# How do you find all zeros of F(x)= x^3 -8x^2 +25x -26?

Sep 19, 2015

There are three solutions:
${x}_{0} = 2$
${x}_{1} = 3 + 2 i$
${x}_{2} = 3 - 2 i$

#### Explanation:

The rational root theorem tells us that rational roots to a polynomial equation with integer coefficients can be written in the form $\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.

The polynomial equation is $1 \cdot {x}^{3} - 8 {x}^{2} + 25 x - 26 = 0$.
The integer factors of the constant $- 26$ are $\pm 26 , \pm 13 , \pm 2 , \pm 1$.
The integer factors of the leading coefficient $1$ are $\pm 1$.

The rational solutions must therefore be among the following: $\pm \frac{26}{1} , \pm \frac{13}{1} , \pm \frac{2}{1} , \pm \frac{1}{1}$

Try each answer by substituting them into $x$ in the equation until you find the solution. You will find that $x = 2$ solves the equation. This is the first solution.

Now we know that $\left(x - 2\right)$ is a factor of the polynomial, meaning that ${x}^{3} - 8 {x}^{2} + 25 x - 26 = \left(x - 2\right) \cdot g \left(x\right)$ where $g \left(x\right)$ is some polynomial of a lower degree. Use polynomial division to find:

 g(x) = (x^3 - 8x^2 + 25x - 26) / (x-2)=x^2−6x+13

(Polynomial divison might need its own explanation for those who haven't learned it. If there's a separate Socratic thread on it, it should be linked here.)

With the factorized polynomial (x-2)(x^2−6x+13) = 0, now solve for $x$ when (x^2−6x+13) = 0

A quadratic equation $a {x}^{2} + b x + c = 0$ can be solved using the quadratic formula: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

Plug $a = 1$, $b = - 6$ and $c = 13$ into it and we get:

$x = 3 \pm \sqrt{- 4}$
$= 3 \pm \sqrt{4 \cdot - 1}$
$= 3 \pm \sqrt{4} \sqrt{- 1}$
$= 3 \pm 2 i$

Now we have found all three solutions:
${x}_{0} = 2$
${x}_{1} = 3 + 2 i$
${x}_{2} = 3 - 2 i$