Question

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

$f\left(x\right)=x^4+1.2x^3-7.46x^2-4.692x+15.2881$

Short answer

The real zeros offafter rounding it two decimal places are$-2.3and1.7.$

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)=x^4+1.2x^3-7.46x^2-4.692x+15.2881.$

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 of f is $1$ with$a_3=1.2,a_2=-7.46,a_1=-4.692,and{a}_0=15.2881.$

Step 3. Evaluate the expressions using the formula 

So,

$$\begin{gathered} Max,\{1,\left|1.2\right|+\left|-7.46\right|+\left|-4.692\right|+\left|15.2881\right|\} \\ =Max\{1,28.6401\} \\ =28.6401 \\ 1+Max\{\left|1.2\right|,\left|-7.46\right|,\left|-4.692\right|,\left|15.2881\right|\} \\ =1+15.2881 \\ =16.2881 \end{gathered}$$

The smaller of two numbers, $16.2881$is the bound.

Every real zero of f liesbetween$-16.2881and16.2881.$

Step 4. Sketch the graph of the polynomial 

From the graph, we conclude that f touches the x-axis at $1.7and-2.3.$

And$f(1.7)=0,f(-2.3)=0$