Chapter 3.4, Problem 21E

To determine

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

Answer to Problem 21E

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

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

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^3-4x^2+x+6$ .

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 $6$ and the leading coefficient is $1$ , so

  $\text{possible rational zero of }P=\frac{\text{factor of 6}}{\text{factor of 1}}$

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

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

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

Simplify the fractions, so the all possible rational zeros:

  $\pm 1,\pm 2,\pm 3,\pm 6$

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

  $\begin{array}{l}P\left(x\right)=x^3-x^2-8x+12\\ P\left(-1\right)=0\\ P\left(1\right)\ne 0\\ P\left(2\right)=0\\ P\left(-2\right)\ne 0\\ P\left(3\right)=0\\ P\left(-3\right)\ne 0\\ P\left(6\right)\ne 0\\ P\left(-6\right)\ne 0\end{array}$

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

So the factored polynomial is ,

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