Chapter 3.4, Problem 7E

To determine

To find:the all possible rational zeros by rational zeros theorem.

Answer to Problem 7E

The rational zeros are $\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2}$ .

Explanation of Solution

Given information:

The polynomial is $R\left(x\right)=2x^5+3x^3+4x^2-8$ .

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 $R$ are of the form,

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

The constant term is $8$ and the leading coefficient is $2$ , so

  $\text{possible rational zero of }R=\frac{\text{factor of 8}}{\text{factor of 2}}$

The factors of $8$ are $\pm 1,\pm 2,\pm 4,\pm 8$ and the factors of $2$ are $\pm 1,\pm 2$ .

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

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1},\pm \frac{8}{1},\pm \frac{1}{2},\pm \frac{2}{2},\pm \frac{4}{2},\pm \frac{8}{2}$

Simplify the fractions and eliminating duplicates, so the all possible rational zeros:

  $\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2}$

Hence, the all possible zeros are $\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2}$ .