Question

find all real and imaginary solutions to w^4+216w=0

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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