Question

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

$f\left(x\right)=x^3+42.2x^2-664.8x+1490.4$

Short answer

The real zeros offafter rounding it two decimal places are$9.85,-54.82,and2.75.$

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)=x^3+42.2x^2-664.8x+1490.4$

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$a_2=42.2,a_1=-664.8,and{a}_0=1490.4$

Step 4. Evaluate the expressions using the formula 

So,

$$\begin{gathered} Max,\{1,\left|42.2\right|+\left|-664.8\right|+\left|1490.4\right|\} \\ =Max,\{1,2197.4\} \\ =2197.4 \\ 1+Max,\{\left|42.2\right|,\left|-664.8\right|,\left|1490.4\right|\} \\ =1+1490.4 \\ =1491.4 \end{gathered}$$

The smaller of two numbers, $1491.4$is the bound.

Every real zero of f liesbetween$-1491.4and1491.4$.

Step 5. Sketch the graph of the polynomial 

From the graph, we conclude that f appears to have three x-intercepts: $9.85,-54.82,and2.75.$