# Math Help - Finding complex roots of polynomials

1. ## Finding complex roots of polynomials

Find all solutions of the equation:

x^4 + 2x^2 - 8 = 0

I know the roots are either complex or irrational because none of the rational possibilities work out. Is there a way to find the other roots without using a graphing calculator? Thanks for any help.

2. start by saying $a = x^2$ then solve $a^2 + 2a - 8 = 0$

what do you get?

3. Let $u = x^2$

$u^2 + 2u - 8 = (u-2)(u+4) = 0$

$u = x^2 = 2 \: \implies x = \pm \sqrt{2}$

OR

$u = x^2 = -4 \: \implies x = \pm 2i \: \text{ where } i^2 = -1$

4. Also, by Descartes Rule of Signs, we know there are 1 positive and 1 negative solution; hence, we have a pair of complex numbers.

If you use the formula $\displaystyle \frac{P}{Q}$, you can use synthetic division to find your 2 real solutions and then use $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to find the complex pair.

5. Originally Posted by dwsmith
Also, by Descartes Rule of Signs, we know there are 1 positive and 1 negative solution; hence, we have a pair of complex numbers.

If you use the formula $\displaystyle \frac{P}{Q}$, you can use synthetic division to find your 2 real solutions and then use $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to find the complex pair.
Doesn't the formula P/Q only find the rational real roots and not the irrational?

6. You equation doesn't contain irrational values.

7. The equation has 2 complex roots and 2 irrational roots. I tried using P/Q and none of the rational possibilties were roots.

8. Why not tackle it head on?

Factor it: $(x^2+4)(x^2-2)$.

From that the roots pop right out: $x=\pm 2i~\&~\pm\sqrt2~.$

9. Originally Posted by Rae
The equation has 2 complex roots and 2 irrational roots. I tried using P/Q and none of the rational possibilties were roots.
You are correct. I really didn't look at the question. haha

10. Thanks, I understand how to find the solutions now.