Question

Find all the zeros of the function.

f(x)=x^4+4x^3-21x^2

Answer 1

using synthetic division? @KellyN

Answer 2

Factor

Answer 3

\(f(x) = x^4+4x^3-21x^2 = x^2\left(x^2+4x-21\right)\). Now set this equal to 0.
\(x^2\left(x^2+4x-21\right) = 0\). So \(x^2 = 0 \) and \(x^2-4x-21 = 0\).
\(x^2 = 0\) -> x = 0 multiplicity 2
\(x^2-4x-21 = 0\) -> \(\left(x+7\right)\left(x-3\right) = 0\) -> x = -7, x = 3.
So the zeros of f(x) are 0, 0, -7, and 3.