2.5 HCF(GCD) and LCM of Algebraic terms:
In the case of numbers, we have understood what HCF
(GCD) of two or more numbers is. It is the highest among all common factors of
the numbers.
Let us take an example of 4 numbers 4, 8, 20, 16.
We know by inspection that 2 and 4 are common factors of all the four numbers
and highest among them is 4.
Therefore HCF of 4, 8, 20, and 16 is 4.
HCF is useful in
simplifying fractions.
Let us examine how we simplify the fraction 30/48.
First, we find HCF of 30 and 48 which happens to be
6.Then we express both numerator and denominator as a product of this HCF.
(30 = 6*5 and 48 = 6*8)
Next, we cancel this common factor from both
numerator and denominator to get the proper fraction
(I.e. 30/48 = 6*5/6*8 = 5/8)
What is LCM?
It is the lowest number among all the common multiples of the numbers.
Let us take the same example of numbers 4, 8, 20,
16.
We notice that 80, 160, 320
are all common
multiples of the above numbers. The least among these common multiples is 80
and 80 is called LCM
LCM is useful in adding fractions.
Let us add 1/4, 1/8,
1/20
LCM of 4,8,20 is 40
1/4 = 10/40
1/8 = 5/40
1/20 = 2/40
_{}1/4+1/8+1/20 = 10/40+5/40+2/40 = (10+5+2)/40 = 17/40
For finding HCF and LCM of algebraic terms, we
follow same method that we use in the case of finding HCF and LCM of numbers.
Let us recollect Finding HCF and LCM of numbers.
Finding
HCF
Step |
Procedure |
1 |
List
all the numbers |
2 |
On
the left side write the smallest common divisor for all of these numbers |
3 |
On
the 2^{nd} line write the quotients |
4 |
On
the left side write the common divisor for all of these quotients as got in the step2 |
5 |
Repeat
these steps till there are no more common factors for
all the numbers |
Product of the common divisors (appearing on the left
hand side) is HCF of the numbers
Find HCF of
16,24,20
2 | 16,24,20
2 | 8,12,10
4, 6, 5
We stop further divisions as the remaining numbers(all) ( 4 6 and 5) do not have common factors other
than 1
Therefore (2*2) = 4 is HCF of 16,24,20
Finding
LCM
Step |
Procedure |
1 |
List
all the numbers |
2 |
On
the left side write the common divisor for any of
these numbers |
3 |
On
the 2^{nd} line write the quotients |
4 |
On
the left side write the common divisor for any
of these quotients as got in the step2 |
5 |
Repeat
these steps till there are no more common factors among any two numbers |
Product of the common divisors and the remaining
numbers on the last line is LCM
Find LCM of
16,24,20
2 | 16,24,20
2 | 8,12,10
2 | 4,6,5
| 2,3,5
We stop here further divisions as none of the remaining numbers (2,3,5) have common factors other than 1
Thus LCM = (2*2*2)*(2*3*5) = 240
Also observe that, product of HCF and LCM of
2 numbers = Product of 2 numbers.
This relationship holds good for algebraic terms also, hence if two terms and their HCF or LCM
are given we can find their LCM or HCF respectively.
2.5 Problem 1 : Find HCF of 16a^{4}b^{3}x^{3}, 24b^{2}m^{3}n^{4}y, 20a^{2}b^{3}nx^{3}
Solution:
1.HCF of co-efficients
(16,24,20) is 4
The variables are a^{4}b^{3}x^{3},
b^{2}m^{3}n^{4}y, a^{2}b^{3}nx^{3}
and we notice that b is a common factor of the variables
4b | 16a^{4}b^{3}x^{3}, 24b^{2}m^{3}n^{4}y,
20a^{2}b^{3}nx^{3 }( We start dividing by 4b as it
is the common factor for all the
terms)
b | 4a^{4}b^{2}x^{3}, 6bm^{3}n^{4}y, 5a^{2}b^{2}nx^{3} (b is a common factor of all terms)^{}
|4a^{4}bx^{3},
6m^{3}n^{4}y, 5a^{2}bnx^{3}
We need to
stop further division as there are no more common factors among all the terms
_{}4b*b= 4b^{2
} is HCF of the term
HCF is useful for taking common
factors out in addition/subtractions and in simplifying expressions
Let us simplify 16a^{4}b^{3}x^{3}+24b^{2}m^{3}n^{4}y- 20a^{2}b^{3}nx^{3}
16a^{4}b^{3}x^{3}+24b^{2}m^{3}n^{4}y- 20a^{2}b^{3}nx^{3}
=4b^{2}(4a^{4}bx^{3}+6m^{3}n^{4}y-
5a^{2}bnx^{3})
2.5 Problem 2 : Find HCF and LCM of 6x^{2}y^{3}, 8x^{3}y^{2}, 12x^{4}y^{3}, 10x^{3}y^{4}
Solution:
1.HCF of
co-efficients (6,8,12,10) is 2
The variables are x^{2}y^{3}, x^{3}y^{2},
x^{4}y^{3}, x^{3}y^{4}and we notice that x is a
common factor of the variables
a)
Finding HCF
2x | 6x^{2}y^{3}, 8x^{3}y^{2},
12x^{4}y^{3}, 10x^{3}y^{4}^{ }(2x is a common factor of all the terms)
x | 3xy^{3}, 4x^{2}y^{2}, 6x^{3}y^{3}, 5x^{2}y^{4 }
y |3y^{3}, 4xy^{2}, 6x^{2}y^{3}, 5xy^{4}
y |3y^{2}, 4xy,
6x^{2}y^{2}, 5xy^{3}
3y, 4x, 6x^{2}y, 5xy^{2}
We need to stop further division as there are no
more common factors among all the terms
_{}2x*x*y*y = 2x^{2}y^{2} is HCF of variable
Use of HCF:
Let us simplify 6x^{2}y^{3}+8x^{3}y^{2}-12x^{4}y^{3}+10x^{3}y^{4}
6x^{2}y^{3}+8x^{3}y^{2}-12x^{4}y^{3}+10x^{3}y^{4}
= 2x^{2}y^{2}(3y+4x-6x^{2}y+5xy^{2})
a)
Finding LCM
2x | 6x^{2}y^{3}, 8x^{3}y^{2},
12x^{4}y^{3}, 10x^{3}y^{4}^{ }(2x is a common factor of all the terms)
x | 3xy^{3}, 4x^{2}y^{2}, 6x^{3}y^{3}, 5x^{2}y^{4}
y |3y^{3}, 4xy^{2}, 6x^{2}y^{3}, 5xy^{4}
y |3y^{2}, 4xy,
6x^{2}y^{2}, 5xy^{3}
Y |3y,
4x, 6x^{2}y^{ }, 5xy^{2}
x_{ }_{ }| 3, 4x 6x^{2}, 5xy^{}
2 | 3,
4 6x 5y
_{ญญญญญญญญญญญญญญ }3_{ }| 3, 2
3x 5y
| 1,
2 x 5y
We stop further division, as there are no more
common factors among any two terms.
Thus LCM = ( 2x*x*y*Y)*(Y*x*2*3*2*x*5y) =2x^{2}y^{2}* 60x^{2}y^{2} = 120x^{4}y^{4}^{}
Use of LCM:
Simplify (1/6x^{2}y^{3})+(1/8x^{3}y^{2})-(1/12x^{4}y^{3}
)+(1/10x^{3}y^{4})
Note
(1/6x^{2}y^{3}) = (20x^{2}y/120x^{4}y^{4})
(1/8x^{3}y^{2}) = (15xy^{2}/120x^{4}y^{4})
(1/12x^{4}y^{3}) = (10y/120x^{4}y^{4})
(1/10x^{3}y^{4}) = (12x/120x^{4}y^{4})
_{}(1/6x^{2}y^{3})+(1/8x^{3}y^{2})-(1/12x^{4}y^{3}
)+(1/10x^{3}y^{4})
= (20x^{2}y+15xy^{2}-10y+12x)๗(120x^{4}y^{4})
2.5 Summary of learning
No |
Points studied |
1 |
Finding
HCF and LCM of algebraic terms by division method |