91Ó°ÊÓ

In this case, the following information related to the polynomial is given:

Zeros: -4 (multiplicity 1), 0 (multiplicity 3), 2 (multiplicity 1); degree: 5; contains the point (-2. 64).

The general expression of a polynomial containing 3 zeros with multiplicities is

$f\left(x\right)=a{\left(x-z_1\right)}^{m_1}{\left(x-z_2\right)}^{m_2}{\left(x-z_3\right)}^{m_3}--->\left(1\right)$where

$a:$constant of proportionality

$z_1,z_2,z_3:$zeros of the polynomial

$m_1,m_2,m_3$: multiplicity of the zeros.

In this case, the zero $z_1=0$has a multiplicity of $m_1=3$and the other zeros $z_2=-4$and $z_3=2$has a multiplicity of 1.

Substituting the values into (1), we get that

$f\left(x\right)=a{x}^3\left(x+4\right)\left(x-2\right)--->\left(2\right)$

Since the polynomial contains the point (-=2, 64), by substituting $x=-2$, we get the following:

$$\begin{gathered} f\left(-2\right)=64=a{\left(-2\right)}^3\left(-2+4\right)\left(-2-2\right) \\ 64a=64 \\ a=\frac{64}{64}=\boxed{1} \end{gathered}$$

So, the expression of the polynomial will be

$f\left(x\right)=x^3\left(x+4\right)\left(x-2\right)$