Chapter 3.4, Problem 30E

To determine

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

Answer to Problem 30E

The rational zeros are $-3,3,-\frac{1}{2},1$ .

Factored form of polynomial is $P\left(x\right)=\left(x+3\right)\left(2x+1\right)\left(2x-3\right)\left(x-1\right)$

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=2x^4-x^3-19x^2+9x+9$ .

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 $9$ and the leading coefficient is $2$ , so

  $\text{possible rational zero of }P=\frac{\text{factor of 9}}{\text{factor of 2}}$

The factors of $9$ are $\pm 1,\pm 3,\pm 9$ and the factors of $2$ are $\pm 1,\pm 2$ .

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

  $\pm \frac{1}{1},\pm \frac{3}{1},\pm \frac{9}{1},\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{9}{2}$

Simplify the fractions, so the all possible rational zeros:

  $\pm 1,\pm 3,\pm 9,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{9}{2}$

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

  $\begin{array}{l}P\left(x\right)=2x^4-x^3-19x^2+9x+9\\ P\left(-1\right)\ne 0\\ P\left(1\right)=0\\ P\left(\frac{1}{2}\right)\ne 0\\ P\left(-\frac{1}{2}\right)=0\\ P\left(3\right)=0\\ P\left(-3\right)=0\\ P\left(-\frac{3}{2}\right)\ne 0\\ P\left(\frac{3}{2}\right)\ne 0\\ P\left(-9\right)\ne 0\\ P\left(9\right)\ne 0\\ P\left(-\frac{9}{2}\right)\ne 0\\ P\left(\frac{9}{2}\right)\ne 0\end{array}$

The rational zeros are $-3,3,-\frac{1}{2},1$

So the factored polynomial is ,

  $P\left(x\right)=\left(x+3\right)\left(2x+1\right)\left(2x-3\right)\left(x-1\right)$