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

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

Step 1

Simplify the numerator.

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

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

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 $b = 3$ .

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

Step 2

Factor $x$ out of $x^{2} - 3 x$ .

Factor $x$ out of $x^{2}$ .

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

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

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

Factor $x$ out of $x \cdot x + x \cdot - 3$ .

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

Step 3

Factor $x^{2} + x - 12$ 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 $1$ .

$- 3, 4$

Write the factored form using these integers.

$\frac{(x + 3) (x - 3)}{x^{2}} \cdot \frac{x (x - 3)}{(x - 3) (x + 4)}$

Step 4

Simplify terms.

Combine.

$\frac{(x + 3) (x - 3) (x (x - 3))}{x^{2} ((x - 3) (x + 4))}$

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{(x + 3) (x - 3) (x (x - 3))}{x^{2} ((x - 3) (x + 4))}$

Rewrite the expression.

$\frac{(x + 3) (x (x - 3))}{x^{2} (x + 4)}$

Cancel the common factor of $x$ and $x^{2}$ .

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

$\frac{x ((x + 3) (x - 3))}{x^{2} (x + 4)}$

Cancel the common factors.

Factor $x$ out of $x^{2} (x + 4)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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