Chapter 3.4, Problem 8E

To determine

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

Answer to Problem 8E

The rational zeros are $\pm 1,\pm 2,\pm 4,\pm 8,\pm 6,\pm 12,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{1}{6}$ .

Explanation of Solution

Given information:

The polynomial is $S\left(x\right)=6x^4-x^2+2x+12$ .

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 $S$ are of the form,

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

The constant term is $12$ and the leading coefficient is $6$ , so

  $\text{possible rational zero of }S=\frac{\text{factor of 12}}{\text{factor of 6}}$

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

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

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

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

  $\pm 1,\pm 2,\pm 4,\pm 8,\pm 6,\pm 12,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{1}{6}$

Hence, the all possible zeros are $\pm 1,\pm 2,\pm 4,\pm 8,\pm 6,\pm 12,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{1}{6}$ .