Factor (2x+7)^2-(7x+5)^2

${(2 x + 7)}^{2} - {(7 x + 5)}^{2}$

Step 1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x + 7$ and $b = 7 x + 5$ .

$(2 x + 7 + 7 x + 5) (2 x + 7 - (7 x + 5))$

Step 2

Simplify.

Add $2 x$ and $7 x$ .

$(9 x + 7 + 5) (2 x + 7 - (7 x + 5))$

Add $7$ and $5$ .

$(9 x + 12) (2 x + 7 - (7 x + 5))$

Factor $3$ out of $9 x + 12$ .

Factor $3$ out of $9 x$ .

$(3 (3 x) + 12) (2 x + 7 - (7 x + 5))$

Factor $3$ out of $12$ .

$(3 (3 x) + 3 (4)) (2 x + 7 - (7 x + 5))$

Factor $3$ out of $3 (3 x) + 3 (4)$ .

$3 (3 x + 4) (2 x + 7 - (7 x + 5))$

Apply the distributive property.

$3 (3 x + 4) (2 x + 7 - (7 x) - 1 \cdot 5)$

Multiply $7$ by $- 1$ .

$3 (3 x + 4) (2 x + 7 - 7 x - 1 \cdot 5)$

Multiply $- 1$ by $5$ .

$3 (3 x + 4) (2 x + 7 - 7 x - 5)$

Subtract $7 x$ from $2 x$ .

$3 (3 x + 4) (- 5 x + 7 - 5)$

Subtract $5$ from $7$ .

$3 (3 x + 4) (- 5 x + 2)$

Step 3

Factor.

Factor $- 1$ out of $- 5 x + 2$ .

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

$3 (3 x + 4) (- (5 x) + 2)$

Rewrite $2$ as $- 1 (- 2)$ .

$3 (3 x + 4) (- (5 x) - 1 (- 2))$

Factor $- 1$ out of $- (5 x) - 1 (- 2)$ .

$3 (3 x + 4) (- (5 x - 2))$

Remove unnecessary parentheses.

$3 (3 x + 4) \cdot - 1 (5 x - 2)$

Step 4

Combine exponents.

Factor out negative.

$- (3 (3 x + 4) (5 x - 2))$

Multiply $3$ by $- 1$ .

$- 3 ((3 x + 4) (5 x - 2))$

Step 5

Remove unnecessary parentheses.

$- 3 (3 x + 4) (5 x - 2)$

Factor (2x+7)^2-(7x+5)^2

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