Chapter 3, Problem 61RE

To determine

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

Answer to Problem 61RE

the zeros of the polynomial is $-3,1,5$ (all zeros has one multiplicities).

Explanation of Solution

Given information:

the polynomial is $P\left(x\right)=x^3-3x^2-13x+15$

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 $15$ and the leading coefficient is 1 so

  $\text{possible rational zero of }P=\frac{\text{factor of 15}}{\text{factor of 1}}$

The factors of $15$ are $\pm 1,\pm 3,\pm 5,\pm 15$ and the factors of 1 are $\pm 1$ .

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

  $\pm \frac{1}{1},\pm \frac{3}{1},\pm \frac{5}{1},\pm \frac{15}{1}$

the possible zeros of $P$ are $\pm 1,\pm 3,\pm 5,\pm 15$ .

Using synthetic division,

-3	1	-3	-13	15
		-3	18	-15
	1	-6	5	0

1	1	-3	--13	15
		1	-2	-1
	1	-2	-15	0

5	1	-3	-13	15
		5	10	-15
	1	2	-3	0

The zeros of the polynomial is $-3,1,5$ .