Q. 58 - Precalculus Enhanced with Graphing Utilities

Question

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

$f\left(x\right)=x^3+3.2x^2-7.25x-6.3$

Step-by-Step Solution

Answer

The real zeros of f after rounding it two decimal places are $2,-4.5, and -0.7.$

Step 1. Given Information

The given function is $f\left(x\right)=x^3+3.2x^2-7.25x-6.3.$

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 $a_2=3.2, a_1=-7.25, and a_0=-6.3.$

Step 4. Evaluate the expressions using the formula

So,

$$\begin{gathered} Max,\{1,\left|3.2\right|+\left|-7.25\right|+\left|-6.3\right|\} \\ =Max\{1,16.75\} \\ =16.75 \\ 1+Max\{\left|3.2\right|,\left|-7.25\right|,\left|-6.3\right|\} \\ =1+7.25 \\ =8.25 \end{gathered}$$

The smaller of two numbers, $16.75,$ is the bound.

Every real zero of f lies between $-16.75 and 16.75.$

Step 5. Sketch the graph of the polynomial

From the graph, we conclude that f  appears to have three x-intercepts: one between $0 and -1,$ one between $-4 and -5,$ and one near $2.$

Step 6. Use the Factor Theorem

$$\begin{gathered} f\left(2\right)={\left(2\right)}^3+3.2{\left(2\right)}^2-7.25\left(2\right)-6.3 \\ f\left(2\right)=8+12.8-14.5-6.3 \\ f\left(2\right)=0 \end{gathered}$$

Therefore, $2$ is the zero of the polynomial.

Using Zero we find that the remaining zeros are $-4.5 and -0.7,$ rounded to two decimal places.