How do you find all the real and complex roots of x^4 - 2x^3 - 4x^2 - 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 f(x) = x^4-2x^3-4x^2-8x-32

By the rational root theorem, the only possible rational roots of f(x) = 0 are expressible in the form p/q 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)

x^4-2x^3-4x^2-8x-32

= (x+2)(x^3-4x^2+4x-16)

= (x+2)(x-4)(x^2+4)

= (x+2)(x-4)(x-2i)(x+2i)