How do you find all the real and complex roots of x4−2x3−4x2−8x−32=0?

1 Answer
George C.
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 f(x)=x4−2x3−4x2−8x−32

By the rational root theorem, the only possible rational roots of f(x)=0 are expressible in the form pq for some integers p and q where p is a divisor of the constant term −32 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational roots are factors of 32:

±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 x=−2 and x=4 are roots and f(x) has corresponding factors (x+2) and (x−4)

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)

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