GCD OF ALGEBRAIC EXPRESSIONS

Find the GCD of the following

Example 1:

c² - d², c(c - d)

Solution :

c²-d², c(c-d)

By using algebraic identity

a² - b²  = (a + b)(a - b)

c² - d² = (c + d) (c - d)

c(c - d)  =  c(c - d)

Common factor is (c-d).

So, greatest common divisor is (c - d).

Example 2 :

x⁴ - 27a³x, (x - 3a)²

Solution :

x⁴-27a³x, (x-3a)² 

x⁴-27a³x  =  x(x³-27a³)

=  x(x³-(3a)³)

By using algebraic identity

a³-b³  =  (a-b)(a²+ab+b²)

x³-(3a)³  =  (x-3a)(x²+3ax+9a²)

x(x³-(3a)³)  =  x(x-3a)(x²+3ax+9a²) ----(1)

(x-3a)²  =  (x-3a)(x-3a) ----(2)

Common factors of (1) and (2) are (x-3a).

So, greatest common divisor is (x-3a).

Example 3 :

m² - 3m - 18,  m² + 5m + 6

Solution :

m² - 3 m - 18,  m² + 5m + 6

By doing factorization, we get

m² - 3m - 18 = (m - 6) (m + 3)

m² + 5m + 6 = (m + 2) (m + 3)

Common factor is (m + 3).

So, greatest common divisor is (m + 3).

Example 4 :

x² + 14x + 33, x³ + 10x² - 11x

Solution :

x² + 14x + 33, x³ + 10x² - 11x

x² + 14x + 33  = (x + 11)(x + 3)  ---(1)

x³ + 10x² - 11x = x(x² + 10x - 11)

  =  x(x + 11)(x - 1)  ----(2)

Common factor is (x + 11).

So, greatest common divisor is (x + 11).

Example 5 :

x² + 3xy + 2y², x² + 5xy + 6y²

Solution :

x² + 3xy + 2y², x² + 5xy + 6y²

x² + 3xy + 2y² = x² + xy + 2xy + 2y²

= x(x + y) + 2y(x + y)

=  (x + 2y)(x + y)  ---(1)

x² + 5xy + 6y²  =  x² + 2xy + 3xy + 6y²

=  x(x + 2y) + 3y(x + 2y)

=  (x + 2y) (x + 3y)----(2)

Greatest common factor of (1) and (2) is (x+2y).

Example 6 :

2x² - x -1, 4x² + 8x + 3

Solution :

2x² - x - 1, 4x² + 8x + 3

2x²-x-1  =  (2x+1) (x-1)

4x²+8x+3  =  (2x+3)(2x+1)

Common factor is (2x+1). 

So, the greatest common divisor is (2x+1).

Example 7 :

x² - x - 2, x² + x - 6, 3x² - 13x + 14

Solution :

x² - x - 2, x² + x - 6, 3x² - 13x + 14

x² - x - 2  =  (x - 2) (x + 1)

x² + x - 6 = (x + 3)(x - 2)

3x² - 13x + 14  = (x - 2)(3x - 7)

Common factor is (x - 2).

So, the greatest common divisor is (x - 2).

Example 8 :

x³ - x² + x - 1, x⁴ - 1

Solution :

x³ - x² + x - 1, x⁴ - 1

=  (x - 1)(x² + 1)  ----(1)

x⁴ - 1  =  (x²)² - (1²)²

=  (x² + 1) (x² - 1)

=  (x² + 1)(x + 1)(x - 1) ----(2)

Common factors in (1) and (2) are (x - 1) and (x² + 1)    

So, the greatest common divisor is (x - 1)(x²+1).

Example 9 :

24 (6x⁴ - x³ - 2x²), 20(2x⁶ + 3x⁵ + x⁴)

Solution :

24 (6x⁴ - x³ - 2x²), 20(2x⁶ + 3x⁵ + x⁴)

24 (6x⁴ - x³ - 2x²)  =  2³ ⋅ 3 x² (6x² - x - 2)

=  2³ ⋅ 3 x² (3x - 2) (2x + 1)

20(2x⁶ + 3x⁵ + x⁴)  =  5 ⋅ 2² x⁴ (2x² + 3x + 1)

=  5 ⋅ 2² x⁴ (x + 1) (2x + 1)

Common factors are 2² x² (2x + 1)

So, the greatest common divisor is 4x²(2x + 1).

Example 10 :

(a - 1)⁵ (a  + 3)², (a - 2)² (a - 1)³ (a + 3)⁴

Solution :

(a - 1)⁵ (a + 3)², (a - 2)² (a - 1)³ (a + 3)⁴

The common factors are (a - 1)³(a + 3)²

So, the greatest common divisor is (a - 1)³(a + 3)².

Example 11 :

x² - 2xy + y², x² - xy, x² - y²

Solution :

x² - 2xy + y² 

It looks like a² - 2ab + b², 

x² - 2xy + y² = (x - y)² -----(1)

x² - xy = x(x - y) -----(2)

 x² - y²

looks like a² - b², using the algebraic identity

= (x + y)(x - y) -----(3)

Comparing (1), (2) and (3), we get the common factor as (x - y). 

So, the highest common factor is (x - y).

Example 12 :

cd - bd, c² - 2bc + b², 2c² - 3bc + b²

Solution :

cd - bd

Factoring d from the above terms, we get

cd - bd = d(c - b) ------(1)

c² - 2bc + b²

Looks like an algebraic identity, a² - 2ab + b²

c² - 2bc + b² = (c - b)² ------(2)

2c² - 3bc + b²

= 2c² - 2bc - bc + b²

Factoring 2c and factoring b, we get 

= 2c(c - b) - b(c - b)

= (2c - b) (c - b)

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