Chapter 5.6, Problem 31PPS

a.

To determine

To calculate: the depressed polynomial.

Answer to Problem 31PPS

The depressed polynomial is $-9x^4+50x^3+51x^2-150x-72$ .

Explanation of Solution

Given information:

  $f(x)=-9x^5+104x^4-249x^3-456x^2+828x+432$

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 ,

  $f(x)=-9x^5+104x^4-249x^3-456x^2+828x+432$

Since, the binomial $x-6$ is a factor of the polynomial by using Synthetic division we get,

  $\begin{array}{l}\overline{)6}\begin{array}{cccc}-9& 104& -\text{249}& \begin{array}{cc}-\text{456}& \begin{array}{cc}\text{ 828}& \text{ 43}2\end{array}\end{array}\end{array}\\ \begin{array}{cccc}-\text{5}4& 300& 306& \begin{array}{cc}-900& -432\end{array}\end{array}\\ \overline{\begin{array}{cccc}-9& 50& 51& \begin{array}{cc}-150& -72\end{array}\end{array}\hspace{.2em}|}\begin{array}{c}\text{ 0}\end{array}\end{array}$

Thus, the remainder is zero. The polynomial $-9x^5+104x^4-249x^3-456x^2+828x+432$ can be factored as $(x-6)(-9x^4+50x^3+51x^2-150x-72)$ .

The polynomial $-9x^4+50x^3+51x^2-150x-72$ is the depressed polynomial.

b.

To determine

To calculate: the values for the depressed polynomial.

Answer to Problem 31PPS

Values of x	Value
$-5$	$-9,922$
$-4$	$-4,160$
$-3$	$-1,242$
$-2$	$-112$
$-1$	$70$
$0$	$-72$
$1$	$-130$
$2$	$88$
$3$	$558$
$4$	$1,040$
$5$	$1,078$
$6$	$0$

Explanation of Solution

Given information:

  $f(x)=-9x^5+104x^4-249x^3-456x^2+828x+432$

  $-9x^4+50x^3+51x^2-150x-72$ is the depressed polynomial.

  $-5\le x\le 6$ value of x .

Formula used:

Direct Substitution :- The method is used to find the value of a function.

Calculations:

Values of x	Depressed polynomial $-9x^4+50x^3+51x^2-150x-72$	Value
$-5$	$\begin{array}{l}$ =-9{(-5)}^4+50{(-5)}^3+51{(-5)}^2-150(-5)-72$=-5625-6250+1275+750-72\\ =-9922\end{array}$	$-9,922$
$-4$	$\begin{array}{l}$ =-9{(-4)}^4+50{(-4)}^3+51{(-4)}^2-150(-4)-72$=-2304-3200+816+600-72\\ =-4160\end{array}$	$-4,160$
$-3$	$\begin{array}{l}$ =-9{(-3)}^4+50{(-3)}^3+51{(-3)}^2-150(-3)-72$=-729-1350+459+450-72\\ =-1242\end{array}$	$-1,242$
$-2$	$\begin{array}{l}$ =-9{(-2)}^4+50{(-2)}^3+51{(-2)}^2-150(-2)-72$=-144-400+204+300-72\\ =-112\end{array}$	$-112$
$-1$	$\begin{array}{l}$ =-9{(-1)}^4+50{(-1)}^3+51{(-1)}^2-150(-1)-72$=-9-50+51+150-72\\ =70\end{array}$	$70$
$0$	$\begin{array}{l}$ =-9{(0)}^4+50{(0)}^3+51{(0)}^2-150(0)-72$=-0-0+0+0-72\\ =-72\end{array}$	$-72$
$1$	$\begin{array}{l}$ =-9{(1)}^4+50{(1)}^3+51{(1)}^2-150(1)-72$=-9+50+51-150-72\\ =-130\end{array}$	$-130$
$2$	$\begin{array}{l}$ =-9{(2)}^4+50{(2)}^3+51{(2)}^2-150(2)-72$=-144+400+204-300-72\\ =88\end{array}$	$88$
$3$	$\begin{array}{l}$ =-9{(3)}^4+50{(3)}^3+51{(3)}^2-150(3)-72$=-729+1350+459-450-72\\ =558\end{array}$	$558$
$4$	$\begin{array}{l}$ =-9{(4)}^4+50{(4)}^3+51{(4)}^2-150(4)-72$=-2304+3200+816-600-72\\ =1040\end{array}$	$1,040$
$5$	$\begin{array}{l}$ =-9{(5)}^4+50{(5)}^3+51{(5)}^2-150(5)-72$=-5625+6250+1275-750-72\\ =1078\end{array}$	$1,078$
$6$	$\begin{array}{l}$ =-9{(6)}^4+50{(6)}^3+51{(6)}^2-150(6)-72$=-11664+10800+1836-900-72\\ =0\end{array}$	$0$

The table shows the value of the depressed polynomial .

c.

To determine

To comment: about the locations of the other zeros based on table.

Explanation of Solution

Given information:

Values of x	Value
$-5$	$-9,922$
$-4$	$-4,160$
$-3$	$-1,242$
$-2$	$-112$
$-1$	$70$
$0$	$-72$
$1$	$-130$
$2$	$88$
$3$	$558$
$4$	$1,040$
$5$	$1,078$
$6$	$0$

Formula used:

Direct Substitution :- The method 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 $P(r)=0$ .

Proof:

Observing the table we concluded that -

With the change in sign there exists a zero between values of $x=-2\text{ and }x=-1$ ; $x=-1\text{ and }x=0$ . Also at value of $x=6$ the value is zero.

d.

To determine

To graph: the depressed polynomial.

Explanation of Solution

Given information:

  $f(x)=-9x^5+104x^4-249x^3-456x^2+828x+432$

Graph:

  

Figure 1.

Interpretation:

The above graph shows the conclusion made in part (b) and (c) both. It defines the zero and the curves defines the change in sign of the value of the function.