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

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.

Ithink its 126 hope its

Option a : week- amuont paid