HOW TO FIND HCF OF ALGEBRAIC TERMS

Find the HCF of :

Example 1 :

4x and 12

Solution :

Factor of 4x and 12

4x  =  2 ⋅ 2 ⋅ x

12  =  2 ⋅ 2 ⋅ 3 

=  2 ⋅ 2

=  4

HCF of 4x and 12 is 4.

Example 2 :

3x and 6

Solution :

Factor of 3x and 6

3x  =  3 ⋅ x

6  =  2 ⋅ 3 

=  3

HCF of 3x and 6 is 3.

Example 3 :

3ab and 6b

Solution :

Factor of 3ab and 6b

3ab  =  3 ⋅ a ⋅ b

6b  =  2 ⋅ 3 ⋅ b

=  3b

HCF of 3ab and 6b is 3b.

Example 4 :

4y and 4xy

Solution :

Factor of 4y and 4xy

4y  =  2 ⋅ 2 ⋅ y

4xy  =  2 ⋅ 2 ⋅ x ⋅ y

=  2 ⋅ 2 ⋅ y

=  4y 

HCF of 4y and 4xy is 4y.

Example 5 :

6x² and 2x

Solution :

Factor of 6x² and 2x

6x²  =  2 ⋅ 3 ⋅ x ⋅ x

2x  =  2 ⋅ x

=  2x 

HCF of 6x² and 2x is 2x.

Example 6  :

3y and 9y²

Solution :

Factor of 3y and 9y²

3y  =  3 ⋅ y

9y²  =  3 ⋅ 3 ⋅ y ⋅ y

=  3y 

HCF of 3y and 9y² is 3y.

Example 7 :

2(x – 1) and 3(x – 1)

Solution :

Factor of 2(x – 1) and 3(x – 1)

 2(x – 1)  =  2 ⋅ (x – 1)

3(x – 1)  =  3 ⋅ (x – 1)

=  (x – 1)

HCF of 2(x – 1) and 3(x – 1) is (x – 1).

Example 8 :

4(x + 2) and (x + 2)

Solution :

Factor of 4(x + 2) and (x + 2)

 4(x + 2)  =  2 ⋅ 2 ⋅ (x + 2)

x + 2  =  (x + 2)

=  x + 2

HCF of 4(x + 2) and x + 2 is (x + 2).

Example 9 :

3(x + 1) and (x + 1) (x – 4)

Solution :

Factor of 3(x + 1) and (x + 1) (x – 4)

 3(x + 1)  =  3 ⋅ (x + 1)

(x + 1) (x – 4)  =  (x + 1)  ⋅ (x - 4)

=  x + 1

HCF of 3(x + 1) and (x + 1) (x – 4) is (x + 1).

Example 10 :

(x - 2)² and 2(x - 2) (x – 5)

Solution :

Factor of (x - 2)² and 2(x - 2) (x – 5)

 (x - 2)²  =  (x - 2) ⋅ (x - 2)

2(x - 2) (x – 5)  =  2 ⋅ (x – 2) ⋅ (x - 5)

=  (x - 2)

HCF of (x - 2)² and 2(x - 2) (x – 5) is (x - 2).

Example 11 :

2x(5 - x) and x²(5 - x)

Solution :

Factor of 2x(5 - x) and x²(5 - x)

 2x(5 - x)  =  2 ⋅ x ⋅ (5-x)

x²(5 - x)  =  x ⋅ x ⋅ (5-x)

=  x . (5 – x)

=  x(5 – x)

HCF of 2x(5 – x) and x²(5 - x) is x(5 – x).

Example 12 :

(x + 2)² and 5(x - 4) (x + 2)

Solution :

Factor of (x + 2)² and 5(x - 4) (x + 2)

 (x + 2)²  =  (x + 2) ⋅ (x + 2)

5(x - 4) (x + 2)  =  5 ⋅ (x - 4) ⋅ (x + 2)

=  (x + 2)

HCF of (x + 2)² and 5(x - 4) (x + 2) is (x + 2).

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