Chapter 3.4, Problem 57E

a.

To determine

To find: the all rational zeros of polynomial .

Answer to Problem 57E

The rational zeros are $-2,2,3,$ .

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^3-3x^2-4x+12$ .

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 $12$ and the leading coefficient is $1$ , so

  $\text{possible rational zero of }P=\frac{\text{factor of 12}}{\text{factor of 1}}$

The factors of $12$ are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$ and the factors of 1 are $\pm 1$ .

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

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{4}{1},\pm \frac{6}{1},\pm \frac{12}{1}$

Simplify the fractions, so the all possible rational zero

  $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$

Using synthetic division,

3	1	-3	-4	12
		3	0	-12
	1	0	-4	0

2	1	-3	-4	12
		2	-2	-12
	1	-1	-6	0

-2	1	-3	-4	12
		-2	10	-12
	1	-5	6	0

By using synthetic division, 2, -2 , 3 is a zero then P factors are,

  $x^3-3x^2-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)$

The rational zeros are $-2,2,3,$

b.

To determine

To graph: the polynomial.

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^3-3x^2-4x+12$ .

Graph:

  

Interpretation:

From the above graph it is observed that the zeros of polynomial is $-2,2,3,$ .