# How do you find all zeros of the function P(x)=2x^3-5x^2+6x-2 given the zero 1+i?

Jul 16, 2016

Zeros: $1 + i$, $1 - i$, $\frac{1}{2}$

#### Explanation:

$P \left(x\right) = 2 {x}^{3} - 5 {x}^{2} + 6 x - 2$

We are told that $x = 1 + i$ is a zero.

Since the coefficients of $P \left(x\right)$ are all Real, the Complex conjugate $1 - i$ must be a zero too. So $P \left(x\right)$ must be divisible by:

$\left(x - \left(1 - i\right)\right) \left(x - \left(1 + i\right)\right)$

$= \left(\left(x - 1\right) - i\right) \left(\left(x - 1\right) + i\right)$

$= {\left(x - 1\right)}^{2} - {i}^{2}$

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

We find:

$2 {x}^{3} - 5 {x}^{2} + 6 x - 2 = \left({x}^{2} - 2 x + 2\right) \left(2 x - 1\right)$

So the remaining zero is $x = \frac{1}{2}$