HCF of Algebraic Expressions

HCF of Algebraic Expressions

 

HCF is the highest common factor in all the given algebraic expressions. So, HCF can exactly divide each of the given algebraic expressions. 

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For example, 3x²y and 6xy² are two algebraic expressions. In these expressions, 3xy is the highest common factor in both expressions. So, HCF of 3x²y and 6xy² is 3xy.

 

While finding HCF, we should use the following steps:

 

Steps:

 

1.     Find the factors of the given algebraic expressions.

2.     Take common factors of the given algebraic expressions.

3.     Express the common factors in the product form which is the required HCF.

4.     If there is no common factor in the given expressions, then HCF is 1 since 1 is the least factor of every expression.

 

The concept of HCF will be more clear from the following worked-out examples.

Worked Out Examples

Example 1: Find HCF of 9a³b² and 15b³c²d.

 

Solution:

 

Here,

 

1^st expression = 9a³b²

       = 3 × 3 × a × a × a × b × b

 

2^nd expression = 15b³c²d

       = 3 × 5 × b × b × b × c × c × d

 

∴ HCF = 3 × b × b = 3b²

 

 

Example 2: Find HCF of 3(a² – b²) and 12a²b(a² + 2ab + b²)

 

Solution:

 

Here,

 

1^st expression = 3(a² – b²)

                  = 3(a + b)(a – b)

 

2^nd expression = 12a²b(a² + 2ab + b²)

                  = 2 × 2 × 3 × a × a × b(a + b)²

                  = 2 × 2 × 3 × a × a × b(a + b)(a + b)

 

∴ HCF = 3(a + b)

Example 3: Find HCF of a³ – b³, 3a² + 2ab – 5b² and a⁴ – b⁴

 

Solution:

 

Here,

 

1^st expression = a³ – b³

                     = (a – b)(a² + ab + b²)

 

2^nd expression = 3a² + 2ab – 5b²

                      = 3a² + (5 – 3)ab – 5b²

                      = 3a² + 5ab – 3ab – 5b²

                      = a(3a + 5b) – b(3a + 5b)

                      = (3a + 5b)(a – b)

 

3^rd expression = a⁴ – b⁴

                     = (a²)² – (b²)²

                     = (a² + b²)(a² – b²)

                     = (a² + b²)(a + b)(a – b)

 

∴ HCF = (a – b)

Example 4: Find HCF of x² – 5x + 6, x² + 4x – 12 and x² – 2x

 

Solution:

 

Here,

 

1^st expression = x² – 5x + 6

                     = x² – (3 + 2)x + 6

                     = x² – 3x – 2x + 6

                     = x(x – 3) – 2(x – 3)

                     = (x – 3)(x – 2)

 

2^nd expression = x² + 4x – 12

                      = x² + (6 – 2)x – 12

                      = x² + 6x – 2x - 12

                      = x(x + 6) – 2(x + 6)

                      = (x + 6)(x – 2)  

 

3^rd expression = x² – 2x

                     = x(x – 2)

 

∴ HCF = (x – 2)

Example 5: Find HCF of a² – b² – 2bc – c², b² – c² – 2ca – a² and c² – a² – 2ab – b²

 

Solution:

 

Here,

 

1^st expression = a² – b² – 2bc – c²

                     = a² – (b² + 2bc + c²)

                     = a² – (b + c)²

                     = (a + b + c)(a – b – c)

 

2^nd expression = b² – c² – 2ca – a²

                      = b² – (c² + 2ca + a²)

                      = b² – (c + a)²

                      = (b + c + a)(b – c – a)

                      = (a + b + c)(b – c – a)

         

3^rd expression = c² – a² – 2ab – b²

                      = c² – (a² + 2ab + b²)

                      = c² – (a + b)²

                      = (c + a + b)(c – a – b)

                      = (a + b + c)(c – a – b)

 

∴ HCF = (a + b + c)

 

 

Example 6: Find HCF of a³ – 2a²b + 2ab² – b³, a⁴ + b⁴ + a²b², 4a⁴b + 4ab⁴

 

Solution:

 

Here,

 

1^st expression = a³ – 2a²b + 2ab² – b³

                     = a³ – b³ – 2a²b + 2ab²

                     = (a – b)(a² + ab + b²) – 2ab(a – b)

                     = (a – b)(a² + ab + b² – 2ab)

                     = (a – b)(a² – ab + b²)

 

2^nd expression = a⁴ + b⁴ + a²b²

                      = (a²)² + 2a²b² + (b²)² – a²b²

                      = (a² + b²)² –(ab)²

                      = (a² + b² + ab)(a² + b² – ab)

                      = (a² + ab + b²)(a² – ab + b²)

         

3^rd expression = 4a⁴b + 4ab⁴

                      = 4ab(a³ + b³)

                      = 4ab(a + b)(a² – ab + b²)

 

∴ HCF = (a² – ab + b²)

Example 7: Find HCF of a² – 18a – 19 + 20b – b², a² + a – b² + b, 4a² – 4b² + 8b - 4

 

Solution:

 

Here,

 

1^st expression = a² – 2.a.9 + 9² – 100 + 20b – b²

                     = (a – 9)² – (10² – 2.10.b + b²)

                     = (a – 9)² – (10 – b)²

                     = {(a – 9) + (10 – b)}{(a – 9) – (10 – b)}

                     = (a – 9 + 10 – b)(a – 9 – 10 + b)

                     = (a – b + 1)(a + b – 19)

 

2^nd expression = a² + a – b² + b

                      = a² – b² + a + b

                      = (a + b)(a – b) + 1(a + b)

                      = (a + b)(a – b + 1)

 

3^rd expression = 4a² – 4b² + 8b - 4

                      = 4(a² – b² + 2b – 1)

                      = 4{a² – (b² – 2b + 1)}

                      = 4{a² – (b – 1)²}

                      = 4(a + b – 1)(a – b + 1)

 

∴ HCF = (a + b – 1)

 

 

Example 8: Find HCF of a(1 + ab) + a(a – b²) – b (1 + b), 5a⁴b – 5ab⁴ and 2ab³ – 2a³b

 

Solution:

 

Here,

 

1^st expression = a(1 + ab) + a(a – b²) – b(1 + b)

                     = a + a²b + a² – ab² – b – b²

                     = a² – b² + a – b + a²b – ab²

                     = (a + b)(a – b) + 1(a – b) + ab(a – b)

                     = (a – b)(a + b +1 + ab)

 

2^nd expression = 5a⁴b – 5ab⁴

                      = 5ab(a³ – b³)

                      = 5ab(a – b)(a² + ab + b²)

 

3^rd expression = 2ab³ – 2a³b

                      = 2ab(b² – a²)

                      = -2ab(a² – b²)

                      = -2ab(a + b)(a – b)

 

∴ HCF = (a – b)

 

 

If you have any question or problems regarding the HCF of algebraic expressions, you can ask here, in the comment section below.

 

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