Chapter 3.4, Problem 80E

To determine

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

Answer to Problem 80E

the zeros of the polynomial is $-2,-\frac{1}{2},\frac{-5-\sqrt{17}}{2},\frac{-5+\sqrt{17}}{2}$ .

Explanation of Solution

Given information:

the polynomial is $P\left(x\right)=2x^4+15x^3+31x^2+20x+4$

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:

The constant term is $4$ and the leading coefficient is 2 so

  $\text{possible rational zero of }P=\frac{\text{factor of 4}}{\text{factor of 2 }}$

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

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

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

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

Using synthetic division,

-2	2	15	31	20	4
		-4	-22	-18	-4
	2	11	9	2	0

- $\frac{1}{2}$	2	15	31	20	4
		-1	-7	-12	-4
	2	14	24	8	0

Two zeros - $-2,-\frac{1}{2}$

Factors of the polynomial ,

  $2x^4+15x^3+31x^2+20x+4=\left(x+2\right)\left(2x+1\right)\left(x^2+5x+2\right)$

For other two zeros use quadratic formula,

  $\begin{array}{l}x=\frac{-5\pm \sqrt{5^2-4·1·2}}{2·1}\\ x=\frac{-5\pm \sqrt{25-8}}{2}\\ x=\frac{-5\pm \sqrt{17}}{2}\end{array}$

The zeros of the polynomial is $-2,-\frac{1}{2},\frac{-5-\sqrt{17}}{2},\frac{-5+\sqrt{17}}{2}$ .