Chapter 4.2, Problem 35AYU

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

47. $f\left(x\right)\text{ = }{x}^3\text{ - 5}{x}^2\text{ - 11}x\text{ + 11}$

To determine

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

Answer to Problem 35AYU

Here $r=\text{ 7}$, the third row of synthetic division contains only numbers that are positive. Therefore $r=\text{ 7}$ is the upper bound. There can be no zeros greater than 7.

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

Explanation of Solution

Given:

$f\left(x\right)=x^3-5x^2-11x+11$

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

$f\left(x\right)=x^3-5x^2-11x+11$

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 $a_n$.

$f\left(x\right)=x^3-5x^2-11x+11$

Here $a_0=\text{ 11}$ and $a_n=1$.

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

Zeros of $11$, $p=\pm 1,\pm 11$.

The potential rational zeros of $f=\pm 1,\pm 11$.

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,

$f\left(x\right)=x^3-5x^2-11x+11$

The coefficients are 1, $-5$, $-11$ and 11.

$r$	Coefficient of $q\left(x\right)$	Remainder
1	1, $-4$, $-15$	$-4$
2	1, $-3$, $-17$	$-23$
3	1, $-2$, $-17$	$-40$
4	1, $-1$, $-15$	$-49$
5	1, 0, $-11$	$-44$
6	1, 1, $-5$	$-19$
7	1, 2, 3	32
$-1$	1, $-6$, $-5$	16
$-2$	1, $-7$, 3	5
$-3$	1, $-8$, 13	$-28$

Here $r=7$, the third row of synthetic division contains only numbers that are positive. Therefore $r=7$ is the upper bound. There can be no zeros greater than 7.

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