Algebra Examples

${(7 x - 1)}^{2} - {(9 x - 2)}^{2}$

Step 1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 7 x - 1$ and $b = 9 x - 2$ .

$(7 x - 1 + 9 x - 2) (7 x - 1 - (9 x - 2))$

Step 2

Simplify.

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Step 2.1

Add $7 x$ and $9 x$ .

$(16 x - 1 - 2) (7 x - 1 - (9 x - 2))$

Step 2.2

Subtract $2$ from $- 1$ .

$(16 x - 3) (7 x - 1 - (9 x - 2))$

Step 2.3

Apply the distributive property.

$(16 x - 3) (7 x - 1 - (9 x) - - 2)$

Step 2.4

Multiply $9$ by $- 1$ .

$(16 x - 3) (7 x - 1 - 9 x - - 2)$

Step 2.5

Multiply $- 1$ by $- 2$ .

$(16 x - 3) (7 x - 1 - 9 x + 2)$

Step 2.6

Subtract $9 x$ from $7 x$ .

$(16 x - 3) (- 2 x - 1 + 2)$

Step 2.7

Add $- 1$ and $2$ .

$(16 x - 3) (- 2 x + 1)$

Step 3

Factor.

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Step 3.1

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

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Step 3.1.1

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

$(16 x - 3) (- (2 x) + 1)$

Step 3.1.2

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

$(16 x - 3) (- (2 x) - 1 (- 1))$

Step 3.1.3

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

$(16 x - 3) (- (2 x - 1))$

Step 3.2

Remove unnecessary parentheses.

$(16 x - 3) \cdot - 1 (2 x - 1)$

Step 4

Factor out negative.

$- ((16 x - 3) (2 x - 1))$

Step 5

Remove unnecessary parentheses.

$- (16 x - 3) (2 x - 1)$