Chapter 3, Problem 37RE

a.

To determine

To find: the all rational zeros of polynomial .

Answer to Problem 37RE

The rational zeros are $\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$ .

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^5-6x^3-x^2+2x+18$ .

Concept used: rational zero theorem,

  $\text{possible rational zero of }P=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}$

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 $18$ and the leading coefficient is $1$ , so

  $\text{possible rational zero of }P=\frac{\text{factor of 18}}{\text{factor of 1}}$

The factors of $18$ are $\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$ and the factors of 1are $\pm 1$ .

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

$\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{6}{1},\pm \frac{9}{1},\pm \frac{18}{1}$

Simplify the fractions, so the all possible rational zero

$\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$

The rational zeros are $\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$

b.

To determine

To find: the all positive and negative real zeros using Descartes rule of signs.

Answer to Problem 37RE

The polynomial has either two or 0 positive and either three or one negative zeros and total number of real zeros either five or one.

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^5-6x^3-x^2+2x+18$ .

Concept used: Descartes’s rule of signs:

Let P be a polynomial with real coef?cients.

1. The number of positive real zeros of $p\left(x\right)$ either is equal to the number of variations in sign in $p\left(x\right)$ or is less than that by an even whole number.

2. The number of negative real zeros of $p\left(x\right)$ either is equal to the number of variations in sign in $p\left(-x\right)$ or is less than that by an even whole number.

Calculation:

The polynomial has two variation in sign, so it has either two or 0 positive zero, now

  $\begin{array}{l}P\left(-x\right)={\left(-x\right)}^5-6{\left(-x\right)}^3-{\left(-x\right)}^2+2\left(-x\right)+18\\ P\left(-x\right)=-x^5+6x^3-x^2-2x+18\end{array}$

So, $p\left(-x\right)$ has three variations in sign.

Thus, $p\left(x\right)$ has either three or one negative zeros. Making a total of either five or one real zeros.