# How do you find all the zeros of x^3-6x^2+13x-10 with its multiplicities?

Apr 17, 2016

$2 , 2 \pm i$

#### Explanation:

Explanation:

We will use the Rational Root Theorem:

If the rational number r/s is a root of a polynomial whose coefficients are integers, then the integer r is a factor of the constant term, and the integer s is a factor of the leading coefficient.

So, the candidates for roots are:

$\pm 1 , \pm 2 , \pm 5 , \pm 10$

We discover than 2 is a root.

Now we divide the polymorph by (x-2) to discover the other roots:

$\frac{{x}^{3} - 6 {x}^{2} + 13 x - 10}{x - 2} = {x}^{2} + \frac{- 4 {x}^{2} + 13 x - 10}{x - 2}$
$= {x}^{2} - 4 x + \frac{5 x - 10}{x - 2} = {x}^{2} - 4 x + 5$

This second degree polynom is solved with the quadratic formula:

$x = \frac{4 \pm \sqrt{{4}^{2} - 4 \cdot 1 \cdot 5}}{2} = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{- 4}}{2} = \frac{4 \pm 2 i}{2} = 2 \pm i$