Chapter 3.4, Problem 11E

a.

To determine

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

Answer to Problem 11E

The rational zeros are $\pm 1,\pm \frac{1}{5}$ .

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=5x^3-x^2-5x+1$ .

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 $1$ and the leading coefficient is $5$ , so

  $\text{possible rational zero of }P=\frac{\text{factor of 1}}{\text{factor of 5}}$

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

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

  $\pm \frac{1}{1},\pm \frac{1}{5}$

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

  $\pm 1,\pm \frac{1}{5}$

Hence, the all possible zeros are $\pm 1,\pm \frac{1}{5}$ .

b.

To determine

To determine the possible rational zeros which actually turns out to be zero.

Answer to Problem 11E

The possible rational zeros which actually turns out to be zero are $\pm 1,+\frac{1}{5}$ .

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=5x^3-x^2-5x+1$ .

  

From the above graph it is observed that the curve of graph cut the x-axis at the $\pm 1,+\frac{1}{5}$ , so the rational zeros which actually turns out to zero are $\pm 1,+\frac{1}{5}$ .