Question

I need help with this question please. Ignore the wording below. Fyi this is a part of a homework practice

Answer 1

So,

Given that the zeros of our polynomial function are:

$3i,2,-4$

We know that there are 2 real zeros, and 2 complex zeros. (3i and -3i).

So, what we're going to do to find the equation of this polynomial, is to multiply all zeros together, such that we obtain an expression that we can simplify. Like this:

$(x-2)(x+4)(x-3i)(x+3i)$

Now, we're going to multiply and distribute:

$\begin{gathered} (x^2+2x-8)(x^2+3xi-3xi-9i^2) \\ \to(x^2+2x-8)(x^2-9i^2) \end{gathered}$

Remember that:

$i=\sqrt[]{-1}\to i^2=-1$

So, we can rewrite:

$(x^2+2x-8)(x^2+9)$

Multiplying these terms, we got that:

$\begin{gathered} (x^2+2x-8)(x^2+9) \\ \to x^4+9x^2+2x^3+18x-8x^2-72 \\ \to x^4+2x^3+x^2+18x-72 \end{gathered}$

Therefore,

$P(x)=x^4+2x^3+x^2+18x-72$