FINDING LCM OF ALGEBRAIC EXPRESSIONS WORKSHEET

Find LCM of the following algebraic expressions.

1)  2x² - 18 y², 5 x²y + 15 xy², x³ + 27y³

2)  (x+4)² (x-3)³, (x-1) (x+4) (x-3)²

3) 10 (9x²+6xy+y²) , 12 (3x²-5xy-2y²), 14 (6x⁴+2x³)

4)  3(a-1), 2(a - 1)² , (a²-1)

5)  5x - 10, 5x² - 20

6)  x⁴ - 1, x² - 2x + 1

7)  x³ - 27, (x - 3)² and x² - 9

8)  (2x² - 3xy)², (4x -6y)³ and (8x³ - 27y³)

9)  2x² - 5x - 3, 4x² - 36

1) Solution :

2x² - 18 y², 5 x²y + 15 xy², x³ + 27y³

2x² - 18 y²  =  2(x² - 9y²)

=  2(x² - (3y)²)

2x² - 18 y²  =  2(x + 3y) (x - 3y) ----(1)

5x²y + 15x  =  5xy (x + 3y) ----(2)

x³ + 27y³  =  x³ + (3y)³

= (x + 3y) (x² + x (3y) + (3y)²)

=  (x + 3y) (x² + 3xy + 9y²)

= 2(x + 3y) 5 ⋅ x ⋅  y ⋅ (x² + 3xy + 9y²)

=  10xy(x + 3y) (x² + 3xy + 9y²)

So, the required least common multiple is

10xy(x + 3y) (x² + 3xy + 9y²)

2) Solution :

(x + 4)² (x - 3)³, (x - 1) (x + 4) (x - 3)²

By comparing (x + 4) and (x + 4)², the highest term is (x + 4)².

By comparing (x - 3)² and (x - 3)³, the highest term is (x - 3)³

The extra term is (x-1).

So, the least common multiple is 

(x - 1)(x + 4)²(x - 3)³

The least common multiple is 

(x - 1)(x + 4)²(x - 3)³

3) Solution :

10 (9x² + 6xy + y²), 12 (3x² - 5xy - 2y²), 14 (6x⁴ + 2x³)

10 (9x² + 6xy + y²) :

10  =  2 ⋅ 5

By factoring 9x² + 6xy + y², we get

9x² + 6xy + y²  =  9x² + 3xy + 3xy + y²

=  3x(3x + y) + y(3x + y)

(9x² + 6xy + y²)  =  (3x + y)(3x + y)

10 (9x² + 6xy + y²)  =   2 ⋅ 5 (9x² + 6xy + y²) ----(1)

12(3x² - 5xy - 2y²) :

12  =  2²⋅ 3

3x² - 5xy - 2y²  =  (3x² - 6xy + xy - 2y²)

=  3x(x - 2y) + y(x - 2y)

=  (3x + y) (x - 2y) ----(2)

14(6x⁴ + 2x³) :

14  =  2 ⋅ 7

6x⁴ + 2x³ =  2x³(3x + 1)

14(6x⁴ + 2x³)  =  2² ⋅ 7⋅ x³ (3x + 1) ----(3)

By comparing (1), (2) and (3), we get

=  2² ⋅ 5 ⋅ 7 ⋅ 3 ⋅ x³ ⋅ (3 x + y)²(3 x + 1)(x - 2y)

=  420 x³ (3 x + y)²(3 x + 1)(x - 2y)

So, the least common multiple is

420 x³ (3 x + y)²(3 x + 1)(x - 2y)

4)  Solution :

3(a - 1), 2(a - 1)² , (a² - 1)

= 3 (a- 1) -------(1)

2 (a - 1)²  =  2(a - 1)(a - 1) -------(2)

(a²-1) = (a + 1) (a - 1) -------(3)

By comparing (1), (2) and (3), we get

= 3 ⋅ 2 (a - 1)² (a + 1)

So, the least common multiple is 

6(a - 1)²(a + 1)

5)  Solution :

5x - 10, 5x² - 20

= 5x - 10

Factoring 5, we get

5x - 10 = 5(x - 2)  ------(1)

5x² - 20

Factoring 5, we get

= 5(x² - 4)

= 5(x² - 2²)

= 5(x + 2) (x - 2)  ------(2)

Comparing (1) and (2), the least common multiple is 

5(x + 2) (x - 2)

6) Solution :

x⁴ - 1, x² - 2x + 1

x⁴ - 1 = (x²)²- (1²)²

= (x² - 1)(x² + 1)

= (x + 1)(x - 1)(x² + 1) -----(1)

x² - 2x + 1 = (x - 1)(x - 1) ------(2)

Comparing (1) and (2), we get

= (x - 1)²(x + 1)(x² + 1)

So, the least common multiple is

(x - 1)²(x + 1)(x² + 1)

7)  Solution :

x³ - 27, (x - 3)² and x² - 9

x³ - 27 = x³ - 3³

Looks like an algebraic formula

a³ - b³ = (a - b)(a² + ab + b²)

 x³ - 3³ = (x - 3)(x² + x(3) + 3²)

= (x - 3)(x² + 3x + 9) ---(1)

x² - 9 = x² - 3²

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

Comparing (1) and (2)

 (x - 3)(x + 3) (x² + 3x + 9)

So, the least common multiple is 

 (x - 3)(x + 3) (x² + 3x + 9)

8) Solution :

(2x² - 3xy)², (4x -6y)³ and (8x³ - 27y³)

(2x² - 3xy)²

Factoring x, we get

x² (2x - 3y)²   ------(1)

(4x -6y)³ 

Factoring 2, we get [2(2x - 3y)]³

[2(2x - 3y)]³ = 2³(2x - 3y)³

= 2³ (2x - 3y)³  ------(2)

(8x³ - 27y³) = (2x)³ - (3y)³

= (2x - 3y)  [ (2x)² + 2(2x)(3y) + (3y)²]

= (2x - 3y)  [ 4x² + 12xy + 9y²]   ------(3)

Least common multiple is 

x²2³ (2x - 3y)³ [ 4x² + 12xy + 9y²]

8x² (2x - 3y)³ [ 4x² + 12xy + 9y²]

9) Solution :

2x² - 5x - 3, 4x² - 36

2x² - 5x - 3 = 2x² - 6x + 1x - 3

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

= (2x + 1)(x - 3) ---(1)

4x² - 36 = 4(x² - 9)

= 4(x² - 3²)

= 4(x - 3)(x + 3)---(2)

Comparing (1) and (2), we get

= 4(x - 3) (x + 3) (2x + 1)

So, the least common multiple is 

4(x² - 9)(2x + 1)

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