# How do you factor the expression 35x^3-61x^2+8x?

Mar 11, 2017

The answer is $= x \left(7 x - 1\right) \left(5 x - 8\right)$

#### Explanation:

We start by factoring $x$

$35 {x}^{3} - 61 {x}^{2} + 8 x = x \left(35 {x}^{2} - 61 x + 8\right)$

We need the roots of

$35 {x}^{2} - 61 x + 8$

We compare this to

$a {x}^{2} + b x + c$

We calculate the discriminant

$\Delta = {b}^{2} - 4 a c = {\left(- 61\right)}^{2} - 4 \cdot 35 \cdot 8 = 2601$

As $\Delta > 0$, there are 2 real roots

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\sqrt{\Delta} = \sqrt{2601} = 51$

${x}_{1} = \frac{61 + 51}{70} = 1.6 = \frac{16}{10} = \frac{8}{5}$

${x}_{2} = \frac{61 - 51}{70} = \frac{10}{70} = \frac{1}{7}$

Therefore,

$35 {x}^{2} - 61 x + 8 = 35 \left(x - \frac{8}{5}\right) \left(x - \frac{1}{7}\right)$

$= \left(7 x - 1\right) \left(5 x - 8\right)$

So,

$35 {x}^{3} - 61 {x}^{2} + 8 x = x \left(35 {x}^{2} - 61 x + 8\right) = x \left(7 x - 1\right) \left(5 x - 8\right)$