Find the integral roots of a given Cubic equation

Last Updated : 15 Jul, 2025

Given 5 integers say A, B, C, D, and E which represents the cubic equation f(x) = A*x^{3} + B*x^{2} + C*x + D = E, the task is to find the integral solution for this equation. If there doesn't exist any integral solution then print "NA".
Examples:

Input: A = 1, B = 0, C = 0, D = 0, E = 27
Output: 3
Input: A = 1, B = 0, C = 0, D = 0, E = 16
Output: NA

Approach: The idea is to use binary search. Below are the steps:

Initialise the start and end variable as 0 & 10⁵ respectively.
Find the middle(say mid) value of start and end check if it satisfy the given equation or not.
If current mid satisfy the given equation then print the mid value.
Else if the value of f(x) is less than E then update start as mid + 1.
Else Update end as mid - 1.
If we can't find any integral solution for the above equation then print "-1".

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the value at x of
// the given equation
long long int check(int A, int B, int C,
                    int D, long long int x)
{

    long long int ans;

    // Find the value equation at x
    ans = (A * x * x * x
           + B * x * x
           + C * x
           + D);

    // Return the value of ans
    return ans;
}

// Function to find the integral
// solution of the given equation
void findSolution(int A, int B, int C,
                  int D, int E)
{

    // Initialise start and end
    int start = 0, end = 100000;

    long long int mid, ans;

    // Implement Binary Search
    while (start <= end) {

        // Find mid
        mid = start + (end - start) / 2;

        // Find the value of f(x) using
        // current mid
        ans = check(A, B, C, D, mid);

        // Check if current mid satisfy
        // the equation
        if (ans == E) {

            // Print mid and return
            cout << mid << endl;
            return;
        }

        if (ans < E)
            start = mid + 1;
        else
            end = mid - 1;
    }

    // Print "NA" if not found
    // any integral solution
    cout << "NA";
}

// Driver Code
int main()
{
    int A = 1, B = 0, C = 0;
    int D = 0, E = 27;

    // Function Call
    findSolution(A, B, C, D, E);
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

// Function to find the value at x of
// the given equation
static long check(int A, int B, int C,
                         int D, long x)
{
    long ans;

    // Find the value equation at x
    ans = (A * x * x * x + 
           B * x * x + C * x + D);

    // Return the value of ans
    return ans;
}

// Function to find the integral
// solution of the given equation
static void findSolution(int A, int B, int C,
                         int D, int E)
{

    // Initialise start and end
    long start = 0, end = 100000;

    long mid, ans;

    // Implement Binary Search
    while (start <= end) 
    {

        // Find mid
        mid = start + (end - start) / 2;

        // Find the value of f(x) using
        // current mid
        ans = check(A, B, C, D, mid);

        // Check if current mid satisfy
        // the equation
        if (ans == E)
        {

            // Print mid and return
            System.out.println(mid);
            return;
        }

        if (ans < E)
            start = mid + 1;
        else
            end = mid - 1;
    }

    // Print "NA" if not found
    // any integral solution
    System.out.println("NA");
}

// Driver Code
public static void main(String args[])
{
    int A = 1, B = 0, C = 0;
    int D = 0, E = 27;

    // Function Call
    findSolution(A, B, C, D, E);
}
}

// This code is contributed by Code_Mech

Python

# Python3 program for the above approach

# Function to find the value at x of
# the given equation
def check(A, B, C, D, x) :

    ans = 0;

    # Find the value equation at x
    ans = (A * x * x * x + 
           B * x * x + C * x + D);

    # Return the value of ans
    return ans;

# Function to find the integral
# solution of the given equation
def findSolution(A, B, C, D, E) :

    # Initialise start and end
    start = 0; end = 100000;

    mid = 0;
    ans = 0;

    # Implement Binary Search
    while (start <= end) :

        # Find mid
        mid = start + (end - start) // 2;

        # Find the value of f(x) using
        # current mid
        ans = check(A, B, C, D, mid);

        # Check if current mid satisfy
        # the equation
        if (ans == E) :

            # Print mid and return
            print(mid);
            return;

        if (ans < E) :
            start = mid + 1;
        else :
            end = mid - 1;

    # Print "NA" if not found
    # any integral solution
    print("NA");

# Driver Code
if __name__ == "__main__" :

    A = 1; B = 0; C = 0;
    D = 0; E = 27;

    # Function Call
    findSolution(A, B, C, D, E);
    
# This code is contributed by AnkitRai01

// C# program for the above approach
using System;
class GFG{

// Function to find the value at x of
// the given equation
static long check(int A, int B, int C,
                         int D, long x)
{
    long ans;

    // Find the value equation at x
    ans = (A * x * x * x + 
           B * x * x + C * x + D);

    // Return the value of ans
    return ans;
}

// Function to find the integral
// solution of the given equation
static void findSolution(int A, int B, int C,
                         int D, int E)
{

    // Initialise start and end
    long start = 0, end = 100000;

    long mid, ans;

    // Implement Binary Search
    while (start <= end) 
    {

        // Find mid
        mid = start + (end - start) / 2;

        // Find the value of f(x) using
        // current mid
        ans = check(A, B, C, D, mid);

        // Check if current mid satisfy
        // the equation
        if (ans == E)
        {

            // Print mid and return
            Console.WriteLine(mid);
            return;
        }

        if (ans < E)
            start = mid + 1;
        else
            end = mid - 1;
    }

    // Print "NA" if not found
    // any integral solution
    Console.Write("NA");
}

// Driver Code
public static void Main()
{
    int A = 1, B = 0, C = 0;
    int D = 0, E = 27;

    // Function Call
    findSolution(A, B, C, D, E);
}
}

// This code is contributed by Code_Mech

JavaScript

<script>

// Javascript program for the above approach

// Function to find the value at x of
// the given equation
function check(A, B, C, D, x)
{
    var ans;

    // Find the value equation at x
    ans = (A * x * x * x + 
           B * x * x + C * x + D);

    // Return the value of ans
    return ans;
}

// Function to find the integral
// solution of the given equation
function findSolution(A, B, C, D, E)
{

    // Initialise start and end
    var start = 0, end = 100000;

    var mid, ans;

    // Implement Binary Search
    while (start <= end) 
    {

        // Find mid
        mid = parseInt(start + (end - start) / 2);

        // Find the value of f(x) using
        // current mid
        ans = check(A, B, C, D, mid);

        // Check if current mid satisfy
        // the equation
        if (ans == E)
        {

            // Print mid and return
            document.write(mid);
            return;
        }

        if (ans < E)
            start = mid + 1;
        else
            end = mid - 1;
    }

    // Print "NA" if not found
    // any integral solution
    document.write("NA");
}

// Driver Code
var A = 1, B = 0, C = 0;
var D = 0, E = 27;

// Function Call
findSolution(A, B, C, D, E);

// This code is contributed by Ankita saini

</script>

Output

Time Complexity: O(log N)

Auxiliary Space: O(1) as constant space for variables is being used

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