# How do you determine the remaining zeroes for g(x)=2x^5-3x^4-5x^3-15x^2-207x+108 if 3i is a zero?

Jul 4, 2015

Use conjugate zeros, factor theorem, division of polynomials, rational zeros theorem, division (again), then solve the resulting quadratic by you favorite method for quadratics.

#### Explanation:

$g \left(x\right) = 2 {x}^{5} - 3 {x}^{4} - 5 {x}^{3} - 15 {x}^{2} - 207 x + 108$

First Two Zeros
Given that $3 i$ is a zero, we observe that the coefficients of $g$ are real, so the Complex Conjugate Zero Theorem tells us that $- 3 i$ is also a zero.
The Factor Theorem tells us that $x - 3 i$ and $x + 3 i$ are factors, so their product is also a factor:
$\left(x - 3 i\right) \left(x + 3 i\right) = {x}^{2} + 9$

Divide $g \left(x\right)$ by ${x}^{2} + 9$ to get:

$2 {x}^{5} - 3 {x}^{4} - 5 {x}^{3} - 15 {x}^{2} - 207 x + 108 = \left({x}^{2} + 9\right) \left(2 {x}^{3} - 3 {x}^{2} - 23 x + 12\right)$

Third Zero

To find another zero, we need a zero of $2 {x}^{3} - 3 {x}^{2} - 23 x + 12$.

Neither of the easiest choices, $\pm 1$, are zeros. So use the Rational Zeros Theorem to search for another zero.

Possible rational zeros are: $\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12 , \pm \frac{1}{2} , \pm \frac{3}{2}$
Test until you find a zero. (Synthetic division is recommended for this testing.)
Depending on the order in which you test, you'll find a rational zero.
I found $- 3$ is a zero first. So (or really because) $x + 3$ is a factor.

$2 {x}^{3} - 3 {x}^{2} - 23 x + 12 = \left(x + 3\right) \left(2 {x}^{2} - 9 x + 4\right)$.

Last Two Zeros
Solve $2 {x}^{2} - 9 x + 4 = 0$ by whatever method you like.

$\left(2 x - 1\right) \left(x - 4\right) = 0$
So the final two zeros of $g$ are $\frac{1}{2}$ and $4$.

List of Zeros

Zeros: $3 i , - 3 i , - 3 , 4 , \frac{1}{2}$