Basic Math Examples

\frac{x^{2} - 6 x - 27}{x^{2} - 6 x + 9}

$\frac{x^{2} - 6 x - 27}{x^{2} - 6 x + 9}$

Step 1

Factor

x^{2} - 6 x - 27

$x^{2} - 6 x - 27$ using the AC method.

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Step 1.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 27

$- 27$ and whose sum is

- 6

$- 6$ .

- 9, 3

$- 9, 3$

Step 1.2

Write the factored form using these integers.

\frac{(x - 9) (x + 3)}{x^{2} - 6 x + 9}

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

\frac{(x - 9) (x + 3)}{x^{2} - 6 x + 9}

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

Step 2

Factor using the perfect square rule.

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Step 2.1

Rewrite

9

$9$ as

3^{2}

$3^{2}$ .

\frac{(x - 9) (x + 3)}{x^{2} - 6 x + 3^{2}}

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

Step 2.2

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

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

Step 2.3

Rewrite the polynomial.

\frac{(x - 9) (x + 3)}{x^{2} - 2 \cdot x \cdot 3 + 3^{2}}

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

Step 2.4

Factor using the perfect square trinomial rule

a^{2} - 2 a b + b^{2} = {(a - b)}^{2}

$a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where

a = x

$a = x$ and

b = 3

$b = 3$ .

\frac{(x - 9) (x + 3)}{{(x - 3)}^{2}}

$\frac{(x - 9) (x + 3)}{{(x - 3)}^{2}}$

\frac{(x - 9) (x + 3)}{{(x - 3)}^{2}}

$\frac{(x - 9) (x + 3)}{{(x - 3)}^{2}}$