Chapter 5.6, Problem 23PPS

To determine

To calculate: the remaining factors of the polynomial.

Answer to Problem 23PPS

The remaining factors are $(x-1)(x+1)(x^2+2x+3)$

Explanation of Solution

Given information:

  $x-1$ is a factor of the polynomial $x^4+2x^3+2x^2-2x-3$

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 ,

  $x^4+2x^3+2x^2-2x-3$

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

  $\begin{array}{l}\overline{)1}\begin{array}{cccc}1& 2& 2& \begin{array}{cc}-2& -3\end{array}\end{array}\\ \begin{array}{cccc}1& 3& 5& 3\end{array}\\ \overline{\begin{array}{cccc}1& 3& 5& 3\end{array}\hspace{.2em}|}\begin{array}{c}\text{ 0}\end{array}\end{array}$

Thus, the remainder is zero. The polynomial $x^4+2x^3+2x^2-2x-3$ can be factored as $(x-1)(x^3+3x^2+5x+3)$ .

The polynomial $x^3+3x^2+5x+3$ is the depressed polynomial.

By hit and trial method the binomial $x+1$ is a factor of the polynomial $x^3+3x^2+5x+3$ by using Synthetic division we get,

  $\begin{array}{l}\overline{)-1}\begin{array}{cccc}1& 3& 5& 3\end{array}\\ \begin{array}{ccc}-1& -2& -3\end{array}\\ \overline{\begin{array}{ccc}1& 2& 3\end{array}\hspace{.2em}|}\begin{array}{c}\text{ 0}\end{array}\end{array}$

Thus, the remainder is zero. The polynomial $x^3+3x^2+5x+3$ can be factored as $(x+1)(x^2+2x+3)$ .

The remaining factors of the polynomial $x^4+2x^3+2x^2-2x-3$ are :-

  $\begin{array}{l}=(x-1)(x^3+3x^2+5x+3)\\ =(x-1)(x+1)(x^2+2x+3)\end{array}$

Hence, the factors of the polynomial are $(x-1)(x-1)(x+1)(x^2+2x+3)$ .