# How do you simplify (c-6)/(3c^2-17c-6)?

May 22, 2018

You break the denominator using middle term break up method

#### Explanation:

$\frac{c - 6}{3 {c}^{2} - 17 c - 6}$

Let's break(factor) the denominator first using middle term break up method.
$3 {c}^{2} - 17 c - 6$

=$3 {c}^{2} - 18 c + c - 6$

=$3 c \left(c - 6\right) + 1 \left(c - 6\right)$

We are taking common factors 3c from $3 {c}^{2} - 18 c$ and +1 from $c - 6$
=$\left(c - 6\right) \left(3 c + 1\right)$

Now let's put it in the math :)
$\frac{c - 6}{\left(c - 6\right) \left(3 c + 1\right)}$

The same factor (c-6) can be divided/cancelles which will result in 1.

$\frac{1}{3 c + 1}$

If you don't know about middle term break up do let me know

May 22, 2018

$\frac{c - 6}{3 {c}^{2} - 17 c - 6} = \frac{1}{3 c + 1}$

#### Explanation:

$\frac{c - 6}{3 {c}^{2} - 17 c - 6}$

Factor the denominator:

$= \frac{c - 6}{3 {c}^{2} - 18 c + c - 6}$

$= \frac{c - 6}{3 c \left(c - 6\right) + \left(c - 6\right)}$

$= \frac{c - 6}{\left(3 c + 1\right) \left(c - 6\right)}$

Simplify by cancelling:

$= \frac{\cancel{\left(c - 6\right)}}{\left(3 c + 1\right) \cancel{\left(c - 6\right)}}$

$= \frac{1}{3 c + 1}$