HCF of Algebraic Expressions

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

HCF in Venn diagram

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 any common factor in the given expressions, then HCF is 1 since 1 is a factor of any expression.

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


Workout Examples

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

Solution: Here,

                        1st expression = 9a3b2 = 3 × 3 × a × a × a × b × b

                        2nd 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,

                        1st expression = 3(a2 – b2)

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

                        2nd 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,

                        1st expression = a3 – b3

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

                        2nd expression = 3a2 + 2ab – 5b2

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

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

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

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

                        3rd 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,

                        1st expression = x2 – 5x + 6

                                               = x2 – (3 + 2)x + 6

                                               = x2 – 3x – 2x + 6

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

                                               = (x – 3)(x – 2)

                        2nd expression = x2 + 4x – 12

                                                = x2 + (6 – 2)x – 12

                                                = x2 + 6x – 2x - 12

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

                                                = (x + 6)(x – 2)          

                        3rd 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,

                        1st expression = a2 – b2 – 2bc – c2

                                               = a2 – (b2 + 2bc + c2)

                                               = a2 – (b + c)2

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

                        2nd 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)          

                        3rd 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,

                        1st 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)

                        2nd 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)

                        3rd 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,

                        1st 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)

                        2nd expression = a2 + a – b2 + b

                                                = a2 – b2 + a + b

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

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

                        3rd 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,

                        1st 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)

                        2nd expression = 5a4b – 5ab4

                                                = 5ab(a3 – b3)

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

                        3rd expression = 2ab3 – 2a3b

                                                = 2ab(b2 – a2)

                                                = -2ab(a2 – b2)

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

                        HCF = (a – b)

 

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