Question

How do you write a polynomial with Zeros: $-2$, multiplicity $2$; $4$, multiplicity $1$; degree $3$?

Answer 1

Evaluate the polynomial:

Given,

We have two unique zeros: $-2$ and $4$.

We know that,

For a polynomial, if $x=a$ is a zero of the function, then $(x-a)$ is a factor of the function.

So,

$$\begin{gathered} \Rightarrow \left(x-(-2)\right)=x+2 \\ \Rightarrow \left(x-\left(4\right)\right)=x-4 \end{gathered}$$

However, $-2$ has a multiplicity of $2$ , [given]

This means that the factor that correlates to a zero of $-2$ is represented in the polynomial twice.

$\Rightarrow \left(x+2\right)\left(x+2\right)\to \left(1\right)$

Similarly,

$4$, multiplicity $1$;

This means that the factor that correlates to a zero of 4 is represented in the polynomial once

$\Rightarrow x-4\to \left(2\right)$

The required polynomial $p\left(x\right)$=product of equation $\left(1\right)\text{and}\left(2\right)$

$$\begin{gathered} p\left(x\right)=(x+2)\left(x+2\right)(x-4) \\ \Rightarrow p\left(x\right)={(x+2)}^2(x-4) \\ \Rightarrow p\left(x\right)=(x^2+2x+4)(x-4)\left[\because {\left(a+b\right)}^2=a^2+2ab+b^2\right] \\ \Rightarrow p\left(x\right)=x^3-12x-16 \end{gathered}$$

Hence, the required polynomial $p\left(x\right)=x^3-12x-16$.