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saveliy_v [14]
3 years ago
5

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

Mathematics
1 answer:
IgorC [24]3 years ago
4 0

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).

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