## anonymous 5 years ago find all real and imaginary solutions to w^4+216w=0

1. anonymous

$w^4+216w=0$ $w(w^3+216)=0$ one solution is 0 and 216 is the cube of 6, so another solution is -6

2. anonymous

now factor $w^3+216$ as the sum of two cubes. it is $(w+6)(w^2-6w+36)$

3. anonymous

the first factor we know give a zero of -6 and the other two we find by using the quadratic formula unless you are supposed to know how to fine the cube roots of -1 using polar form of complex numbers. if not just use the formula.

4. anonymous

$w^2-6w+36=0$ $w=\frac{-6+\sqrt{6^2-4\times 36}}{2}$ $w=\frac{-6+\sqrt{-108}}{2}$ $w=\frac{-6+6\sqrt{3}i}{2}$ $w=-3+3\sqrt{3}i$ and of course $w=-3-3\sqrt{3}i$