How do you find all the real and complex roots of x42x34x28x32=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)=x42x34x28x32

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+1616+1632=0

f(4)=256128643232=0

So x=2 and x=4 are roots and f(x) has corresponding factors (x+2) and (x4)

x42x34x28x32

=(x+2)(x34x2+4x16)

=(x+2)(x4)(x2+4)

=(x+2)(x4)(x2i)(x+2i)