 GeeksforGeeks App
Open App Browser
Continue

## Related Articles

• Write an Interview Experience
• Mathematical Algorithms

# Multiply two polynomials

Given two polynomials represented by two arrays, write a function that multiplies given two polynomials.

Example:

```Input:  A[] = {5, 0, 10, 6}
B[] = {1, 2, 4}
Output: prod[] = {5, 10, 30, 26, 52, 24}

The first input array represents "5 + 0x^1 + 10x^2 + 6x^3"
The second array represents "1 + 2x^1 + 4x^2"
And Output is "5 + 10x^1 + 30x^2 + 26x^3 + 52x^4 + 24x^5"```

A simple solution is to one by one consider every term of the first polynomial and multiply it with every term of the second polynomial. Following is the algorithm of this simple method.

```multiply(A[0..m-1], B[0..n-1])
1) Create a product array prod[] of size m+n-1.
2) Initialize all entries in prod[] as 0.
3) Traverse array A[] and do following for every element A[i]
...(3.a) Traverse array B[] and do following for every element B[j]
prod[i+j] = prod[i+j] + A[i] * B[j]
4) Return prod[].```

The following is the implementation of the above algorithm.

## C++

 `// Simple C++ program to multiply two polynomials``#include ``using` `namespace` `std;`` ` `// A[] represents coefficients of first polynomial``// B[] represents coefficients of second polynomial``// m and n are sizes of A[] and B[] respectively``int` `*multiply(``int` `A[], ``int` `B[], ``int` `m, ``int` `n)``{``   ``int` `*prod = ``new` `int``[m+n-1];`` ` `   ``// Initialize the product polynomial``   ``for` `(``int` `i = 0; i

## Java

 `// Java program to multiply two polynomials``class` `GFG``{`` ` `    ``// A[] represents coefficients ``    ``// of first polynomial``    ``// B[] represents coefficients ``    ``// of second polynomial``    ``// m and n are sizes of A[] and B[] respectively``    ``static` `int``[] multiply(``int` `A[], ``int` `B[], ``                            ``int` `m, ``int` `n) ``    ``{``        ``int``[] prod = ``new` `int``[m + n - ``1``];`` ` `        ``// Initialize the product polynomial``        ``for` `(``int` `i = ``0``; i < m + n - ``1``; i++) ``        ``{``            ``prod[i] = ``0``;``        ``}`` ` `        ``// Multiply two polynomials term by term``        ``// Take ever term of first polynomial``        ``for` `(``int` `i = ``0``; i < m; i++) ``        ``{``            ``// Multiply the current term of first polynomial``            ``// with every term of second polynomial.``            ``for` `(``int` `j = ``0``; j < n; j++) ``            ``{``                ``prod[i + j] += A[i] * B[j];``            ``}``        ``}`` ` `        ``return` `prod;``    ``}`` ` `    ``// A utility function to print a polynomial``    ``static` `void` `printPoly(``int` `poly[], ``int` `n) ``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++) ``        ``{``            ``System.out.print(poly[i]);``            ``if` `(i != ``0``) ``            ``{``                ``System.out.print(``"x^"` `+ i);``            ``}``            ``if` `(i != n - ``1``) ``            ``{``                ``System.out.print(``" + "``);``            ``}``        ``}``    ``}`` ` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// The following array represents ``        ``// polynomial 5 + 10x^2 + 6x^3``        ``int` `A[] = {``5``, ``0``, ``10``, ``6``};`` ` `        ``// The following array represents ``        ``// polynomial 1 + 2x + 4x^2``        ``int` `B[] = {``1``, ``2``, ``4``};``        ``int` `m = A.length;``        ``int` `n = B.length;`` ` `        ``System.out.println(``"First polynomial is n"``);``        ``printPoly(A, m);``        ``System.out.println(``"nSecond polynomial is n"``);``        ``printPoly(B, n);`` ` `        ``int``[] prod = multiply(A, B, m, n);`` ` `        ``System.out.println(``"nProduct polynomial is n"``);``        ``printPoly(prod, m + n - ``1``);``    ``}``}`` ` `// This code contributed by Rajput-Ji`

## Python3

 `# Simple Python3 program to multiply two polynomials`` ` `# A[] represents coefficients of first polynomial``# B[] represents coefficients of second polynomial``# m and n are sizes of A[] and B[] respectively``def` `multiply(A, B, m, n):`` ` `    ``prod ``=` `[``0``] ``*` `(m ``+` `n ``-` `1``);``     ` `    ``# Multiply two polynomials term by term``     ` `    ``# Take ever term of first polynomial``    ``for` `i ``in` `range``(m):``         ` `        ``# Multiply the current term of first ``        ``# polynomial with every term of ``        ``# second polynomial.``        ``for` `j ``in` `range``(n):``            ``prod[i ``+` `j] ``+``=` `A[i] ``*` `B[j];`` ` `    ``return` `prod;`` ` `# A utility function to print a polynomial``def` `printPoly(poly, n):`` ` `    ``for` `i ``in` `range``(n):``        ``print``(poly[i], end ``=` `"");``        ``if` `(i !``=` `0``):``            ``print``(``"x^"``, i, end ``=` `"");``        ``if` `(i !``=` `n ``-` `1``):``            ``print``(``" + "``, end ``=` `"");`` ` `# Driver Code`` ` `# The following array represents``# polynomial 5 + 10x^2 + 6x^3``A ``=` `[``5``, ``0``, ``10``, ``6``];`` ` `# The following array represents ``# polynomial 1 + 2x + 4x^2``B ``=` `[``1``, ``2``, ``4``];``m ``=` `len``(A);``n ``=` `len``(B);`` ` `print``(``"First polynomial is "``);``printPoly(A, m);``print``(``"\nSecond polynomial is "``);``printPoly(B, n);`` ` `prod ``=` `multiply(A, B, m, n);`` ` `print``(``"\nProduct polynomial is "``);``printPoly(prod, m``+``n``-``1``);`` ` `# This code is contributed by chandan_jnu`

## C#

 `// C# program to multiply two polynomials``using` `System;`` ` `class` `GFG ``{`` ` `    ``// A[] represents coefficients ``    ``// of first polynomial``    ``// B[] represents coefficients ``    ``// of second polynomial``    ``// m and n are sizes of A[] ``    ``// and B[] respectively``    ``static` `int``[] multiply(``int` `[]A, ``int` `[]B, ``                            ``int` `m, ``int` `n) ``    ``{``        ``int``[] prod = ``new` `int``[m + n - 1];`` ` `        ``// Initialize the product polynomial``        ``for` `(``int` `i = 0; i < m + n - 1; i++) ``        ``{``            ``prod[i] = 0;``        ``}`` ` `        ``// Multiply two polynomials term by term``        ``// Take ever term of first polynomial``        ``for` `(``int` `i = 0; i < m; i++)``        ``{``            ``// Multiply the current term of first polynomial``            ``// with every term of second polynomial.``            ``for` `(``int` `j = 0; j < n; j++) ``            ``{``                ``prod[i + j] += A[i] * B[j];``            ``}``        ``}`` ` `        ``return` `prod;``    ``}`` ` `    ``// A utility function to print a polynomial``    ``static` `void` `printPoly(``int` `[]poly, ``int` `n)``    ``{``        ``for` `(``int` `i = 0; i < n; i++) {``            ``Console.Write(poly[i]);``            ``if` `(i != 0) {``                ``Console.Write(``"x^"` `+ i);``            ``}``            ``if` `(i != n - 1) {``                ``Console.Write(``" + "``);``            ``}``        ``}``    ``}`` ` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``         ` `        ``// The following array represents ``        ``// polynomial 5 + 10x^2 + 6x^3``        ``int` `[]A = {5, 0, 10, 6};`` ` `        ``// The following array represents``        ``// polynomial 1 + 2x + 4x^2``        ``int` `[]B = {1, 2, 4};``        ``int` `m = A.Length;``        ``int` `n = B.Length;`` ` `        ``Console.WriteLine(``"First polynomial is n"``);``        ``printPoly(A, m);``        ``Console.WriteLine(``"nSecond polynomial is n"``);``        ``printPoly(B, n);`` ` `        ``int``[] prod = multiply(A, B, m, n);`` ` `        ``Console.WriteLine(``"nProduct polynomial is n"``);``        ``printPoly(prod, m + n - 1);``    ``}``}`` ` `// This code has been contributed by 29AjayKumar `

## PHP

 ``

## Javascript

 ``

Output

```First polynomial is 5 + 0x^1 + 10x^2 + 6x^3
Second polynomial is 1 + 2x^1 + 4x^2
Product polynomial is 5 + 10x^1 + 30x^2 + 26x^3 + 52x^4 + 24x^5```

The time complexity of the above solution is O(mn). If the size of two polynomials same, then the time complexity is O(n2).

Auxiliary Space: O(m + n)

Can we do better?
There are methods to do multiplication faster than O(n2) time. These methods are mainly based on divide and conquer. Following is one simple method that divides the given polynomial (of degree n) into two polynomials one containing lower degree terms(lower than n/2) and the other containing higher degree terms (higher than or equal to n/2)

```Let the two given polynomials be A and B.
For simplicity, Let us assume that the given two polynomials are of
same degree and have degree in powers of 2, i.e., n = 2i

The polynomial 'A' can be written as A0 + A1*xn/2
The polynomial 'B' can be written as B0 + B1*xn/2

For example 1 + 10x + 6x2 - 4x3 + 5x4 can be
written as (1 + 10x) + (6 - 4x + 5x2)*x2

A * B  = (A0 + A1*xn/2) * (B0 + B1*xn/2)
= A0*B0 + A0*B1*xn/2 + A1*B0*xn/2 + A1*B1*xn
= A0*B0 + (A0*B1 + A1*B0)xn/2 + A1*B1*xn  ```

So the above divide and conquer approach requires 4 multiplications and O(n) time to add all 4 results. Therefore the time complexity is T(n) = 4T(n/2) + O(n). The solution of the recurrence is O(n2) which is the same as the above simple solution.
The idea is to reduce the number of multiplications to 3 and make the recurrence as T(n) = 3T(n/2) + O(n)

How to reduce the number of multiplications?
This requires a little trick similar to Strassen’s Matrix Multiplication. We do the following 3 multiplications.

```X = (A0 + A1)*(B0 + B1) // First Multiplication
Y = A0B0  // Second
Z = A1B1  // Third

The missing middle term in above multiplication equation A0*B0 + (A0*B1 +
A1*B0)xn/2 + A1*B1*xn can obtained using below.
A0B1 + A1B0 = X - Y - Z  ```

In-Depth Explanation
Conventional polynomial multiplication uses 4 coefficient multiplications:

`(ax + b)(cx + d) = acx2 + (ad + bc)x + bd`

However, notice the following relation:

`(a + b)(c + d) = ad + bc + ac + bd`

The rest of the two components are exactly the middle coefficient for the product of two polynomials. Therefore, the product can be computed as:

```(ax + b)(cx + d) = acx2 +
((a + b)(c + d) - ac - bd )x + bd```

Hence, the latter expression has only three multiplications.
So the time taken by this algorithm is T(n) = 3T(n/2) + O(n)
The solution of the above recurrence is O(nLg3) which is better than O(n2).
We will soon be discussing the implementation of the above approach.
There is an O(nLogn) algorithm also that uses Fast Fourier Transform to multiply two polynomials (Refer to this and this for details)

My Personal Notes arrow_drop_up