Chapter 6.6, Problem 26PPS

To determine

To calculate: the remaining factors of the polynomial.

Answer to Problem 26PPS

The remaining factors are $2(8x^4-4x^3-6x^2-9x-54)$

Explanation of Solution

Given information:

  $2x-3$ is a factor of the polynomial $16x^5-32x^4-81x+162$

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 ,

  $16x^5-32x^4-81x+162$

Since, the binomial $2x-3$ is a factor of the polynomial by using Synthetic division we get,

  $\begin{array}{l}\overline{)\frac{3}{2}}\begin{array}{cccc}16& -32& 0& \begin{array}{cc}0& \begin{array}{cc}-81& 162\end{array}\end{array}\end{array}\\ \begin{array}{cccc}24& -12& -18& \begin{array}{cc}-27& -162\end{array}\end{array}\\ \overline{\begin{array}{cccc}16& -8& -12& \begin{array}{cc}-18& -108\end{array}\end{array}\hspace{.2em}|}\begin{array}{c}\text{ 0}\end{array}\end{array}$

Thus, the remainder is zero. The polynomial $16x^5-32x^4-81x+162$ can be factored as $(2x-3)(16x^4-8x^3-12x^2-18x-108)$ .

The polynomial $16x^4-8x^3-12x^2-18x-108$ is the depressed polynomial.

The remaining factors of the polynomial are :-

  $\begin{array}{l}=16x^4-8x^3-12x^2-18x-108\\ =2(8x^4-4x^3-6x^2-9x-54)\end{array}$

Therefore,

  $\begin{array}{l}=(2x-3)(16x^4-8x^3-12x^2-18x-108)\\ =2(2x-3)(8x^4-4x^3-6x^2-9x-54)\end{array}$

Hence, the factors of the polynomial are $2(2x-3)(8x^4-4x^3-6x^2-9x-54)$ .