Chapter 3.4, Problem 33E

To determine

To find: the all rational zeros of polynomial and write the polynomial in factored form.

Answer to Problem 33E

The rational zeros are $-1,\frac{1}{2},-\frac{1}{2}$ .

Factored form of polynomial is $P\left(x\right)=\left(x+1\right)\left(2x-1\right)\left(2x+1\right)$

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=4x^3+4x^2-x-1$ .

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 $4$ , so

  $\text{possible rational zero of }P=\frac{\text{factor of 1}}{\text{factor of 4}}$

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

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

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

Simplify the fractions, so the all possible rational zeros:

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

For the rational zeros, substitute the x values by the possible rational zeros in the polynomial,

  $\begin{array}{l}P\left(x\right)=4x^3+4x^2-x-1\\ P\left(1\right)\ne 0\\ P\left(-1\right)=0\\ P\left(-\frac{1}{2}\right)=0\\ P\left(\frac{1}{2}\right)=0\\ P\left(\frac{1}{4}\right)\ne 0\\ P\left(-\frac{1}{4}\right)\ne 0\end{array}$

The rational zeros are $-1,\frac{1}{2},-\frac{1}{2}$

So the factored polynomial is ,

  $P\left(x\right)=\left(x+1\right)\left(2x-1\right)\left(2x+1\right)$