Chapter 6.1, Problem 67E

To determine

To find: integral upper bound and lower bound of the zeros of $f\left(x\right)$ by using upper bound and lower bound theorem respectively.

Answer to Problem 67E

Integrals upper bound: $5\text{ and 10}$

Integrals lower bound: $-5\text{ and }-\text{10}$

Explanation of Solution

Given information:

  $f\left(x\right)=x^4-3x^3-2x^2+6x+10$

Calculation:

According to the rational root theorem, $\pm 1,\pm 2,\pm 5$ and $\pm 10$ might be the roots of $f\left(x\right)$ .

Apply synthetic division on $f\left(x\right)$ for different values of roots $\left(r\right)$ as shown below.

$f\left(x\right)=x^4-3x^3-2x^2+6x+10$
$r$	1	$-3$	$-2$	6	10
1	1	$-2$	$-4$	2	12
2	1	$-1$	$-4$	$-2$	6
5	1	2	8	46	240
10	1	7	68	686	6870

Since $10>0$ and all of the signs in the bottom row of the synthetic division are positive. Similarly $5>0$ and all of the signs in the bottom row of the synthetic division are positive. Thus, the upper bound integrals of $f\left(x\right)$ are 5 and 10.

$f\left(-x\right)=x^4+3x^3-2x^2-6x+10$
$r$	1	3	$-2$	$-6$	10
1	1	4	2	$-4$	6
2	1	5	8	10	30
5	1	8	38	184	930
10	1	13	128	1274	12750

The upper bound integrals of $f\left(-x\right)$ are 10 and 5. So the lower bound integrals of $f\left(x\right)$ are $-10$ and $-5$ .