Chapter 5.6, Problem 22PPS

To determine

To calculate: the remaining factors of the polynomial.

Answer to Problem 22PPS

The remaining factors are $\left(2x+1\right)\left(x+7\right)$

Explanation of Solution

Given information:

  $x-4$ is a factor of the polynomial $2x^3+7x^2-53x-28$

Formula used:

Synthetic Substitution :- Synthetic division is used to find the value of a function.

Factor Theorem :- The binomial $x-r$ is a factor of the polynomial $P(x)$ if and only if $P(r)=0$ .

Remainder Theorem:- If a polynomial $P(x)$ is divided by $x-r$ , the remainder s a constant $P(r)$ , and

  $P(x)=Q(x)(x-r)+P(r)$ ,

Where $Q(x)$ is a polynomial with degree one less than $P(x)$ .

Calculation:

Consider ,

  $2x^3+7x^2-53x-28$

Since, the binomial $x-4$ is a factor of the polynomial by using Synthetic division we get,

  $\begin{array}{l}\overline{)4}\begin{array}{cccc}2& 7& -\text{53}& -28\end{array}\\ \begin{array}{ccc}8& \text{ 60}& \text{ 28}\end{array}\\ \overline{\begin{array}{ccc}2& \text{ 15}& 7\end{array}\hspace{.2em}|}\begin{array}{c}\text{ 0}\end{array}\end{array}$

Thus, the remainder is zero. The polynomial $2x^3+7x^2-53x-28$ can be factored as $(x-4)(2x^2+15x+7)$ .

The polynomial $2x^2+15x+7$ is the depressed polynomial.

The remaining factors of the polynomial are :-

  $\begin{array}{l}=2x^2+15x+7\\ =2x^2+14x+x+7\\ =2x(x+7)+1(x+7)\\ =(2x+1)(x+7)\end{array}$

Hence, the factors of the polynomial are $\left(x-4\right)\left(2x+1\right)\left(x+7\right)$ .