How do you find the complex roots of 6c^3-3c^2-45c=0?

Nov 27, 2016

Remove a common factor of $3 c$ from each term.

$6 {c}^{3} - 3 {c}^{2} - 45 c = 0$

$3 c \left(2 {c}^{2} - c - 9\right) = 0$

Checking the discriminant of the resulting quadratic, $\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - 4 \left(2\right) \left(- 9\right) = 73$ we see that the quadratic will have two Real roots for a total of three Real roots (and no Complex roots).

The first zero comes from $3 c = 0$, or $c = 0$.

The second two, through the quadratic formula for $2 {c}^{2} - c - 9 = 0$ are:

$c = \frac{1 \pm \sqrt{1 - 4 \left(2\right) \left(- 9\right)}}{4} = \frac{1 \pm \sqrt{73}}{4}$

We see that our three zeros are $c = 0 , \frac{1 \pm \sqrt{73}}{4}$ and that these are all three Real zeros.