Chapter 3.4, Problem 31E

To determine

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

Answer to Problem 31E

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

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

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=3x^4-10x^3-9x^2+40x-12$ .

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 $P$ 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 $12$ and the leading coefficient is $3$ , so

  $\text{possible rational zero of }P=\frac{\text{factor of 12}}{\text{factor of 3}}$

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

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

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

Simplify the fractions, so the all possible rational zeros:

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

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

  $\begin{array}{l}P\left(x\right)=3x^4-10x^3-9x^2+40x-12\\ P\left(-1\right)\ne 0\\ P\left(1\right)\ne 0\\ P\left(-2\right)=0\\ P\left(2\right)=0\\ P\left(3\right)=0\\ P\left(-3\right)\ne 0\\ P\left(-\frac{1}{3}\right)\ne 0\\ P\left(\frac{1}{3}\right)=0\\ P\left(-4\right)\ne 0\\ P\left(4\right)\ne 0\\ P\left(-6\right)\ne 0\\ P\left(6\right)\ne 0\\ P\left(-12\right)\ne 0\\ P\left(12\right)\ne 0\\ P\left(-\frac{2}{3}\right)\ne 0\\ P\left(\frac{2}{3}\right)\ne 0\\ P\left(-\frac{4}{3}\right)\ne 0\\ P\left(\frac{4}{3}\right)\ne 0\end{array}$

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

So the factored polynomial is ,

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