Question

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

$f\left(x\right)=x^3+19.5x^2-1021x+1000.5$

Short answer

The real zeros offafter rounding it two decimal places are$1,23,and-43.5.$

Step-by-step solution

Step 1. Given Information 

The given function is$f\left(x\right)=x^3+19.5x^2-1021x+1000.5$

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

Step 2. Finding the real zeros 

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

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

Step 3. Determine the bounds on the zeros of f 

The leading coefficient of f is$1$ with$a_2=19.5,a_1=-1021,a_0=1000.5$

Step 4. Evaluate the expressions using the formula 

So,

$$\begin{gathered} Max,\{1,\left|19.5\right|+\left|-1021\right|+\left|1000.5\right|\} \\ =Max,\{1,2041\} \\ =2041 \\ 1+Max,\{\left|19.5\right|,\left|-1021\right|,\left|1000.5\right|\} \\ =1+1021 \\ =1022 \end{gathered}$$

The smaller of two numbers, $1022$is the bound.

Every real zero of f liesbetween$-1022and1022.$

Step 5. Sketch the graph of the polynomial 

From the graph, we conclude that f appears to have three x-intercepts: $1,23,and-43.5.$

And

$$\begin{gathered} f\left(1\right)=0 \\ f\left(23\right)=0 \\ f(-43.5)=0 \end{gathered}$$