Question

In Problems 41– 48, form a polynomial function whose real zeros and degree are given. Answers will vary depending on the choice of a leading coefficient.

$Zeros:-4,0,2,degree3$

Short answer

The polynomial function is $f\left(x\right)=x^3+2x^2-8x$with degree 3.

Step-by-step solution

Step 1. Given Information

The zeros of the polynomial are$-4,0,2$and degree is 3

Step 2. Forming the polynomial function

If $r$is a real zero of polynomial function, then $x-r$is factor of $f.$This means that

$x-\left(-4\right)=x+4$and $\left(x-0\right),\left(x-2\right)$are factors of $f.$ As a result, any polynomial function is of the form$$\begin{gathered} f\left(x\right)=a\left(x+4\right)\left(x-0\right)\left(x-2\right) \\ f\left(x\right)=x\left(x+4\right)\left(x-2\right) \end{gathered}$$

Step 3. Solving Polynomial Function

$$\begin{gathered} f\left(x\right)=x\left(x+4\right)\left(x-2\right) \\ f\left(x\right)=x\left(x^2+2x-8\right) \\ f\left(x\right)=x^3+2x^2-8x \end{gathered}$$