Question

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

$f\left(x\right)=x^4-1.4x^3-33.71x^2+23.94x+292.41$

Short answer

The real zeros offafter rounding it two decimal places are$-3.8and4.5$.

Step-by-step solution

Step 1. Given Information 

The given function is $f\left(x\right)=x^4-1.4x^3-33.71x^2+23.94x+292.41$

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

Step 2. Determine the bounds on the zeros of f  

The leading coefficient off is$1$with$a_3=-1.4,a_2=-33.71,a_1=23.94,a_0=292.41$

Step 3. Evaluate the expressions using the formula 

So,$$\begin{gathered} Max,\{1,\left|-1.4\right|+\left|-33.71\right|+\left|23.94\right|+\left|292.41\right|\} \\ =Max(1,351.46\} \\ =351.46 \\ 1+Max\{\left|-1.4\right|,\left|-33.71\right|,\left|23.94\right|,\left|292.41\right|\} \\ =1+292.41 \\ =293.41 \end{gathered}$$

The smaller of two numbers, $293.41$is the bound.

Every real zero of flies between

Step 4. Sketch the graph of the polynomial 

From the graph, we conclude that f touches the x-axis at $-3.8and4.5$

And$f(-3.8)=0,f(4.5)=0$