Chapter 3, Problem 38RE

a.

To determine

To find: the all rational zeros of polynomial .

Answer to Problem 38RE

The rational zeros are $\pm 1,\pm 2,\pm 4,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{1}{6}$ .

Explanation of Solution

Given information:

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

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 $4$ and the leading coefficient is $6$ , so

  $\text{possible rational zero of }P=\frac{\text{factor of 4}}{\text{factor of 6}}$

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

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

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

Simplify the fractions, so the all possible rational zero

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

The rational zeros are $\pm 1,\pm 2,\pm 4,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{1}{6}$

b.

To determine

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

Answer to Problem 38RE

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

Explanation of Solution

Given information:

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

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 no variation in sign, so it has 0 positive zero, now

  $\begin{array}{l}P\left(-x\right)=6{\left(-x\right)}^4+3{\left(-x\right)}^3+{\left(-x\right)}^2+3\left(-x\right)+4\\ P\left(-x\right)=6x^4-3x^3+x^2-3x+4\end{array}$

So, $p\left(-x\right)$ hasfourvariations in sign.

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