HIGHEST COMMON FACTOR OF ALGERAIC EXPRESSIONS

Find the HCF of :

Problem 1 :

5(a + 7) and (a + 1) (a + 7)

Solution :

Factor of 5(a + 7) and (a + 1) (a + 7)

5(a + 7) = 5 ⋅ (a + 7)

(a + 1) (a + 7) = (a + 1) ⋅ (a + 7)

= (a + 7)

HCF of 5(a + 7) and (a + 1) (a + 7) is (a + 7).

Problem 2 :

3(2 + b)² and 3(6 + b) (2 + b)

Solution :

Factor of 3(2 + b)² and 3(6 + b) (2 + b)

3(2 + b)² = 3 ⋅ (2 + b) ⋅ (2 + b)  

3(6 + b) (2 + b) = 3 ⋅ (6 + b) ⋅ (2 + b) 

= 3 ⋅ (2 + b) 

HCF of 3(2 + b)² and 3(6 + b) (2 + b) is 3(2 + b).

Problem 3 :

x²(x – 4) and x(x – 4)

Solution :

Factor of x²(x – 4) and x(x – 4)

x²(x – 4) = x ⋅ x ⋅ (x – 4)

 x(x – 4) = x ⋅ (x – 4)

= x ⋅ (x – 4)

HCF of x²(x – 4) and x(x – 4) is x(x – 4).

Problem 4 :

15(x – 2)² and 5(x – 2) (x + 3)

Solution :

Factor of 15(x – 2)² and 5(x – 2) (x + 3)

15(x – 2)² = 5 ⋅ 3 ⋅ (x – 2) ⋅ (x – 2)

5(x – 2) (x + 3) = 5 ⋅ (x – 2) ⋅ (x + 3)

= 5 ⋅ (x – 2)

HCF of 15(x – 2)² and 5(x – 2) (x + 3) is 5(x – 2).

Problem 5 :

6(x – 1)² and 16(x – 7) (x - 1)

Solution :

Factor of 6(x – 1)² and 16(x – 7) (x - 1)

6(x – 1)² = 3 ⋅ 2 ⋅ (x – 1) ⋅ (x – 1)

 16(x – 7) (x - 1) = 8 ⋅ 2 ⋅ (x – 7) ⋅ (x – 1)

= 2 ⋅ (x – 1)

HCF of 6(x – 1)² and 16(x – 7)(x - 1) is 2(x – 1).

Problem 6 :

9y(y + 1) and 6y(y + 1)²

Solution :

Factor of 9y(y + 1) and 6y(y + 1)²

9y(y + 1) = 3 ⋅ 3 ⋅ y ⋅ (y + 1)

6y(y + 1)² = 2 ⋅ 3 ⋅ y ⋅ (y + 1) ⋅ (y + 1)

= 3 ⋅ y ⋅ (y + 1)

HCF of 9y(y + 1) and 6y(y + 1)² is 3y(y + 1).