Chapter 3.4, Problem 15E

To determine

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

Answer to Problem 15E

The rational zeros are $1,-2$ .

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

Explanation of Solution

Given information:

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

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 $4$ and the leading coefficient is $1$ , so

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

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

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

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

Simplify the fractions, so the all possible rational zeros:

  $\pm 1,\pm 2,\pm 4$

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

  $\begin{array}{l}P\left(x\right)=x^3+3x^2-4\\ P\left(-1\right)=-1+3-4\\ P\left(-1\right)\ne 0\\ P\left(1\right)=1+3-4\\ P\left(1\right)=0\\ P\left(2\right)=8+12-4\\ P\left(2\right)\ne 0\\ P\left(-2\right)=-8+12-4\\ P\left(-2\right)=0\\ P\left(4\right)=64+48-4\\ P\left(4\right)\ne 0\\ P\left(-4\right)=-64+48-4\\ P\left(-4\right)\ne 0\end{array}$

The rational zeros are $1,-2$ ,

So the factored polynomial is ,

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