Chapter 4.2, Problem 36AYU

In Problems 45-50, find the bounds to the zeros of each polynomial function. Use the bounds to obtain a complete graph of f .

48. $f\left(x\right)\text{ = 2}{x}^3\text{ - }{x}^2\text{ - 11}x\text{ - 6}$

To determine

To find: The bounds to zeros of each polynomial function to complete the graph of $f$.

Answer to Problem 36AYU

Here $r=\text{ 3}$, the third row of synthetic division contains only numbers that are positive. Therefore $r=\text{ 3}$ is the upper bound. There can be no zeros greater than 3.

Here $r=-2$, the third row of synthetic division contains only numbers that are alternating positive. Therefore $r=-2$ is the lower bound. There can be no zeros lesser than $-3$.

Explanation of Solution

Given:

$f\left(x\right)=\text{ 2}{x}^3-x^2-11x-6$

The degree of the polynomial is 3. Therefore the number of real zeros by real zero theorem can be at most $=3$.

$f\left(x\right)=\text{ 2}{x}^3-x^2-11x-6$

Rational zeros theorem provides information about the potential rational zeros of a polynomial function with integer coefficients.

If $\frac{p}{q}$ in its lowest terms is a rational zero of $f$, then $p$ is a factor of $a_0$ and $q$ is the factor of $a_n$.

$f\left(x\right)=\text{ 2}{x}^3-x^2-11x-6$

Here $a_0=\text{ −6}$ and $a_n=2$.

Zeros of 2, $q=\pm 1,\pm 2$.

Zeros of $-6$, $p=\pm 1,\pm 2,\pm 3,\pm 6$.

The potential rational zeros of $f=\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2}$.

To find the upper bound, start with the smallest positive integer potential rational zero $=1$.

To find the lower bound start the largest negative integer potential rational zero $=-1$.

Use repeated Synthetic division,

$f\left(x\right)=\text{ 2}{x}^3-x^2-11x-6$

The coefficients are 2, $-1$, $-11$ and $-6$.

$r$	Coefficient of $q\left(x\right)$	Remainder
1	2, 1, $-10$	$-16$
2	2, 3, $-5$	$-16$
3	2, 5, 4	6
$-1$	2, $-3$, $-8$	2
$-2$	2, $-5$, $-1$	$-4$

Here $r=3$, the third row of synthetic division contains only numbers that are positive. Therefore $r=3$ is the upper bound. There can be no zeros greater than 3.

Here $r=-2$, the third row of synthetic division contains only numbers that are alternating positive. Therefore $r=-2$ is the lower bound. There can be no zeros lesser than $-2$.