# How do you simplify and find the restrictions for (2x^2+11x+5)/(3x^2+17x+10)?

May 21, 2017

See a solution process below:

#### Explanation:

First, factor the numerator and denominator:

$\frac{2 {x}^{2} + 11 x + 5}{3 {x}^{2} + 17 x + 10} \implies \frac{\left(2 x + 1\right) \left(x + 5\right)}{\left(3 x + 2\right) \left(x + 5\right)}$

Now, cancel common terms in the numerator and denominator:

$\frac{\left(2 x + 1\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 5\right)}}}}{\left(3 x + 2\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 5\right)}}}} \implies \frac{2 x + 1}{3 x + 2}$

To find the restrictions the denominator cannot be $0$ therefore we need to solve for:

$3 {x}^{2} + 17 x + 10 =$

Or

$\left(3 x + 2\right) \left(x + 5\right) = 0$

Solution 1)

$3 x + 2 = 0$

$3 x + 2 - \textcolor{red}{2} = 0 - \textcolor{red}{2}$

$3 x + 0 = - 2$

$3 x = - 2$

$\frac{3 x}{\textcolor{red}{3}} = - \frac{2}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} x}{\cancel{\textcolor{red}{3}}} = - \frac{2}{3}$

$x = - \frac{2}{3}$

Solution 2)

$x + 5 = 0$

$x + 5 - \textcolor{red}{5} = 0 - \textcolor{red}{5}$

$x + 0 = - 5$

$x = - 5$

Therefore, the restrictions are:

$x \ne - \frac{2}{3}$ and $x \ne - 5$