91Ó°ÊÓ

The following information related to the polynomial is given:

Zeros: -5 (multiplicity 2), 2 (multiplicity 1), 4 (multiplicity 1); degree: 4; contains the point (3, 128).

The general expression of a polynomial containing three zeros is given as

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

$a$is the constant of proportionality

$z_1,z_2,z_3:$are the three zeros

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

In this case, the multiplicity of the zero -5 is $m_1=2$and the remaining zeros have a multiplicity of 1.

Substituting the values into (1), we get the following equation:

$f\left(x\right)=a{\left(x+5\right)}^2\left(x-2\right)\left(x-4\right)--->\left(2\right)$

The polynomial passes through the point (3,128). Substituting $x=3$in equation (2), we get the following:

$$\begin{gathered} f\left(3\right)=a{\left(3+5\right)}^2\left(3-2\right)\left(3-4\right) \\ 128=-64a \\ a=-\frac{128}{64}=-2 \end{gathered}$$

So, expression of the polynomial will be

$f\left(x\right)=-2{\left(x+5\right)}^2\left(x-2\right)\left(x-4\right)$