Question

Which polynomial will definitely have nonreal zeros?

Hint: Use Descartes's rule of signs.
x3 − 2x2 − x − 2

x3 − 2x2 + x − 2

x4 + 8x2 + 16


x4 − 8x2 + 16

Answer 1

The answer is Letter C - the third option = x4 + 8x2 + 16

Descartes' Rule of Signs is a useful help for finding the zeroes of a polynomial without the use of graphs.

x^4 + 8x^2 + 16 = (x^2 + 4)^2
Solve by square rooting both sides.
x² + 4 = 0
x² = -4
x = ±√-4
x = ±√4i²
x = ±2i
x = 2i or x = -2i

or

f(x) = x^4+8x^2+16 : no changes

f(-x) = (-x)^4 + (-x)^2 + 16
f(-x) = x^4+x^2+16 : no changes

Other choices have changes in their signs, the third option has no changes in sign and no real zero. Using Descartes's rule of signs, x^4 + 8x^2 + 16 will definitely have nonreal zeros.