EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 9780100254145
Author: Chapra
Publisher: YUZU
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Chapter 7, Problem 1P

Divide a polynomial f ( x ) x 4 7.5 x 3 + 14.5 x 2 + 3 x 20 by the monomial factor x 2 . Is x = 2 a root?

Expert Solution & Answer
Check Mark
To determine

To calculate: Whether x=2 is the root to the polynomial f(x)=x47.5x3+14.5x2+3x20 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)=x47.5x3+14.5x2+3x20 and the monomial factor is x2.

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 nth

order polynomial by a monomial factor xt

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

r=a(n)a(n)=0DOFOR  i=n1,0,1    s=a(i)    a(i)=r    r=s+r*tEND DO

Calculation:

Consider the provided expression:

f(x)=x47.5x3+14.5x2+3x20

Consider the monomial factor x2.

For the polynomial provided in the question n=4,a0=20

, a1=3,a2=14.5

, a3=7.5, a4=1 

, and t=2.

Follow the approach provided by following Pseudo code.

r=a(n)a(n)=0DOFOR  i=n1,0,1    s=a(i)    a(i)=r    r=s+r*tEND DO

Here,

r=a4=1a4=0

Iterate the root in the following interval:

i=41=3 to 0

. For i=3, s=a3=7.5

a3=r=1

r=s+rt=7.5+1(2)=5.5

For i=2, s=a2=14.5

a2=r=5.5

r=s+rt=14.55.5(2)=3.5

For i=1, s=a1=3

a1=r=3.5

r=s+rt=3+3.5(2)=10

For i=0, s=a0=20

a0=r=10.

r=s+rt=20+10(2)=0

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:

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 7, Problem 1P

Thus, the remainder is 0

and quotient is x35.5x2+3.5x+10.

Therefore, the term x2 will be a factor of the polynomial f(x)=x47.5x3+14.5x2+3x20.

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