Factor (5x+4)^2-(10x-7)^2

${(5 x + 4)}^{2} - {(10 x - 7)}^{2}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5 x + 4$ and $b = 10 x - 7$ .

$(5 x + 4 + 10 x - 7) (5 x + 4 - (10 x - 7))$

Simplify.

Add $5 x$ and $10 x$ .

$(15 x + 4 - 7) (5 x + 4 - (10 x - 7))$

Subtract $7$ from $4$ .

$(15 x - 3) (5 x + 4 - (10 x - 7))$

Factor $3$ out of $15 x - 3$ .

Factor $3$ out of $15 x$ .

$(3 (5 x) - 3) (5 x + 4 - (10 x - 7))$

Factor $3$ out of $- 3$ .

$(3 (5 x) + 3 (- 1)) (5 x + 4 - (10 x - 7))$

Factor $3$ out of $3 (5 x) + 3 (- 1)$ .

$3 (5 x - 1) (5 x + 4 - (10 x - 7))$

Apply the distributive property.

$3 (5 x - 1) (5 x + 4 - (10 x) - - 7)$

Multiply $10$ by $- 1$ .

$3 (5 x - 1) (5 x + 4 - 10 x - - 7)$

Multiply $- 1$ by $- 7$ .

$3 (5 x - 1) (5 x + 4 - 10 x + 7)$

Subtract $10 x$ from $5 x$ .

$3 (5 x - 1) (- 5 x + 4 + 7)$

Add $4$ and $7$ .

$3 (5 x - 1) (- 5 x + 11)$

Factor.

Factor $- 1$ out of $- 5 x + 11$ .

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

$3 (5 x - 1) (- (5 x) + 11)$

Rewrite $11$ as $- 1 (- 11)$ .

$3 (5 x - 1) (- (5 x) - 1 (- 11))$

Factor $- 1$ out of $- (5 x) - 1 (- 11)$ .

$3 (5 x - 1) (- (5 x - 11))$

Remove unnecessary parentheses.

$3 (5 x - 1) \cdot - 1 (5 x - 11)$

Combine exponents.

Factor out negative.

$- (3 (5 x - 1) (5 x - 11))$

Multiply $3$ by $- 1$ .

$- 3 ((5 x - 1) (5 x - 11))$

Remove unnecessary parentheses.

$- 3 (5 x - 1) (5 x - 11)$

Factor (5x+4)^2-(10x-7)^2

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