Divide ((x^2-x-6)/(x+4))÷(x^2-6x+9)

$\frac{x^{2} - x - 6}{x + 4} \div (x^{2} - 6 x + 9)$

Step 1

Rewrite the division as a fraction.

$\frac{\frac{x^{2} - x - 6}{x + 4}}{x^{2} - 6 x + 9}$

Step 2

Multiply the numerator by the reciprocal of the denominator.

$\frac{x^{2} - x - 6}{x + 4} \cdot \frac{1}{x^{2} - 6 x + 9}$

Step 3

Factor $x^{2} - x - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$\frac{(x - 3) (x + 2)}{x + 4} \cdot \frac{1}{x^{2} - 6 x + 9}$

Step 4

Factor using the perfect square rule.

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

$\frac{(x - 3) (x + 2)}{x + 4} \cdot \frac{1}{x^{2} - 6 x + 3^{2}}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$\frac{(x - 3) (x + 2)}{x + 4} \cdot \frac{1}{x^{2} - 2 \cdot x \cdot 3 + 3^{2}}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

$\frac{(x - 3) (x + 2)}{x + 4} \cdot \frac{1}{{(x - 3)}^{2}}$

Step 5

Cancel the common factor of $x - 3$ .

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

$\frac{(x - 3) (x + 2)}{x + 4} \cdot \frac{1}{(x - 3) (x - 3)}$

Cancel the common factor.

$\frac{(x - 3) (x + 2)}{x + 4} \cdot \frac{1}{(x - 3) (x - 3)}$

Rewrite the expression.

$\frac{x + 2}{x + 4} \cdot \frac{1}{x - 3}$

Step 6

Multiply $\frac{x + 2}{x + 4}$ by $\frac{1}{x - 3}$ .

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

Step 7

Split the fraction $\frac{x + 2}{(x + 4) (x - 3)}$ into two fractions.

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

Divide ((x^2-x-6)/(x+4))÷(x^2-6x+9)

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