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H.C.F. of Polynomials by Division Method | How to find HCF of Polynomials?

Do you want to calculate the highest common factor of polynomials by using the division method? If yes then stay on this page. Here we are providing the detailed step by step procedure to find the G.C.F of polynomials by division method. Along with the steps you can also check some solved example questions from the below sections.

How to find the Highest Common Factor of Polynomials by Division Method?

We are using the division method to find the G.C.F of polynomials when the polynomials have the highest factor and it is difficult to compute the HCF. Then follow the steps and instructions provided below and get the answer easily.

  • Let us take two polynomials f(x), g(x).
  • Divide the polynomials f(x) / g(x) to get f(x) = g(x) * q(x) + r(x). Here the degree of g(x) > degree of r(x).
  • If the remainder r(x) is zer0, then g(x) is the highest common factor of polynomials.
  • If the remainder is not equal to zero, then again divide g(x) by r(x) to obtain g(x) = r(x) * q(x) + r1(x). Here if r1(x) is zero then required H.C.F is r(x).
  • If it is not zero, then continue the process until we get zero as a remainder.

Solved Examples on GCF of Polynomials

Example 1.

Find the H.C.F of x⁴ + 4x³ + x – 10 and x² + 3x – 5 by using the division method?

Solution:

Given polynomials are f(x) = x⁴ + 4x³ + x – 10, g(x) = x² + 3x – 5

Arranging the polynomials in the descending order f(x) = 1x⁴ + 4x³ + 0x² + x – 10

By using the division method,

The remainder is zero.

So, the required H.C.F of x⁴ + 4x³ + x – 10 and x² + 3x – 5 is x² + 3x – 5.

Example 2.

Calculate the greatest common factor of 8x³ – 10x² – x + 3 and x – 1 by using the division method?

Solution:

Given two polynomials are 8x³ – 10x² – x + 3 and x – 1

Let us take f(x) = 8x³ – 10x² – x + 3, g(x) = x – 1

Divide f(x) by g(x)

The remainder is zero.

So, the required H.C.F of 8x³ – 10x² – x + 3 and x – 1 is x – 1.

Example 3.

Find the HCF of the following pairs of polynomials using the division algorithm

2 x³ + 2 x² + 2 x + 2 , 6 x³ + 12 x² + 6 x + 12?

Solution:

Given two polynomials are 2 x³ + 2 x² + 2 x + 2 , 6 x³ + 12 x² + 6 x + 12

Let f(x) = 2 x³ + 2 x² + 2 x + 2 , g(x) = 6 x³ + 12 x² + 6 x + 12

Take 2 common from the first polynomial.

f(x) = 2(x³ + x² + x + 1)

Take 6 common from the second polynomial.

g(x) = 6(x³ + 2x² + x + 2)

Divide f(x) by g(x)

 

The remainder is not zero. So, we have to repeat this long division once again.

Then divide g(x) by r1(x). i.e x³ + 2x² + x + 2 by x² + 1

The remainder is zero.

The H.C.F of 2, 6 is 2.

So, required H.C.F of 2 x³ + 2 x² + 2 x + 2 , 6 x³ + 12 x² + 6 x + 12 is 2 ( x² + 1).

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