# Question #ba224

May 2, 2017

two real roots $x = 3 \mathmr{and} 2$
two imaginary roots $x = \pm i$

#### Explanation:

before searching for the zeros in f(x)
we need to know there are two conditions
1. four real roots
2.two real roots & two imaginary roots
imaginary roots always be pair!

graph{y=x^4-5x^3+7x^2-5x+6 [-8.89, 8.89, -4.44, 4.45]}
let $f \left(x\right) = {x}^{4} - 5 {x}^{3} + 7 {x}^{2} - 5 x + 6$
if x=3 $f \left(3\right) = 81 - 135 + 63 - 15 + 6 = 0$
so $f \left(x\right)$can be divided by $\left(x - 3\right)$

$\frac{{x}^{4} - 5 {x}^{3} + 7 {x}^{2} - 5 x + 6}{x - 3} = {x}^{3} - 2 {x}^{2} + x - 2$

the next step also use by the same method

graph{y=x^3-2x^2+x-2 [-8.89, 8.89, -4.44, 4.45]}
let$g \left(x\right) = {x}^{3} - 2 {x}^{2} + x - 2$
if $x = 2$
$g \left(x\right) = 8 - 8 + 2 - 2 = 0$

$g \frac{x}{x - 2} = {x}^{2} + 1$
exist two imaginary roots $\pm i$ in the equation ${x}^{2} + 1$