# How do you solve 4x ^3 -13x^2 + 11x - 2 = 0?

Lets look if there are any roots among the divisors of $- 2$ which are $\pm 1$ , $\pm 2$.
We can easily find out that $x = 1$ and $x = 2$ are roots.Hence

the original question can be written as follows

$4 {x}^{3} - 13 {x}^{2} + 11 x - 2 = \left(x - 1\right) \cdot \left(x - 2\right) \cdot \left(a \cdot x + b\right)$

By equating the two sides we find out that $a = 4$ , $b = - 1$
hence

$4 {x}^{3} - 13 {x}^{2} + 11 x - 2 = \left(x - 1\right) \cdot \left(x - 2\right) \cdot \left(4 x - 1\right)$

Finally the roots are

${x}_{1} = 1$ , ${x}_{2} = 2$ , ${x}_{3} = \frac{1}{4}$