Chapter 3.4, Problem 47E

To determine

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

Answer to Problem 47E

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

Explanation of Solution

Given information:

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

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 $2$ and the leading coefficient is $1$ , so

  $\text{possible rational zero of }P=\frac{\text{factor of 2}}{\text{factor of 1}}$

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

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

  $\pm \frac{1}{1},\pm \frac{2}{1}$

Simplify the fractions, so the all possible rational zeros:

  $\pm 1,\pm 2$

Using synthetic division,

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

-2	1	4	3	-2
		-2	-4	2
	1	2	-1	0

By using synthetic division, 1 is not the zero but -2 is a zero then P factors are,

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

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

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

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