Chapter 3.4, Problem 84E

To determine

To find:the rational and irrational zeros of the polynomial.

Answer to Problem 84E

the zeros of the polynomial is $\frac{-2+\sqrt{2}}{2},\frac{-2-\sqrt{2}}{2},1,-2,\frac{3}{4}$ .

Explanation of Solution

Given information:

the polynomial is $P\left(x\right)=8x^5-14x^4-22x^3+57x^2-35x+6$

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:

The constant term is 6and the leading coefficient is 8so

  $\text{possible rational zero of }P=\frac{\text{factor of 6}}{\text{factor of 8}}$

The factors of 6are $\pm 1,\pm 2,\pm 3,\pm 6$ and the factors of 8 are $\pm 1,\pm 2,\pm 4,\pm 8$ .

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

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1}\pm \frac{6}{1},\pm \frac{1}{2},\pm \frac{2}{2},\pm \frac{3}{2}\pm \frac{6}{2},\pm \frac{1}{4},\pm \frac{2}{4},\pm \frac{3}{4}\pm \frac{6}{4},\pm \frac{1}{8},\pm \frac{2}{8},\pm \frac{3}{8}\pm \frac{6}{8}$

the possible zeros of $P$ are $\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{1}{8},\pm \frac{3}{8}$ .

Using synthetic division,

1	8	-14	-22	57	-35	6
		8	-6	-28	29	-6
	8	-6	-28	29	-6	0

-2	8	-14	-22	57	-35	6
		-16	60	-76	38	-6
	8	-30	38	-19	3	0

The zeros $1,-2,\frac{3}{4}$ .

Factors:

  $8x^5-14x^4-22x^3+57x^2-35x+6=\left(x-1\right)\left(4x-3\right)\left(x+2\right)\left(2x^2+4x+1\right)$

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

The zeros of the polynomial is $\frac{-2+\sqrt{2}}{2},\frac{-2-\sqrt{2}}{2},1,-2,\frac{3}{4}$ .