Chapter 6.6, Problem 6CYU

To determine

To calculate: the remaining factors of the polynomial.

Answer to Problem 6CYU

The remaining factors are $(x+1)(x+12)$

Explanation of Solution

Given information:

  $x-1$ is a factor of the polynomial $3x^3+10x^2-x-12$

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 ,

  $3x^3+10x^2-x-12$

Since, the binomial $x-1$ is a factor of the polynomial by using Synthetic division we get,

  $\begin{array}{l}\overline{)1}\begin{array}{cccc}3& 10& -1& -12\end{array}\\ \begin{array}{ccc}3& \text{ 13}& \text{ 12}\end{array}\\ \overline{\begin{array}{ccc}3& 13& 12\end{array}\hspace{.2em}|}\begin{array}{c}\text{ 0}\end{array}\end{array}$

Thus, the remainder is zero. The polynomial $3x^3+10x^2-x-12$ can be factored as $(x-1)(3x^2+13x+12)$ .

The polynomial $x^2+13x+12$ is the depressed polynomial.

The remaining factors of the polynomial are :-

  $\begin{array}{l}=x^2+13x+12\\ =x^2+12x+x+12\\ =x(x+12)+1(x+12)\\ =(x+1)(x+12)\end{array}$

Hence, the factors of the polynomial are $(x-1)(x+1)(x+12)$ .