# How do you factor 6c^2 + 17c-14=0?

Oct 28, 2015

$6 {c}^{2} + 17 c - 14 = 6 \left(c + \frac{7}{2}\right) \left(c - \frac{2}{3}\right)$

#### Explanation:

First of all, try to solve it using the classical formula

$\setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

and in your case $a = 6$, $b = 17$ and $c = - 14$.

So, the formula becomes

$\setminus \frac{- 17 \setminus \pm \setminus \sqrt{{17}^{2} - 4 \cdot 6 \cdot \left(- 14\right)}}{2 \cdot 6} = \setminus \frac{- 17 \setminus \pm \setminus \sqrt{289 - \left(- 336\right)}}{12}$

$= \setminus \frac{- 17 \setminus \pm \setminus \sqrt{625}}{12} = \setminus \frac{- 17 \setminus \pm 25}{12}$

Which leads us to the two solutions

${c}_{1} = \frac{- 17 - 25}{12} = - \frac{7}{2}$ and
${c}_{2} = \frac{- 17 + 25}{12} = \frac{2}{3}$.

Now, use the fact that a polynomial can be written in terms of his solutions, as the product of factors of the form $x - {x}_{n}$, where ${x}_{n}$ is the $n$-th solutions, multiplied by the coefficient of the higher power. In your case,

$6 {c}^{2} + 17 c - 14 = 6 \left(c + \frac{7}{2}\right) \left(c - \frac{2}{3}\right)$