This short code allows you to enter B and C in a trinomial and then outputs it in factored form, and gives you the solutions.
Note: this only works if a = 1
(ax^2 + Bx + C)
Example:
Enter B: 7
Enter C: 10
Factored Form: (x+5)(x+2)
Solutions: x=-5, x=-2
Code:#include<iostream.h> #include<vector> using namespace std; int check(vector<int> vec, int b, int c) { for(vector<int>::iterator x = vec.begin(); x!=vec.end(); x++) { for(vector<int>::iterator y = vec.end() - 1; y!=vec.begin(); y--) { if((*x + *y) == b && (*x)*(*y)== c) { cout<<"Factored form is: "<<"(x "; if(*x>0) cout<<"+ "<<*x<<")(x "; else cout<<*x<<")(x "; if(*y>0) cout<<"+ "<<*y<<")"; else cout<<*y<<") "; cout<<endl<<"Solutions: "<<"x = "<<-1*(*x)<<", x = "<<-1*(*y); return 1; } } } cout<<"This problem can't be solved. No factors of C add up to B"<<endl; return 0; } int main() { vector<int> vec; int b, c; cout<<"This can solve equations in the following format:"<<endl; cout<<"x^2 + Bx + C "<<endl; cout<<"Enter B: "; cin>>b; cout<<"Enter C: "; cin>>c; for(int a = c * -1; a<c; a++) { if(c%a == 0) { vec.push_back(a); } if(a==(-1))//to make it skip zero. a++; } check(vec, b, c); cin.get(); cin.get(); return 0; }
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Thread: [C++]Trinomial Factoring
- 19 Jun. 2010 03:04pm #1
[C++]Trinomial Factoring
LG's Dyke. Enough. Said.