Chapter 3.4, Problem 52E

To determine

To find: the all rational zeros of polynomial and using the quadratic formula.

Answer to Problem 52E

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

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^5-4x^4-x^3+10x^2+2x-4$ .

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 $1$ are $\pm 1$ .

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

2	1	-4	-1	10	2	-4
		2	-4	-10	0	4
	1	-2	-5	0	2	0

By using synthetic division, 2,-1 is a zero but graph of this polynomial get a curve on -1 so its two zeros are same then P factors are,

  $x^5-4x^4-x^3+10x^2+2x-4=\left(x+1\right)\left(x+1\right)\left(x-2\right)\left(x^2+4x+1\right)$

Now we use the quadratic formula to obtain the two remaining zeros,

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

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