Chapter 3.4, Problem 43E

To determine

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

Answer to Problem 43E

The rational zeros are $-3,-2,1,3$ .

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

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^5+3x^4-9x^3-31x^2+36$ .

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 $36$ and the leading coefficient is $1$ , so

  $\text{possible rational zero of }P=\frac{\text{factor of 36}}{\text{factor of 1}}$

The factors of $36$ are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 9,\pm 12,\pm 18,\pm 36$ and the factors of $1$ are $\pm 1$ .

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

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{4}{1},\pm \frac{6}{1},\pm \frac{9}{1},\pm \frac{12}{1},\pm \frac{18}{1},\pm \frac{36}{1}$

Simplify the fractions, so the all possible rational zeros:

  $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 9,\pm 12,\pm 18,\pm 36$

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

  $\begin{array}{l}P\left(x\right)=x^5+3x^4-9x^3-31x^2+36\\ P\left(1\right)=0\\ P\left(-1\right)\ne 0\\ P\left(2\right)\ne 0\\ P\left(-2\right)=0\\ P\left(-3\right)=0\\ P\left(3\right)=0\\ P\left(4\right)\ne 0\\ P\left(-4\right)\ne 0\\ P\left(-6\right)\ne 0\\ P\left(6\right)\ne 0\\ P\left(9\right)\ne 0\\ P\left(-9\right)\ne 0\\ P\left(12\right)\ne 0\\ P\left(-12\right)\ne 0\\ P\left(18\right)\ne 0\\ P\left(-18\right)\ne 0\\ P\left(36\right)\ne 0\\ P\left(-36\right)\ne 0\end{array}$

The rational zeros are $-3,-2,1,3$

So the factored polynomial is ,

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