# How do you find all the zeros of x^4 − 4x^3 + 14x^2 − 4x + 13 with the zero 2-3i?

Feb 27, 2016

$x = i , - i , 2 + 3 i , 2 - 3 i$

#### Explanation:

Since the function has a zero of $2 - 3 i$, and all the coefficients of the polynomial are real, you know that the complex conjugate of the zero will also be a zero.

Thus $2 - 3 i$ and $2 + 3 i$ are zeros of the polynomial.

Since $2 - 3 i$ and $2 + 3 i$ are roots, we know that two of the polynomial's factors are $\left(x - \left(2 - 3 i\right)\right)$ and $\left(x - \left(2 + 3 i\right)\right)$.

$\left(x - \left(2 - 3 i\right)\right) \left(x - \left(2 + 3 i\right)\right)$

$= \left(x - 2 + 3 i\right) \left(x - 2 - 3 i\right)$

$= \left(\left(x - 2\right) + 3 i\right) \left(\left(x - 2\right) - 3 i\right)$

$= {\left(x - 2\right)}^{2} - {\left(3 i\right)}^{2}$

$= {x}^{2} - 4 x + 4 + 9$

$= {x}^{2} - 4 x + 13$

The remaining factors of the polynomial can be found through

$\frac{{x}^{4} - 4 {x}^{3} + 14 {x}^{2} - 4 x + 13}{{x}^{2} - 4 x + 13} = {x}^{2} + 1$

Thus the remaining zeros can be solved through

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

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

$x = \pm i$

The function has four imaginary zeros and never crosses the $x$-axis.

graph{x^4-4x^3+14x^2-4x+13 [-5, 5, -21.47, 120]}