Solve by Factoring (x^2-16)/(x-4)=x+4

$\frac{x^{2} - 16}{x - 4} = x + 4$

Step 1

Move all the expressions to the left side of the equation.

Subtract $x$ from both sides of the equation.

$\frac{x^{2} - 16}{x - 4} - x = 4$

Subtract $4$ from both sides of the equation.

$\frac{x^{2} - 16}{x - 4} - x - 4 = 0$

Step 2

Simplify $\frac{x^{2} - 16}{x - 4} - x - 4$ .

Simplify each term.

Simplify the numerator.

Rewrite $16$ as $4^{2}$ .

$\frac{x^{2} - 4^{2}}{x - 4} - x - 4 = 0$

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

$\frac{(x + 4) (x - 4)}{x - 4} - x - 4 = 0$

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$\frac{(x + 4) (x - 4)}{x - 4} - x - 4 = 0$

Divide $x + 4$ by $1$ .

$x + 4 - x - 4 = 0$

Combine the opposite terms in $x + 4 - x - 4$ .

Subtract $x$ from $x$ .

$0 + 4 - 4 = 0$

Add $0$ and $4$ .

$4 - 4 = 0$

Subtract $4$ from $4$ .

$0 = 0$

Step 3

Since $0 = 0$ , the equation will always be true.

Always true