# How do you find the roots of x^3-6x^2+13x-10=0?

Sep 15, 2016

$x = 2$

#### Explanation:

${x}^{3} - 6 {x}^{2} + 13 x - 10 = 0$

${x}^{3} - 3 {\left(x\right)}^{2} \left(2\right) + 3 {\left(2\right)}^{2} x + x - {2}^{3} - 2 = 0$

$\left({x}^{3} - 3 {\left(x\right)}^{2} \left(2\right) + 3 x {\left(2\right)}^{2} - {2}^{3}\right) + x - 2 = 0$

We can factorize using the polynomial identity that follows:
${\left(a - b\right)}^{3} = {a}^{3} - 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$

where in our case $a = x$ and $b = 2$

So,
${\left(x - 2\right)}^{3} + \left(x - 2\right) = 0$ taking $x - 2$ as common factor
$\left(x - 2\right) \left({\left(x - 2\right)}^{2} + 1\right) = 0$
$\left(x - 2\right) \left({x}^{2} - 4 x + 4 + 1\right) = 0$
$\left(x - 2\right) \left({x}^{2} - 4 x + 5\right) = 0$

$x - 2 = 0$ then $x = 2$
Or
${x}^{2} - 4 x + 5 = 0$
$\delta = {\left(- 4\right)}^{2} - 4 \left(1\right) \left(5\right) = 16 - 20 = - 4 < 0$
$\delta < 0 \Rightarrow$ no root in R