Chapter 4.2, Problem 57AYU

In Problems 69-74, find the real zeros of $f$. If necessary, round to two decimal places.

$f\left(x\right)=x^3+3.2x^2-16.83x-5.31$

To determine

To find: The real zeros of $f\left(x\right)=x^3+3.2x^2-16.83x-5.31$.

Answer to Problem 57AYU

$\text{The zero is rounded to 2}\text{.99 of 2 decimal places}$.

Explanation of Solution

Given:

$f\left(x\right)=x^3+3.2x^2-16.83x-5.31$

$\text{F}$ is a polynomial function.

Step 1: The degree of the polynomial function is 3. There are at most 3 real zeros.

From Descartes’ rule of signs

$f\left(x\right)=x^3+3.2x^2-16.83x-5.31$

There will be 1 positive real zero.

$f\left(-x\right)=x^3+3.2x^2+16.83x-5.31$

There will be 2 or 0 negative real zeros.

Step 2: Since all coefficients are non integers, Rational zeros theorem does not apply.

Step 3: Determine the bounds on the zeros of $f$.

Using synthetic division, beginning with $\pm 1$,

$\text{Upper bound = 3, lower bound = 2}$

$X_{max}=3,X_{min}=2$

$X\text{-intercepts}$ are near $-\text{3}$, one between 0 and $-\text{1}$ and another between 0 and 1.

$f\left(2.99\right)={\left(2.99\right)}^3+3.2{\left(2.99\right)}^2-16.83\left(2.99\right)-5.31=-0.3017$

The zeros are between to $2.99$ and $3.00$, the zero is rounded to $2.99$ of 2 decimal places.