Question

In the third degree polynomial function f(x) =x^3 + 4x, why are 2i and -2i zeros?

Answer 1

Answer:

We just need to evaluate and get f(2i)=0, f(-2i)=0.

Step-by-step explanation:

Since $i^2=-1$, then $i^3=i^2i=-i$, and we can apply this when we evaluate $f(x) =x^3 + 4x$ for 2i and -2i.

First we have:

$f(2i) =(2i)^3 + 4(2i)=2^3i^3+8i=8(-i)+8i=0$

Which shows that 2i is a zero of f(x).

Then we have:

$f(-2i) =(-2i)^3 + 4(-2i)=(-2)^3i^3-8i=-8(-i)-8i=8i-8i=0$

Which shows that -2i is a zero of f(x).