# How do you simplify (2x^2+6x+4)/(3x^2+9x+6) and then find the excluded values?

##### 1 Answer
Apr 4, 2017

$\frac{2}{3}$

#### Explanation:

The first step is to factorise the numerator/denominator.

Both have a $\textcolor{b l u e}{\text{common factor}}$ that can be taken out.

$\frac{\textcolor{b l u e}{2} \left({x}^{2} + 3 x + 2\right)}{\textcolor{b l u e}{3} \left({x}^{2} + 3 x + 2\right)}$

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

Excluded values are values of x that make the function $\textcolor{b l u e}{\text{ undefined}}$

The denominator cannot equal zero as this would make the function undefined. Equating the denominator to zero and solving gives the values that x cannot be.

$\text{solve } {x}^{2} + 3 x + 2 = 0$

$\Rightarrow \left(x + 1\right) \left(x + 2\right) = 0 \Rightarrow x = - 1 \text{ or } x = - 2$

$\text{Thus " x=-1" and " x=-2" are the excluded values}$