Chapter 9.5, Problem 65STP

To determine

To calculate:

The remaining factors of the given polynomial

Answer to Problem 65STP

The remaining factors of the polynomial $2x^3-5x^2-28x+15$ are $\left(x+3\right)\left(2x-1\right)$

All the factors of the polynomial $2x^3-5x^2-28x+15$ are $\left(x-5\right)\left(x+3\right)\left(2x-1\right)$

Explanation of Solution

Given information:

The polynomial is $2x^3-5x^2-28x+15$

The factor is $x-5$

Calculation:

The binomial $x-5$ is a factor of the polynomial if 5 is a zero of the related polynomial function.

Use the factor theorem and synthetic division.

The synthetic division can be calculated as,

  $\begin{array}{l}5\overline{)\begin{array}{l}2-5-2815\\ \downarrow \text{ 10 25 }-15\end{array}}\\ \text{ 2 5 }-3\overline{)\text{0}}\end{array}$

Because the remainder is 0, $x-5$ is a factor of the polynomial.

The polynomial $2x^3-5x^2-28x+15$ can be factored as $\left(x-5\right)\left(2x^2+5x-3\right)$

The polynomial $\left(2x^2+5x-3\right)$ is a depressed polynomial. Check to see if this polynomial can be factored.

  $\begin{array}{l}\left(2x^2+5x-3\right)\\ =\left(x+3\right)\left(2x-1\right)\left[\text{Factor the trinomial}\right]\end{array}$

Therefore, the remaining factors of the polynomial $2x^3-5x^2-28x+15$ are $\left(x+3\right)\left(2x-1\right)$

So, $2x^3-5x^2-28x+15=\left(x-5\right)\left(x+3\right)\left(2x-1\right)$