 1)$
$8 {(3 x - 2 y)}^{2} - 6 x + 4 y - 1$
$= 8 {(3 x - 2 y)}^{2} - 2 (3 x - 2 y) - 1$
Let $3 x - 2 y = a$
$8 a^{2} - 2 a - 1$
$= 8 a^{2} - 4 a + 2 a - 1$
$= - 4 a (2 a - 1) + 1 (2 a - 1)$
$= (2 a - 1) (4 a + 1)$
Putting the value of $a$ ,
$[2 (3 x - 2 y) - 1] [4 (3 x - 2 y) + 1]$
$= (6 x - 4 y - 1) (12 x - 8 y + 1)$

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Similar questions

$8 (3 x - 2 y)^{2} - 6 x + 4 y - 1$

9 x^{2} + 4 y^{2} + 1 + 12 x y + 4 y + 6 x

3 x + 4 y = 6 & 6 x + 8 y = 12

3 x + 4 y - z = s, 6 x + 8 y - 2 z = 15

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