How do you find all the real and complex roots of x4−2x3−4x2−8x−32=0?
1 Answer
Jan 28, 2016
Use the rational root theorem, then divide by the factors found to find the remaining roots.
x=−2 ,x=4 ,x=±2i
Explanation:
Let
By the rational root theorem, the only possible rational roots of
That means that the only possible rational roots are factors of
±1 ,±2 ,±4 ,±8 ,±16 ,±32
Trying each of these in turn, we find:
f(−2)=16+16−16+16−32=0
f(4)=256−128−64−32−32=0
So
x4−2x3−4x2−8x−32
=(x+2)(x3−4x2+4x−16)
=(x+2)(x−4)(x2+4)
=(x+2)(x−4)(x−2i)(x+2i)