Pre-Algebra Examples

$\frac{\frac{\frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 16}}{x + 1}}{x^{2} - 16}$

Step 1

Multiply the numerator by the reciprocal of the denominator.

$\frac{\frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 16}}{x + 1} \cdot \frac{1}{x^{2} - 16}$

Step 2

Multiply the numerator by the reciprocal of the denominator.

$\frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 16} \cdot \frac{1}{x + 1} \frac{1}{x^{2} - 16}$

Step 3

Factor using the perfect square rule.

Tap for more steps...

Step 3.1

Rewrite $1$ as $1^{2}$ .

$\frac{x^{2} + 2 x + 1^{2}}{x^{2} - 8 x + 16} \cdot \frac{1}{x + 1} \frac{1}{x^{2} - 16}$

Step 3.2

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

$2 x = 2 \cdot x \cdot 1$

Step 3.3

Rewrite the polynomial.

$\frac{x^{2} + 2 \cdot x \cdot 1 + 1^{2}}{x^{2} - 8 x + 16} \cdot \frac{1}{x + 1} \frac{1}{x^{2} - 16}$

Step 3.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

$\frac{{(x + 1)}^{2}}{x^{2} - 8 x + 16} \cdot \frac{1}{x + 1} \frac{1}{x^{2} - 16}$

Step 4

Factor using the perfect square rule.

Tap for more steps...

Step 4.1

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

$\frac{{(x + 1)}^{2}}{x^{2} - 8 x + 4^{2}} \cdot \frac{1}{x + 1} \frac{1}{x^{2} - 16}$

Step 4.2

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

$8 x = 2 \cdot x \cdot 4$

Step 4.3

Rewrite the polynomial.

$\frac{{(x + 1)}^{2}}{x^{2} - 2 \cdot x \cdot 4 + 4^{2}} \cdot \frac{1}{x + 1} \frac{1}{x^{2} - 16}$

Step 4.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

$\frac{{(x + 1)}^{2}}{{(x - 4)}^{2}} \cdot \frac{1}{x + 1} \frac{1}{x^{2} - 16}$

Step 5

Cancel the common factor of $x + 1$ .

Tap for more steps...

Step 5.1

Factor $x + 1$ out of ${(x + 1)}^{2}$ .

$\frac{(x + 1) (x + 1)}{{(x - 4)}^{2}} \cdot \frac{1}{x + 1} \frac{1}{x^{2} - 16}$

Step 5.2

Cancel the common factor.

$\frac{(x + 1) (x + 1)}{{(x - 4)}^{2}} \cdot \frac{1}{x + 1} \frac{1}{x^{2} - 16}$

Step 5.3

Rewrite the expression.

$\frac{x + 1}{{(x - 4)}^{2}} \cdot \frac{1}{x^{2} - 16}$

Step 6

Simplify the denominator.

Tap for more steps...

Step 6.1

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

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

Step 6.2

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 $b = 4$ .

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

Step 7

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

Tap for more steps...

Step 7.1

Multiply $\frac{x + 1}{{(x - 4)}^{2}}$ by $\frac{1}{(x + 4) (x - 4)}$ .

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

Step 7.2

Multiply ${(x - 4)}^{2}$ by $x - 4$ by adding the exponents.

Tap for more steps...

Step 7.2.1

Move $x - 4$ .

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

Step 7.2.2

Multiply $x - 4$ by ${(x - 4)}^{2}$ .

Tap for more steps...

Step 7.2.2.1

Raise $x - 4$ to the power of $1$ .

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

Step 7.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 7.2.3

Add $1$ and $2$ .

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

Step 8

Split the fraction $\frac{x + 1}{{(x - 4)}^{3} (x + 4)}$ into two fractions.

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