Algebra 1 Practice

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H.C.F. (Highest common factor) and L.C.M. (Lowest common factor) is the part of algebra 1.
 
Methods to solve H.C.F and L.C.M.: -
·         Division method
·         Factorization method.

 
Let’s practice few problems of algebra 1:-
 

Example: - Find out the L.C.M. and H.C.F. by factorization
i)             X^2 + x, x^3 – x
ii)            X^3 + 2 x^2, x^3 + 3 x^2 + 2 x
 

Solution: -
i)             1st expression = x^2 + x = x ( x + 1)

2nd expression = x^3 – x = x ( x^2 -1) = x (x + 1) (x – 1)
The common factors of the two expressions are x and (x +1).
Therefore H.C.F. = x (x + 1)
 

(x – 1) is the extra factor in the 2nd expression.
Hence the required L.C.M. = x (x + 1) (x – 1)
 

ii)            1st expression = x^3 + 2 x^2= x^2 (x + 2) = x * x * ( x +2)

2nd expression = x^3 + 3 x^2 + 2 x = x(x^2+3x+ 2)=x (x^2 + 2 x + x + 2)
                        =x {x ( x + 2 ) + 1 ( x + 2 )}
                        =x (x + 2) (x + 1)


In both the expressions, the common factors are x and x + 2
Therefor H.C.F. = x (x + 2)


The extra factors are x in the 1st expression and x + 1 in the 2nd expression.
Therefore L.C.M. = x ( x + 2) x (x + 1) = x^2 (x+1)(x+2)
                                                     
 

 

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