Finding the root of a polynomial

I can find the root of a polinomial in a segment [A,B]. I can find it if root == A, root == B or if root is between A and B. And all of this using 2 methods not one.

But for the life of me I cannot determine if the root is not in the segment. For example, let's say that the polynomial is a simple one

Code:

x-5=0

The root is obviously 5. But if A and B are both bigger (A=10, B=14) or both smaller (A=-1, B=4) than the root (5 here), then I am stuck!!

EDIT: EUREKA!!! I figured it out. If both A and B are of the same sign (both positive or negative) than the root simply isn't in the segment [A,B].

EDIT 2: IDIOT!!! why shouldn't A and B be both positive, A= 3, B=10. A and B don't matter. There values once inputed in the polynomial matter.

so if we transform

Code:

x-5=0

into

Code:

f(x) = x-5

then f(A) and f(B) must be of the same sign.

Could you post your code?

so you are given a string like this one
"x-5=0"

and you want to solve it?
i couldnt understand in details what is your goal?

Ok, this is my code. I half the segment if the root is in it until the segment length reaches an error tolerance.

Code:

#include <stdio.h>
#include <math.h>

float f(float x)
{
	return (x-5);
}

void Halfing ()
{
	float a,b,s,e;
	int n = 0;
	int test = 0;
	
	do
	{
		printf("Enter the beginning of the segment -> ");
		scanf("%f",&a);
		printf("Enter the end of the segment -> ");
		scanf("%f",&b);
		if ((f(a) > 0 && f(b) > 0) || (f(a) < 0 && f(b) < 0))
		{
			printf("The root is not inside the segment [%4.2f,%4.2f]",a,b);
			printf("\nChoose other endpoints\n");
		}
		else test = 1;
	} while  (test != 1);
	
	printf("Enter error tolerance -> ");
	scanf("%f",&e);
	
	if (f(a) == 0) printf("The root is at an endpoint and it is %6.4f\n",a);
	else if (f(b) == 0) printf("The root is at an endpoint and it is %6.4f\n",b); 
	else	
	{
		do 
		{
			n++;
			s = (a+b)/2;
			printf("%d. pass x = %6.4f\n",n,s);
			if (f(a)*f(s)<0)   b = s;
			else  a = s;
		} while (fabsf(a-b)>e);	
		
		printf("\nRoot = %6.4f", (a+b)/2);
	}
}


int main (int argc, const char * argv[]) 
{
	Halfing ()

	return 0;
}

You probably can't just look at the values of the endpoints. Consider

f(x) = x^2 - 3,

with A = -2 and B = 2.

Then f(A) = 1 and f(B) = 1. However, even though they have the same sign, there are two roots in the domain, sqrt(3) and -sqrt(3). If you are limited to just linear functions, your algorithm is fine, but I don't think there's any foolproof way for a general polynomial (although there is a way for 2nd order, and maybe some higher orders). You can make it better by checking many points for sign changes, though, instead of just two.