In Problems 57– 62, find the real zeros of f. If necessary, round to two decimal places.

$f\left(x\right)=x^3+3.2x^2-16.83x-5.31$

The given function is $f\left(x\right)=x^3+3.2x^2-16.83x-5.31.$

We have to find the real zeros and round them to two decimal places.

The degree of the polynomial is $3$, thus it has three roots.

Since there are noninteger coefficients, the Rational Zeros Theorem doesn't apply.

The leading coefficient offis$1$with$a_2=3.2,a_1=-16.83and{a}_0=-5.31.$

So,

$$\begin{gathered} Max\{1,\left|-5.31\right|+\left|-16.83\right|+\left|3.2\right|\} \\ =Max\{1,25.34\} \\ =25.34 \\ 1+Max\{\left|-5.31\right|,\left|-16.83\right|,\left|3.2\right|\} \\ =1+16.83 \\ =17.83 \end{gathered}$$

The smaller of two numbers, $17.83,$is the bound.

Every real zero of f liesbetween$-17.83and17.83.$

From the graph, we conclude that f

appears to have three x-intercepts: one between$-5and-6$, one between$0and1,$and one near$3.$

$$\begin{gathered} f\left(3\right)=3^3+3.2{\left(3\right)}^2-16.83\left(3\right)-5.31 \\ f\left(3\right)=27+28.8-50.49-5.31 \\ f\left(3\right)=0 \end{gathered}$$

Therefore, $3$is the zero of the polynomial.

Using Zero we find that the remaining zeros arerole="math" localid="1646122746338" $-5.90and-0.30,$rounded to two decimal places.