Chapter 7, Problem 1P

Divide a polynomial $f\left(x\right)-x^4-7.5x^3+14.5x^2+3x-20$ by the monomial factor $x-2$. Is $x=2$ a root?

To determine

To calculate: Whether $x=2$ is the root to the polynomial $f(x)=x^4-7.5x^3+14.5x^2+3x-20$ or not.

Answer to Problem 1P

Solution: Yes, $x=2$ is a root to the provided polynomial.

Explanation of Solution

Given Information:

Provided Polynomial is $f(x)=x^4-7.5x^3+14.5x^2+3x-20$ and the monomial factor is $x-2$.

Formula used:

One of the approaches to find the roots of a polynomial is represented by the pseudo code. The pseudo code performs the division of an $n^{\text{th}}$

order polynomial by a monomial factor $x-t$

:, if the monomial is one of the root of the equation then the remainder becomes $r$ zero.

$\begin{array}{l}r=a\left(n\right)\\ a\left(n\right)=0\\ \text{DOFOR }i=n-1,0,-1\\ s=a\left(i\right)\\ a\left(i\right)=r\\ r=s+r*t\\ \text{END DO}\end{array}$

Calculation:

Consider the provided expression:

$f(x)=x^4-7.5x^3+14.5x^2+3x-20$

Consider the monomial factor $x-2$.

For the polynomial provided in the question $n=4,a_0=-20$

, $a_1=3,a_2=14.5$

, $a_3=-7.5$, $a_4=1$

, and $t=2$.

Follow the approach provided by following Pseudo code.

$\begin{array}{l}r=a\left(n\right)\\ a\left(n\right)=0\\ \text{DOFOR }i=n-1,0,-1\\ s=a\left(i\right)\\ a\left(i\right)=r\\ r=s+r*t\\ \text{END DO}\end{array}$

Here,

$\begin{array}{l}r=a_4=1\\ a_4=0\end{array}$

Iterate the root in the following interval:

$\begin{array}{c}i=4-1\\ =3\text{ to 0}\end{array}$

. For $i=3$, $\begin{array}{c}s=a_3\\ =-7.5\end{array}$

$\begin{array}{c}{a}_3=r\\ =1\end{array}$

$\begin{array}{c}r=s+rt\\ =-7.5+1\left(2\right)\\ =-5.5\end{array}$

For $i=2$, $\begin{array}{c}s=a_2\\ =14.5\end{array}$

$\begin{array}{c}{a}_2=r\\ =-5.5\end{array}$

$\begin{array}{c}r=s+rt\\ =14.5-5.5\left(2\right)\\ =3.5\end{array}$

For $i=1$, $\begin{array}{c}s=a_1\\ =3\end{array}$

$\begin{array}{c}{a}_1=r\\ =3.5\end{array}$

$\begin{array}{c}r=s+rt\\ =3+3.5\left(2\right)\\ =10\end{array}$

For $i=0$, $\begin{array}{c}s=a_0\\ =-20\end{array}$

$\begin{array}{c}{a}_0=r\\ =10\end{array}$.

$\begin{array}{c}r=s+rt\\ =-20+10\left(2\right)\\ =0\end{array}$

The remainder of zero. Thus, 2 will be a root of the provided polynomial.

To verify the pseudo code, write a MATLAB program and execute it:

clc

a=[-20 3 14.5 -7.5 1];

%specify the values of the co-efficent

n=4;

%specify the order of the polynomila

t=2;

%Arbitrary/monomial root consideration

r=a(n+1);

a(n+1)=0;

for i=4:-1:1

s=a(i);

a(i)=r;

r=s+r*t;

%remainder

end

display(a);

display(r);

Execute the code following result will be obtained:

Thus, the remainder is $\text{0}$

and quotient is $x^3-5.5x^2+3.5x+10$.

Therefore, the term $x-2$ will be a factor of the polynomial $f(x)=x^4-7.5x^3+14.5x^2+3x-20$.