Factor (2x-3)^2-(3x+2)^2

${(2 x - 3)}^{2} - {(3 x + 2)}^{2}$

Step 1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x - 3$ and $b = 3 x + 2$ .

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

Step 2

Simplify.

Add $2 x$ and $3 x$ .

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

Add $- 3$ and $2$ .

$(5 x - 1) (2 x - 3 - (3 x + 2))$

Apply the distributive property.

$(5 x - 1) (2 x - 3 - (3 x) - 1 \cdot 2)$

Multiply $3$ by $- 1$ .

$(5 x - 1) (2 x - 3 - 3 x - 1 \cdot 2)$

Multiply $- 1$ by $2$ .

$(5 x - 1) (2 x - 3 - 3 x - 2)$

Subtract $3 x$ from $2 x$ .

$(5 x - 1) (- x - 3 - 2)$

Subtract $2$ from $- 3$ .

$(5 x - 1) (- x - 5)$

Step 3

Factor.

Factor $- 1$ out of $- x - 5$ .

Factor $- 1$ out of $- x$ .

$(5 x - 1) (- (x) - 5)$

Rewrite $- 5$ as $- 1 (5)$ .

$(5 x - 1) (- (x) - 1 (5))$

Factor $- 1$ out of $- (x) - 1 (5)$ .

$(5 x - 1) (- (x + 5))$

Remove unnecessary parentheses.

$(5 x - 1) \cdot - 1 (x + 5)$

Step 4

Factor out negative.

$- ((5 x - 1) (x + 5))$

Step 5

Remove unnecessary parentheses.

$- (5 x - 1) (x + 5)$

Factor (2x-3)^2-(3x+2)^2

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