**Vedic Maths Trick to find the HCF of Algebraic Expressions**

To appreciate the Vedic Maths process of finding the HCF you first need to know the other methods taught in school. I am giving you two other methods to compare with.

Example 1: Find the H.C.F. of x^2 + 5x + 4 and x^2 + 7x + 6.

Example 1: Find the H.C.F. of x^2 + 5x + 4 and x^2 + 7x + 6.

1. Factorization method:x^2 + 5x + 4 = (x + 4) (x + 1)

x^2 + 7x + 6 = (x + 6) (x + 1)

H.C.F. is ( x + 1 ).

2. Continuous division process.

x^2 + 5x + 4 ) x^2 + 7x + 6 ( 1

x^2 + 5x + 4___________2x + 2 ) x^2 + 5x + 4 ( ½x

x^2 + x__________4x + 4 ) 2x + 2 ( ½2x + 2______0

Thus 4x + 4 i.e., ( x + 1 ) is H.C.F.

Now see Vedic Maths way of finding HCF of 2 algebraic expressions.

i.e. x+1 is the HCF

Isn’t it much simpler than the above 2 methods.

Now see some more examples –

In example no. 2 . -1) 2x + 3 . If we take -1 is common factor from -2x + 3 we get 2x – 3 as a H.C.F. Please let me know am I correct or not . As I have solved the ex. by Factorization and Continuous Division Method . Please let me know the correct answer .

its very much interesting.i think its more easy m,ethod for calculation

finding h.c.f of numbers with variables how can be done

ex: 12xy , 24x and like that its urgent please reply

this is in ref to Ex -2

______

-1) -2x +3

_______

2x -3

please see if what I’m trying to point out is correct !!

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Thanks for the nice trick. I feel it will be useful for me in CAT Preparation. Do others agree