# How do you find all the zeros of f(x)=x^4-8x^3+17x^2-8x+16 ?

May 1, 2016

Zeros at $x \in \left\{4 , i , - i\right\}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{1} {x}^{4} - 8 {x}^{3} + 17 {x}^{2} - 8 x + \textcolor{b l u e}{16}$

By the rational root theorem, we know that any rational roots must be factors of $\frac{\textcolor{b l u e}{16}}{\textcolor{red}{1}} = 16$
which implies they must be in the set $\left\{\pm 1 , \pm 2 , \pm 4 , \pm 16\right\}$

Testing each of these possibilities give a zero at (an only at) $x = 4$

If we divide ${x}^{4} - 8 {x}^{3} + 17 {x}^{2} - 8 x + 15$ by $\left(x - 4\right)$ (using synthetic or long polynomial division)
we get factors $\left(x - 4\right) \left({x}^{3} - 4 {x}^{2} + x - 4\right)$

Again, applying the factor theorem of x^3-4x^2+x-4)
we find the only zero at $x = 4$ (again).

Dividing ${x}^{3} - 4 {x}^{2} + x - 4$ by $\left(x - 4\right)$
we can re-factor as $\left(x - 4\right) \left(x - 4\right) \left({x}^{2} + 1\right)$

$\left({x}^{2} + 1\right)$ has no Real zeros but can be factored as $\left(x + i\right) \left(x - i\right)$ to get complex zeros at $\pm i$