# How do you find all the zeros of x^3-3x^2-x+3 with 3 as a zero?

The zeros are $x = 3$ and $x = - 1$ and $x = + 1$

#### Explanation:

By synthetic division, arrange the numerical coefficients according to degree(from highest to lowest)

${x}^{3} \text{ " " " " "x^2" " " " " "x^1" " " " " } {x}^{0}$

$1 \text{ " " "-3" " " "-1" " " " " " "3" " " " " } \lfloor \underline{3}$
underline(" " " " " " " 3" " " " " " " 0" " " " "" " -3
$1 \text{ " " " " "0" " " " " -1" " " " " " " 0" " } \leftarrow$the remainder

Our depressed equation is now

${x}^{2} + 0 \cdot x - 1 = 0$

which is reduced by 1 degree from degree 3.
Solve for the remaining roots

${x}^{2} - 1 = 0$

${x}^{2} = 1$

$\sqrt{{x}^{2}} = \pm \sqrt{1}$

$x = - 1$ and $x = + 1$

God bless...I hope the explanation is useful