FIND LCM OF ALGEBRAIC TERMS WORKSHEET

Find the LCM of the following

1)  x³ y², xyz 

2) 3x²yz, 4x³ y³

3)  a² b c, b² c a , c² a b

4) 66 a⁴ b² c³ , 44 a³ b⁴ c² , 24 a² b³ c⁴

5)  a^(m+1), a^(m+2), a^(m+3)

6)  x² y + xy², x² + xy

7)  33x², 9x y²

8)  12xy², 39 y³

9)  30x, 40x³ y

10)  36 m⁴, 9 m², 18 mn²

11)  32x², 24xy², 16x² y

12) 12xy, 8y², 8x²

13)  21 b, 45 ab

14)  w² - 9, 9 w², w² - 6w + 9

15)  8x - 4, 6x² + x - 2

1) 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 required LCM is x³ y² z.

2) 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.

3) Solution :

a²bc, b²ca , c²a b

By comparing the given terms, the least common multiple is

a² b²c²

4) Solution :

66 a⁴ b² c³, 44 a³ b⁴ c², 24 a² b³ c⁴

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.

5) 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).

6) 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) 

7)  Solution :

33x², 9x y²

33x² = 3 ⋅ 11 ⋅ x² 

9x y² = 3 ⋅ 3 ⋅ x ⋅ y²

= 3² x ⋅ y²

• Common factor is 3 in which the highest exponent is 3²
• Common factor is x in which the highest exponent is x² 
• y² and 11 are not in common, but when we select least common multiple, we have to include that also.

least common multiple = 3² ⋅ 11 ⋅ x² ⋅ y²

= 99x² y²

8)  Solution :

12xy², 39 y³

12xy² = 2 ⋅ 2 ⋅ 3 ⋅ x ⋅ y²

39 y³ = 3⋅ 13 ⋅ y³

• Common factor is 3
• Common factor is y², in which highest factor is y³. 
• terms that we see extra are 2², 13, x, y³.

least common multiple = 3 ⋅ 2²⋅ 13 ⋅ x ⋅ y³

= 156x y³

9)  Solution :

30x, 40x³ y

30x = 2⋅ 5 ⋅ 3 ⋅ x

40x³ y = 2 ⋅ 2 ⋅ 2 ⋅ x³ y

= 2³ ⋅ x³ y

Common factor is 2, in which 2³

Extra factors are 5, 3 x³ y

Least common multiple = 2³ ⋅ 5 3 x³ y

= 120x³ y

10) Solution :

36 m⁴, 9 m², 18 mn²

36 m⁴ =  2⋅ 2 ⋅ 3 ⋅ 3 ⋅ m⁴

= 2² ⋅ 3² ⋅ m⁴ -----(1)

9 m² = 3 ⋅ 3 ⋅ m²

= 3² ⋅ m² -----(2)

18 mn² = 2 ⋅ 3 ⋅ 3 ⋅ n ⋅ m²

= 2 ⋅ 3² ⋅ n ⋅ m² -----(3)

Comparing (1), (2) and (3)

Highest exponent terms =  2², 3²

Terms we find extra = n ⋅ m²

Combining everything, we get

= 2² ⋅ 3² n ⋅ m²

= 36 n m²

So, the least common multiple is 36 n m².

11)  Solution :

32x², 24xy², 16x² y

32x² = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ x²

= 2⁵⋅ x²

24xy² =  2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ x ⋅ y²

= 2³ ⋅ 3 x y²

16x² y = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ x² ⋅ y

= 2⁴ x² y

In common factor, highest exponent = 2⁵

factors that we find extra are = 3x² y²

Least common multiple = 2⁵ (3x² y²)

= 96x² y²

12)  Solution :

12xy, 8y², 8x²

12xy = 2 ⋅ 2 ⋅ 3 ⋅ x ⋅ y

= 2²⋅ 3 ⋅ x ⋅ y

8y² =  2 ⋅ 2 ⋅ 2 ⋅ y²

= 2³ ⋅ y²

8x² = 2 ⋅ 2 ⋅ 2 ⋅ x² 

= 2³ x² 

In common factor, highest exponent = 2³

factors that we find extra are = 3x² y²

least common multiple = 2³ (3x² y²)

= 24x² y²

13)  Solution :

21 b, 45 ab

21 b = 3 ⋅ 7 ⋅ b

45 ab =  3 ⋅ 3 ⋅ 5 ⋅ a ⋅ b

= 3² ⋅ 5 ⋅ a ⋅ b

In common factor, highest exponent = 3²

factors that we find extra are = 5 (7) ab

= 35 ab

Least common multiple = 3² (35 ab)

= 315 ab

14)  Solution :

w² - 9, 9 w², w² - 6w + 9

w² - 9 = w² - 3²

= (w + 3)(w - 3) -----(1)

9 w² = 3 ⋅ 3 w²

= 3²  w² -----(2)

 w² - 6w + 9 = (w + 3)(w + 3) -------(3)

Comparing (1), (2) and (3), we get

= 9w² (w + 3)(w - 3)

15)  Solution :

8x - 4, 6x² + x - 2

8x - 4

Factoring 4, we get

= 4(2x - 1) -------(1)

6x² + x - 2 = 6x² + 4x - 3x - 2

= 2x(3x + 2) - 1(3x + 2)

= (2x - 1)(3x + 2) -------(2)

Least common multiple = 4 (2x - 1)(3x + 2)

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