Chapter 3.4, Problem 63E

a.

To determine

To find: the all rational zeros of polynomial .

Answer to Problem 63E

The rational zeros are $-1,2,2+\sqrt{3},2-\sqrt{3}$ .

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^5-x^4-5x^3+x^2+8x+4$ .

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 $4$ and the leading coefficient is 1 so

  $\text{possible rational zero of }P=\frac{\text{factor of 4}}{\text{factor of 1}}$

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

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

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

Simplify the fractions, so the all possible rational zero

  $\pm 1,\pm 2,\pm 4$

Using synthetic division,

-1	1	-1	-5	1	8	4
		-1	2	3	-4	-4
	1	-2	-3	4	4	0

2	1	-1	-5	1	8	4
		2	2	-6	-10	-4
	1	1	-3	-5	-2	0

Factors :

  $x^5-x^4-5x^3+x^2+8x+4={\left(x+2\right)}^2\left(x+1\right)\left(x^2-4x+1\right)$

Use quadratic formula for other zeros,

  $\begin{array}{l}x=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4·\left(1\right)·\left(1\right)}}{2\left(1\right)}\\ x=\frac{4\pm \sqrt{16-4}}{2}\\ x=2\pm \sqrt{3}\end{array}$

The rational zeros are $-1,2,2+\sqrt{3},2-\sqrt{3}$

b.

To determine

To graph: the polynomial.

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^5-x^4-5x^3+x^2+8x+4$ .

Graph:

  

Interpretation:

From the above graph it is observed that the zeros of polynomial is $-1,2,2+\sqrt{3},2-\sqrt{3}$ .