In this chapter we will learn to find HCF of given polynomials using factorization method.

While finding HCF, we are basically looking for highest factor which will completely divide all the given polynomials.

## Finding HCF of polynomials

To find the HCF, **follow the below steps;**

(a)** Factorize each of the given polynomials individually** into smaller components.

(b)** Find the common factors **present in each polynomial and then **select the one with lowest power**.

(c) **Combine all the common factors with lowest power **and you will get the HCF.

I hope you understood the above three steps. Let us now solve some problems for further clarity.

**Example 01**

Find HCF of given polynomials.

\mathtt{\Longrightarrow \ x^{2} -4}\\\ \\ \mathtt{\Longrightarrow \ x( x-2)^{2} +y^{2}( x-2)^{2}}

**Solution**

(a) **Factorize each of the polynomial into smaller components**.

(i) \mathtt{ \ \ x^{2} -4}

Using the formula;

\mathtt{a^{2} -b^{2} =( a-b) \ ( a+b)}

Factorizing the polynomial, we get;

\mathtt{\Longrightarrow \ x^{2} -4}\\\ \\ \mathtt{\Longrightarrow \ ( x-2)( x+2)}

Factorizing second polynomial

(ii) \mathtt{ \ x( x-2)^{2} +y^{2}( x-2)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( x-2)^{2}\left( x+y^{2}\right)}

**(b) Find the common factors and select the one with lowest power.**

Given below is the factorized form of given polynomials.

\mathtt{\Longrightarrow \ ( x-2)( x+2)} \\\ \\ \mathtt{\Longrightarrow \ ( x-2)^{2}\left( x+y^{2}\right)}

Note that (x – 2) is the common factor present in both the factorization.

So we will select the factor (x – 2) with lowest power.

Lowest power of (x – 2) ⟹ 1

Hence, (x – 2) is the HCF of given polynomials.

**Example 02**

Find the HCF of below polynomials.

\mathtt{\Longrightarrow \ 9\ +\ 27x}\\\ \\ \mathtt{\Longrightarrow \ 6+18x+\ x^{2} +3x^{3} \ } **Solution**

(a) **Factorizing each of the polynomial into smaller components**.

(i) \mathtt{ \ 9\ +\ 27x}\\\ \\ \mathtt{\Longrightarrow \ 9\ ( 1+3x)}

(ii) \mathtt{\ 6+18x+\ x^{2} +3x^{3} \ }\\\ \\ \mathtt{\Longrightarrow \ 6( 1+3x) +x^{2}( 1+3x)}\\\ \\ \mathtt{\Longrightarrow \ ( 1+3x)\left( 6+x^{2}\right)}

**(b) Find the common factor and select its lowest power.**

The calculated factors are;

\mathtt{\Longrightarrow \ ( 1+3x)\left( 6+x^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ 9\ ( 1+3x)}

The factor (1+3x) is the common among given polynomial.

Lowest power of (1 + 3x) ⟹ 1.

Hence, **(1+3x) is the HCF of given polynomials.**

**Example 03**

Find the HCF of given polynomials

\mathtt{\Longrightarrow \ \left( x^{2} +10x+25\right)( x+3)}\\\ \\ \mathtt{\Longrightarrow \ ( x+5)^{3}( 5x+2)}

**Solution**

(a) **Factorizing each of the polynomial into smaller components.**

(i) \mathtt{\left( x^{2} +10x+25\right)( x+3)}

Referring to the formula;

\mathtt{( a+b)^{2} =a^{2} +b^{2} +2ab}

Using the formula;

\mathtt{\Longrightarrow \ \left( x^{2} +2.x.5\ +25\right) \ ( x+3)}\\\ \\ \mathtt{\Longrightarrow \ ( x+5)^{2}( x+3)}

(ii) \mathtt{( x+5)^{3}( 5x+2)}

This polynomial is already factorized. So we don’t need to do anything further.

**(b) Find the common factor and select its lowest power**.

Given below are factorized polynomials;

\mathtt{\Longrightarrow \ ( x+5)^{2}( x+3) \ }\\\ \\ \mathtt{\Longrightarrow \ ( x+5)^{3}( 5x+2)}

Here (x + 5) is the common factor present in both the polynomials.

Lowest power of (x+5) ⟹ 2

Hence, \mathtt{( x+5)^{2}} is the HCF of given polynomials.

**Example 04**

Find HCF of below polynomials

\mathtt{\Longrightarrow \ \left( x^{4} -1\right)}\\\ \\ \mathtt{\Longrightarrow \ x^{2} -2x+1}\\\ \\ \mathtt{\Longrightarrow \ x^{3} -x^{2} +4x\ -\ 4} **Solution****Factorize each of the polynomial into small components.**

(i) \mathtt{\left( x^{4} -1\right)}\\\ \\ \mathtt{\Longrightarrow \ \left(\left( x^{2}\right)^{2} -1\right)}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2} -1\right)\left( x^{2} +1\right)}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)( x+1)\left( x^{2} +1\right)}

(ii) \mathtt{x^{2} -2x+1}

Referring the formula;

\mathtt{( a-b)^{2} =a^{2} +b^{2} -2ab}

Applying the formula, we get;

\mathtt{\Longrightarrow \ ( x-1)^{2}}

(iii) \mathtt{ \ x^{3} -x^{2} +4x\ -\ 4}\\\ \\ \mathtt{\Longrightarrow \ x^{2}( x-1) +4( x-1)}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)\left( x^{2} +4\right)}

**Find common factor and select its lowest power.**

Given below are the factorized polynomials.

\mathtt{\Longrightarrow \ ( x-1)( x+1)\left( x^{2} +1\right)}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)\left( x^{2} +4\right)}

Note that (x – 1) is the common factor present in all three polynomials.

Lowest factors of (x – 1) is 1.

Hence, (x – 1) is the HCF of given polynomial.

**Example 05**

Find HCF of given polynomials

\mathtt{\Longrightarrow \ \left( 9x^{2} -81\right)}\\\ \\ \mathtt{\Longrightarrow \ 3.\left( x^{2} +6x+9\right)}

**Solution**

Factorize each of the polynomial into smaller components.

(i) \mathtt{\left( 9x^{2} -81\right)} \\\ \\ \mathtt{\Longrightarrow \ 9\ \left( x^{2} -9\right)}\\\ \\ \mathtt{\Longrightarrow \ 9\ ( x-3) \ ( x+3)}

\mathtt{3.\left( x^{2} +6x+9\right)}

Referring the formula;

\mathtt{( a+b)^{2} =a^{2} +b^{2} +2ab}

Using the formula, we get;

\mathtt{\Longrightarrow \ 3\ \left( x+3.2.x+3^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ 3\ ( x+3)^{2}}

**Find the common factors among the polynomial and select the one with lowest power**.

Given below is the factorized form of above polynomials

\mathtt{\Longrightarrow \ 9\ ( x-3) \ ( x+3)}\\\ \\ \mathtt{\Longrightarrow \ 3\ ( x+3)^{2}}

Here both constants and variable are common among polynomials.

To get the HCF, find the common factors of constant and variables independently.**HCF of constants.**

HCF (3, 9) = 3.**HCF of variables **

Here (x +3) is the common factor among the polynomials.

The lowest power of (x + 3) is 1.

Hence, HCF of variables is (x + 3)**Combining HCF of constants and variable we get 3(x + 3).**

Hence, **3(x + 3) is the solution**.