Chapter 3.4, Problem 62E

a.

To determine

To find: the all rational zeros of polynomial .

Answer to Problem 62E

The rational zeros are $2,2+\sqrt{2},2-\sqrt{2}$ .

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=-x^4+10x^2+8x-8$ .

Concept used: rational zero theorem,

  $\text{possible rational zero of }P=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}$

Quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

By the rational zeros theorem the rational zeros of $P$ are of the form,

  $\text{possible rational zero of }P=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}$

The constant term is $8$ and the leading coefficient is 1 so

  $\text{possible rational zero of }P=\frac{\text{factor of 8}}{\text{factor of 1}}$

The factors of $8$ are $\pm 1,\pm 2,\pm 4,\pm 8$ and the factors of 1are $\pm 1$ .

Thus, the possible rational zeros of $P$ are,

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1},\pm \frac{8}{1}$

Simplify the fractions, so the all possible rational zero

  $\pm 1,\pm 2,\pm 4,\pm 8$

Using synthetic division,

-2	-1	0	10	8	-8
		2	-4	-12	8
	-1	2	6	-4	0

Factors :

  $-x^4+10x^2+8x-8={\left(x+2\right)}^2\left(-x^2+4x-2\right)$

Use quadratic formula for other zeros,

  $\begin{array}{l}x=\frac{-4\pm \sqrt{4^2-4·\left(-1\right)·\left(-2\right)}}{2\left(-1\right)}\\ x=\frac{-4\pm \sqrt{16-8}}{-2}\\ x=2\mp \sqrt{2}\end{array}$

The rational zeros are $2,2+\sqrt{2},2-\sqrt{2}$

b.

To determine

To graph: the polynomial.

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=-x^4+10x^2+8x-8$ .

Graph:

  

Interpretation:

From the above graph it is observed that the zeros of polynomial is $2,2+\sqrt{2},2-\sqrt{2}$ .