# How do you find all the zeros of P(x) = 6x^3 − 13x^2 + 14x − 2 with 1+i as a root?

Jun 12, 2016

All the zeros of the given polynomial $P \left(x\right) = 6 \cdot {x}^{3} - 13 \cdot {x}^{2} + 14 \cdot x - 2$
are $\frac{1}{6} , 1 + i , 1 - i$.

#### Explanation:

Let us remember that the complex roots occur in conjugate pairs. Now one such root is given to be$1 + i$, so, the root has to be $1 - i$.

Since $P \left(x\right)$ is a cubic polynomial (poly.), $P \left(x\right)$ must have 3 roots, out which 2 roots are now known to us, namely, $1 \pm i$.

Let the 3rd root be $\alpha$.

Now, by Vieta's Rule, $\alpha$+$\left(1 + i\right)$+$\left(1 - i\right)$ = $- \left(- \frac{13}{6}\right)$ = $\frac{13}{6.}$
$\alpha$$\left(1 + i\right)$+$\left(1 + i\right)$$\left(1 - i\right)$+$\alpha$$\left(1 - i\right)$ = $\frac{14}{6.}$
$\alpha$$\left(1 + i\right)$$\left(1 - i\right)$=$- \left(- \frac{2}{6}\right)$ = $\frac{2}{6.}$

From the 1st equation (eqn.), we get $\alpha$ = $\frac{1}{6.}$
We verify that this value of $\alpha$ satisfy the remaining 2 eqns.
Hence the roots $P \left(x\right)$ are $\frac{1}{6} , 1 + i , 1 - i$.