# How do you simplify and find the excluded value of (2x^2+6x+4)/(4x^2-12x-16)?

May 26, 2017

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

#### Explanation:

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

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

$\Rightarrow \frac{1}{2} \cdot \frac{x \left(x + 1\right) + 2 \left(x + 1\right)}{x \left(x - 4\right) + 1 \left(x - 4\right)}$

$\Rightarrow \frac{1}{2} \cdot \frac{\left(x + 1\right) \left(x + 2\right)}{\left(x + 1\right) \left(x - 4\right)}$

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

May 26, 2017

$= \frac{x + 2}{2 \left(x - 4\right)}$

This means that $x \ne 4$

#### Explanation:

The first step in algebraic fractions is to factor as far as possible:

(2x^2+6x+4)/(4x^2-12x-16)" "(larr"common factor")/(larr "common factor")

$= \frac{2 \left({x}^{2} + 3 x + 2\right)}{4 \left({x}^{2} - 3 x - 4\right)} \left(\leftarrow \text{quadratic trinomial")/(larr "quadratic trinomial}\right)$

$= \frac{2 \left(x + 2\right) \left(x + 1\right)}{4 \left(x - 4\right) \left(x + 1\right)}$

$= \frac{\cancel{2} \left(x + 2\right) \cancel{\left(x + 1\right)}}{{\cancel{4}}^{2} \left(x - 4\right) \cancel{\left(x + 1\right)}} \text{ } \leftarrow$ cancel like factors

$= \frac{x + 2}{2 \left(x - 4\right)}$