Simplify (x^3-9x)/(x^2-7x+12)

$\frac{x^{3} - 9 x}{x^{2} - 7 x + 12}$

Step 1

Simplify the numerator.

Factor $x$ out of $x^{3} - 9 x$ .

Factor $x$ out of $x^{3}$ .

$\frac{x \cdot x^{2} - 9 x}{x^{2} - 7 x + 12}$

Factor $x$ out of $- 9 x$ .

$\frac{x \cdot x^{2} + x \cdot - 9}{x^{2} - 7 x + 12}$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 9$ .

$\frac{x (x^{2} - 9)}{x^{2} - 7 x + 12}$

Rewrite $9$ as $3^{2}$ .

$\frac{x (x^{2} - 3^{2})}{x^{2} - 7 x + 12}$

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 $b = 3$ .

$\frac{x (x + 3) (x - 3)}{x^{2} - 7 x + 12}$

Step 2

Factor $x^{2} - 7 x + 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $12$ and whose sum is $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$\frac{x (x + 3) (x - 3)}{(x - 4) (x - 3)}$

Step 3

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{x (x + 3) (x - 3)}{(x - 4) (x - 3)}$

Rewrite the expression.

$\frac{x (x + 3)}{x - 4}$

Simplify (x^3-9x)/(x^2-7x+12)

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