# How do you simplify (3+x)/(2x^2+5x-3)?

Aug 21, 2015

$\frac{1}{2 x - 1} \text{ }$, with $x \ne \frac{1}{2}$

#### Explanation:

First, recognize that you're dealing with a rational expression, which means that you need to figure out what values of $x$ will make the denominator equal to zero.

In your case, you need to find the values of $x$ that will get you

$2 {x}^{2} + 5 x - 3 = 0$

To do that, use the quadratic formula to find the two roots of this equation

${x}_{1 , 2} = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \cdot 2 \cdot \left(- 3\right)}}{2 \cdot 2}$

${x}_{1 , 2} = \frac{- 5 \pm \sqrt{49}}{4}$

${x}_{1 , 2} = \frac{- 5 \pm 7}{4} = \left\{\begin{matrix}{x}_{1} = \frac{- 5 - 7}{4} = - 3 \\ {x}_{2} = \frac{- 5 + 7}{4} = \frac{1}{2}\end{matrix}\right.$

For a general form quadratic equation

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

you can use the two roots to factor the equation

$\textcolor{b l u e}{a {x}^{2} + b x + c = a \cdot \left(x - {x}_{1}\right) \cdot \left(x - {x}_{2}\right)}$

In your case, you have $a = 2$ and can thus write

$2 {x}^{2} + 5 x - 3 = 2 \cdot \left(x - \left(- 3\right)\right) \cdot \left(x - \frac{1}{2}\right)$

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

This means that your rational expression can be written as

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

Notice that you still need to have $x \ne \frac{1}{2}$ in order to avoid having the denominator equal to zero.

Finally, rewrite the expression in a simpler form

$\frac{1}{2 \cdot \left(x - \frac{1}{2}\right)} = \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot \frac{\left(2 x - 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}} = \textcolor{g r e e n}{\frac{1}{2 x - 1} , \text{ } x \ne \frac{1}{2}}$