Chapter 3.4, Problem 49E

To determine

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

Answer to Problem 49E

The rational zeros are $-1,4,\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}$ .

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^4-6x^3+4x^2+15x+4$ .

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	-6	4	15	4
		-1	7	-11	-4
	1	-7	11	4	0

4	1	-6	4	15	4
		4	-8	-16	-4
	1	-2	-4	-1	0

By using synthetic division, -1,4 is a zero then P factors are,

  $x^4-6x^3+4x^2+15x+4=\left(x+1\right)\left(x-4\right)\left(x^2-3x-1\right)$

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

  $\begin{array}{l}x=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4·1·\left(-1\right)}}{2·1}\\ x=\frac{3\pm \sqrt{9+4}}{2}\\ x=\frac{3\pm \sqrt{13}}{2}\end{array}$

The rational zeros are $-1,4,\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}$