LCM OF ALGEBRAIC TERMS

Find the least common multiple of the following algebraic terms. 

Example 1 :

x³ y² , xyz 

Solution :

x³ y²  =  x ⋅ x ⋅ x ⋅ y ⋅ y

xyz  =  x ⋅ y ⋅ z

Comparing x terms (LCM) is x³

Comparing y terms (LCM) is y²

So, the least common multiple for the algebraic terms is x³ y² z.

Example 2 :

3x²yz, 4x³ y³

Solution :

3x²yz, 4x³ y³

3x²yz  =  3 ⋅ x ⋅ x ⋅ y ⋅ z

4x³ y³  =  4 ⋅ x ⋅ x ⋅ x ⋅ y ⋅ y⋅ y

Comparing x terms (LCM) is x³

Comparing y terms (LCM) is y³

So, the required LCM is 12x³ y³z.

Example 3 :

a²bc, b²ca , c²a b

Solution :

a²bc, b²ca , c²a b

By comparing the given terms, the least common multiple is

a² b²c²

Example 4 :

66 a⁴b²c³ , 44 a³b⁴c² , 24 a²b³c⁴

Solution :

66 a⁴b²c³, 44 a³b⁴c², 24 a²b³c⁴

66  =  2 ⋅ 3 ⋅ 11

44  =  2² ⋅ 11

24  =  2³ ⋅ 3

Highest common factor of 66, 44 and 24 is 2³ ⋅ 11 ⋅ 3

  =  264

Highest common factor of a⁴b²c³, a³b⁴c² and a²b³c⁴

=  a⁴b⁴c⁴

So, the required LCM is 264a⁴b⁴c⁴.

Example 5 :

 a^(m+1), a^(m+2), a^(m+3)

Solution :

a^(m+1), a^(m+2), a^(m+3)

a^(m+1)  =  a^m ⋅ a

a^(m+2)  =  a^m ⋅ a²

a^(m+3)  =  a^m ⋅ a³

We find a^m in common for all and highest "a" term is a³.

=  a^m⋅ a³

=  a^(m+3)

So, the required LCM is a^(m+3).

Example 6 :

x²y + xy², x² + xy

Solution :

x²y + xy², x² + xy

x²y + xy²  =  xy(x + y)

x² + xy  =  x(x + y)

By comparing the factors, the least common multiple is 

xy(x+y) 

Example 7 :

30 xy, 24 xy² 

Solution :

30 xy = 2 ⋅ 5 ⋅ 3 ⋅ x ⋅ y

24 xy² = 2³ ⋅ 3 ⋅ x ⋅ y² 

To find lcm, we have to select highest term in common terms and terms that we find extra should also be included.

least common multiple  = 2³ ⋅ 5 ⋅ 3 ⋅ x ⋅ y² 

= 120 x y² 

Example 8 :

30ab², 50b² 

Solution :

30 a b² = 2 ⋅ 5 ⋅ 3 ⋅ a ⋅ b²

50 b² = 5 ⋅ 5 ⋅ 2 ⋅ b²

= 5²⋅ 2 ⋅ b²

To find lcm, we have to select highest term in common terms and terms that we find extra should also be included.

least common multiple  = 5² ⋅ 2 ⋅ 3 ⋅ a ⋅ b² 

= 150 a b²

Example 9 :

38x², 18x 

Solution :

38x² = 2 ⋅ 19 ⋅ x ⋅ x

=  2 ⋅ 19 ⋅ x²

18x = 2 ⋅ 3 ⋅ 3 ⋅ x

= 3²⋅ 2 ⋅ x

To find lcm, we have to select highest term in common terms and terms that we find extra should also be included.

least common multiple  = 3² ⋅ 2 ⋅ 19 ⋅ x² 

= 342 x²

Example 10 :

21 b, 45 a b

Solution :

21 b = 3 ⋅ 7 ⋅ b

45 a b = 3 ⋅ 3 ⋅ 5 ⋅ a ⋅ b

= 3²⋅ 5 ⋅ a ⋅ b

To find lcm, we have to select highest term in common terms and terms that we find extra should also be included.

least common multiple  = 3² ⋅ 5 ⋅ 7  a b 

= 315 ab

Example 11 :

32 x², 24 x²y, 16x² y 

Solution :

32 x² = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ x²

= 2⁵ ⋅ x²

24 x² y = 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ x² ⋅ y

= 2³ ⋅ 3 ⋅ x² ⋅ y

16x² y = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ x² ⋅ y

=  2⁴  ⋅ x² ⋅ y

To find lcm, we have to select highest term in common terms and terms that we find extra should also be included.

least common multiple  = 2⁵ ⋅ x² ⋅ y

= 32 x² y

Example 12 :

(y - 2)(y + 2) and (y + 2)²

Solution :

(y - 2)(y + 2) ------(1)

(y + 2)² = (y +2) (y + 2) ------(2)

Comparing (1) and (2), we get 

(y + 2) and (y - 2) is extra.

By including everything, we get (y + 2)(y - 2)

So, the least common multiple is (y + 2)(y - 2).

Example 13 :

t, t² - 1 and t² + 5t - 6

Solution :

= t ------(1)

t² - 1

Using algebraic identity for a² - b²

t² - 1 = (t + 1) (t - 1) ------(2)

t² + 5t - 6

By factoring the quadratic, we get

t² + 5t - 6 = (t - 1)(t + 6) ------(3)

Comparing (1), (2) and (3), there is no terms in common.

= t (t - 1) (t + 1) (t + 6)

= t(t² - 1)(t + 6)

So, the least common multiple is t(t² - 1)(t + 6)

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