# How do you factor 6r^3-35r-11r^2?

May 16, 2015

First, we must recognize all the common elements in your three parts, which, in this case, is just $r$, as follows:

$6 \cdot \textcolor{g r e e n}{r} \cdot r \cdot r \cdot - 35 \cdot \textcolor{g r e e n}{r} - 11 \cdot \textcolor{g r e e n}{r} \cdot r$

So, we take it out an mutiply the rest by this common factor:

$r \left(6 {r}^{2} - 35 - 11 r\right)$

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2} a$
$\frac{11 \pm \sqrt{121 - 4 \cdot 6 \cdot \left(- 35\right)}}{2 \cdot 6}$
$\frac{11 \pm \sqrt{961}}{12}$
$\frac{11 \pm 31}{12}$

${x}_{1} = \frac{7}{2}$, which is the same as $2 {x}_{1} - 7 = 0$.
${x}_{2} = - \frac{5}{3}$, which is the same as $3 {x}_{2} + 5 = 0$.

So, now we have our factors, using our roots:

$r \left(2 x - 7\right) \left(3 x + 5\right)$