Chapter 9.5, Problem 66STP

To determine

To calculate:

The remaining factors of the given polynomial

Answer to Problem 66STP

The remaining factors of the polynomial $3x^3+10x^2-x-12$ are $\left(3x+4\right)\left(x-1\right)$

All the factors of the polynomial $3x^3+10x^2-x-12$ are $\left(x+3\right)\left(3x+4\right)\left(x-1\right)$

Explanation of Solution

Given information:

The polynomial is $3x^3+10x^2-x-12$

The factor is $x+3$

Calculation:

The binomial $x+3$ is a factor of the polynomial if $-3$ is a zero of the related polynomial function.

Use the factor theorem and synthetic division.

The synthetic division can be calculated as,

  $\begin{array}{l}-3\overline{)\begin{array}{l}\text{3 10 }-1-12\\ \downarrow \text{ 9 }-3\text{ 12}\end{array}}\\ \text{ 3 1 }-4\overline{)\text{0}}\end{array}$

Because the remainder is 0, $x+3$ is a factor of the polynomial.

The polynomial $3x^3+10x^2-x-12$ can be factored as $\left(x+3\right)\left(3x^2+x-4\right)$

The polynomial $\left(3x^2+x-4\right)$ is a depressed polynomial. Check to see if this polynomial can be factored.

  $\begin{array}{l}\left(3x^2+x-4\right)\\ =\left(3x+4\right)\left(x-1\right)\left[\text{Factor the trinomial}\right]\end{array}$

Therefore, the remaining factors of the polynomial $3x^3+10x^2-x-12$ are $\left(3x+4\right)\left(x-1\right)$

So, $3x^3+10x^2-x-12=\left(x+3\right)\left(3x+4\right)\left(x-1\right)$