Solutions to $z^3 - z^2- z =15 $ Find in the form $a+bi$, all the solutions to the equation
$$z^3 - z^2- z =15 $$
I have no idea what to do - am I meant to factor out z to get $z(z^2-z-1)=15$ or should I plug in $a+bi$ to z? 
Please help!!!!!!!  
 A: You are on the right track.
$$z^3 - z^2 - z -15 = 0$$
$$(z-3)(z^2 + 2z + 5) = 0$$
The first factor gives you a real root $r_1 = 3$
Now let's calculate the roots of the quadratic factor:
$\Delta = b^2 - 4ac = -16 = 16i^2$
Since it is negative, you know that you have 2 complex conjugate roots of the form:
$\frac{-b - \sqrt{\Delta}}{2a}$
Thus
$r_2= -1 -2i$
$r_3 = -1+2i$
A: Hint: First find the one real root, say $r$ (it's not hard). Then you can factor out $z-r$ out of $z^3-z^2-z-15=0$, and find the remaining roots using the quadratic formula.
A: Real solution:
$$z^3 - z^2- z =15<=> $$
$$z^3-z^2-z-15=0<=>$$
$$(z-3)(z^2+2z+5)=0<=>$$
So:
$$z-3=0<=>z=3$$
Or:
$$z^2+2z+5=0<=>z^2+2z=-5<=>z^2+2z+1=-4<=>(z+1)^2=-4$$
($(z+1)^2=-4$ hasno solution since for all $z$ on the real line, $(z+1)^2$ is bigger or even to zero and $-4$ is smaller than zero)
Complex solution:
$$z^3 - z^2- z =15<=> $$
$$z^3 - z^2- z -15=0<=>$$
$$(z-3)(z^2+2z+5)=0<=>$$
So:
$$z-3=0<=>z=3$$
Or:
$$z^2+2z+5=0<=>z^2+2z=-5<=>z^2+2z+1=-4<=>(z+1)^2=-4<=>$$
(the square root of both sides)
$$z+1=2i<=>z=-1+2i$$
Or:
$$z+1=-2i<=>z=-1-2i$$
