Factorize the below equation:
${(3 x - 2 y)}^{2} + 3 (3 x - 2 y) - 10$

(3 x - 2 y + 5) (3 x - 2 y - 2)

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(x - 2 y + 5) (3 x - 2 y - 2)

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(3 x - y + 5) (3 x - y - 2)

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(x - 2 y + 5) (x - y - 2)

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Solution

The correct option is B $(3 x - 2 y + 5) (3 x - 2 y - 2)$
${(3 x - 2 y)}^{2} + 3 (3 x - 2 y) - 10$
$= {(3 x - 2 y)}^{2} + 5 (3 x - 2 y) - 2 (3 x - 2 y) - 10$
$= (3 x - 2 y) ((3 x - 2 y) + 5) - 2 ((3 x - 2 y) + 5)$
$= ((3 x - 2 y) + 5) ((3 x - 2 y) - 2)$

Suggest Corrections

Similar questions

(3 x - 2 y)^{2} - 5 (3 x - 2 y) - 24

(3 x - 2 y - 8) (3 x - 2 y + 3)

x \neq 0, y \neq 0 :

\frac{1}{2 (x + 2 y)} + \frac{5}{3 (3 x - 2 y)} = \frac{- 3}{2} \frac{5}{4 (x + 2 y)} - \frac{3}{5 (3 x - 2 y)} = \frac{61}{60}

Solve for x and y:

$\frac{1}{2 (3 x + y)} + \frac{5}{3 (3 x - y)} = \frac{- 3}{2} \frac{5}{4 (x + 2 y)} - \frac{3}{5 (3 x - 2 y)} = \frac{61}{60} .$

\frac{1}{2 (x + 2 y)} + \frac{5}{3 (3 x - 2 y)} = \frac{- 3}{2}

\frac{5}{4 (x + 2 y)} - \frac{3}{5 (3 x - 2 y)} = \frac{61}{60}

\frac{1}{2 (x + 2 y)} + \frac{5}{3 (3 x - 2 y)} = - \frac{3}{2}, \frac{5}{4 (x + 2 y)} - \frac{3}{5 (3 x - 2 y)} = \frac{61}{60}

x + 2 y \neq 0 a n d 3 x - 2 y \neq 0

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