Divide ((x^2+2x+1)/(x^2+4x+3))÷((x^2-1)/(x^2+2x+1))

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

Step 1

To divide by a fraction, multiply by its reciprocal.

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

Step 2

Factor using the perfect square rule.

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

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

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

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

Rewrite the polynomial.

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

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

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

Step 3

Factor $x^{2} + 4 x + 3$ 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 $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

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

Step 4

Cancel the common factor of ${(x + 1)}^{2}$ and $x + 1$ .

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

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Step 5

Factor using the perfect square rule.

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

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

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

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

Rewrite the polynomial.

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

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

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

Step 6

Simplify the denominator.

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

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

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 $b = 1$ .

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

Step 7

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 8

Multiply $\frac{1}{x + 3}$ by $\frac{{(x + 1)}^{2}}{x - 1}$ .

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

Divide ((x^2+2x+1)/(x^2+4x+3))÷((x^2-1)/(x^2+2x+1))

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