Simplify (4x^2+4x+1)/(4x^2-1)

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

Step 1

Factor using the perfect square rule.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

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

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

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

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

$4 x = 2 \cdot (2 x) \cdot 1$

Rewrite the polynomial.

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

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

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

Step 2

Simplify the denominator.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

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

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

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

Step 3

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

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

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

$\frac{2 x + 1}{2 x - 1}$