Algorithm of Synthetic Division:
Example of Synthetic Division: x2 + 5x+ 6 by x – 1
Given a polynomial of form p(x) = anxn
+ an-1xn-1+…+ a1x+ a0, we can divide
it by a linear factor x-r, where ‘r’ is a constant, using following steps.
a.
Write the r on the left side of a vertical bar
and the coefficients in decreasing degree.
r| an an-1 ……. a1 a0
b.
Pass the coefficient of the highest degree below
the horizontal line as it is
r| an an-1 ……. a1 a0
|.
.
| an
c.
Multiply the passed coefficient with ‘r’ and add
to the next coefficient
r| an an-1 ……. a1 a0
|. an
.
| an an-1+ an
d.
Continue the step ‘c’ until all coefficients are
covered.
e.
The numbers below an ……. a1
give the coefficients of the quotient whose degree is 1 less than the
polynomial, the number below a0 is the remainder.
Example of Synthetic Division: x2 + 5x+ 6 by x – 1
Writing the coefficients of the polynomial and dividing by the linear
factor.
1|1 5 6
| 1 6
|1 6 12
Therefore,
the quotient Q(x) = x + 6 and remainder R(x) = 12.
Program:
#include<stdio.h>
#include<conio.h>
void main(){
clrscr();
int
poly[6], m, r, i, q[6];
printf("\t\tSYNTHETIC
DIVISION");
printf("\n
Enter the highest degree of the equation (max 5): ");
scanf("%d",&m);
for(i=0;i<=m;i++){
printf("\n
Coefficient x[%d] = ", m-i);
scanf("%d",&poly[i]);
}
printf("\n
Enter the value of constant in (x-r) : ");
scanf("%d",&r);
q[0]
= poly[0];
for(i=1;i<=m;i++){
q[i]
= (q[i-1]*r)+poly[i];
}
printf("\n
Coefficients of quotient are: \n");
for(i=0;i<m;i++){
printf("\t%d",q[i]);
}
printf("\n
Remainder is: %d", q[m]);
getch();
}
Sample Output:
Sample Output:
SYNTHETIC DIVISION
Enter the highest degree of the equation (max 5): 2
Coefficient x[0] = 1
Coefficient x[1] = 5
Coefficient x[2] = 6
Enter the value of constant in (x—r) : 1
Coefficients of quotient are:
1 6
Remainder is: 12
Enter the highest degree of the equation (max 5): 2
Coefficient x[0] = 1
Coefficient x[1] = 5
Coefficient x[2] = 6
Enter the value of constant in (x—r) : 1
Coefficients of quotient are:
1 6
Remainder is: 12
©Dixit Bhatta 2013
thank you for sharing this algorithm to solving synthetic division
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