GCD OF ALGEBRAIC EXPRESSIONS

Find the greatest common divisor of following algebraic terms.

Example 1 :

7 x² y z⁴, 21 x² y⁵ z³

Solution :

7 x² y z⁴, 21 x² y⁵ z³

7 x² y z⁴   =  7 ⋅ x² ⋅ y ⋅ z ⋅ z³

21 x² y⁵ z³  =  3 ⋅ 7 ⋅ x² ⋅ y ⋅ z ⋅ z³

Common factors are 7, x²,  y and z³

Multiplying common factors, we get

=  7x²yz³

So, greatest common divisor of the given algebraic terms is  7x²yz^3.

Example 2 :

x² y, x³ y , x² y²

Solution :

x² y, x³ y , x² y²

x² y  =  x² ⋅ y

 x³ y  =  x² ⋅ x ⋅ y

 x² y²  =  x² ⋅ y ⋅ y

Common factors in the above terms are x² and y.

By multiplying common factors, we get x²y.

So, greatest common divisor of the given algebraic terms is x²y.

Example 3 :

25 b c⁴ d³ , 35 b² c⁵ , 45 c³ d

Solution :

25 b c⁴ d³, 35 b² c⁵, 45 c³ d

25 b c⁴ d³  =  5 ⋅ 5 ⋅ b ⋅ c³ ⋅ d³

35 b² c⁵  =  7 ⋅ 5 ⋅ b ⋅ b ⋅ c³ ⋅ c²

45 c³ d  =  3² ⋅ 5 ⋅ c³ ⋅ d

Common factors in the above terms are 5 and c³.

By multiplying common factors, we get 5c³.

So, greatest common divisor of the given algebraic terms is  5c³.

Example 4 :

35 x⁵ y³ z⁴ , 49 x² y z³ , 14 xy² z²

Solution :

35 x⁵ y³ z⁴ , 49 x² y z³ , 14 xy² z²

35 x⁵ y³ z⁴  =  7 ⋅ 5 ⋅ x ⋅ x ⋅ x ⋅ x ⋅ x ⋅ y ⋅ y ⋅ y ⋅ z² ⋅ z²

49 x² y z³  =   7 ⋅ 7 ⋅ x ⋅ x ⋅ y ⋅ z² ⋅ z

14 xy² z²  =   7 ⋅ 2 ⋅ x ⋅ y ⋅ y ⋅ z²

Common factors in the above terms are 7, x, y and z².

By multiplying common factors, we get 7xyz².

So, greatest common divisor of the given algebraic terms is 7xyz².

Factorise the following expressions and find the highest common factors for each set.

Example 5 :

4x² y, 4x³ - 2xy

Solution :

4x² y = 2 ⋅ 2 ⋅ x ⋅ x ⋅ y

4x³ - 2xy = 2x (2x² - y)

Comparing these two set of factors, we see 2x in common. So, the highest common factor is 2x.

Example 6 :

6p + 3pq, 6p²

Solution :

6p + 3pq, from this we can factor out 3p.

6p + 3pq = 3p(2 + q)

6p²= 2 ⋅3 ⋅ p ⋅ p

By comparing the above set of factors, we see 3 and p are in common. So, the highest common factor is 3p.

Example 7 :

8pq, 4p² q + 16pq²

Solution :

8pq = 2 ⋅ 4 ⋅ p ⋅ q

4p² q + 16pq² = 4pq (p + 4q)

By comparing these two set of factors, we get 4pq in common in both. So, the highest common factor is 4pq.

Example 8 :

x² - 4xy + 4y², x²  - 4y²

Solution :

x² - 4xy + 4y² 

Using the expansion of the formula for (a - b)², we get

(a - b)² = a² - 2ab + b²

= x² - 2 x (2y)  + 4y² 

= (x - 2y)² -----(1)

x²  - 4y² = x²  - (2y)²

Using the algebraic identity, (a - b)² = (a + b) (a - b)

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

Comparing (1) and (2), we get

(x - 2y)

So, the highest common factor is x - 2y.

Example 9 :

4x² - 1,  4x² - 4x + 1

Solution :

4x² - 1 = (2x)² - 1²

Using the algebraic identity, (a - b)² = (a + b) (a - b)

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

  4x² - 4x + 1 = (2x)² - 2(2x)(1) + 1²

= (2x - 1)² ------(2)

Comparing (1) and (2), we get the common factor.

(2x - 1)

So, the greatest common factor is 2x - 1.

Example 10 :

x² - 1,  x² - 3x + 2

Solution :

x² - 1 = x² - 1²

= (x +1)(x - 1) ------(1)

x² - 3x + 2 = (x - 1)(x + 2) ------(2)

Greatest common factor = (x - 1)

Example 11 :

d - bd, 1 - 2b + b²,  1 - b²

Solution :

d - bd

Factor d, we get

= d(1 - b) -------(1)

1 - 2b + b²

Writing in the form of a²- 2ab + b²

= 1 - 2(1) b + b²

= (1 - b)² -------(2)

1 - b² = (1 + b) (1 - b) -------(3)

So, the greatest common factor is 1 - b.

Example 12 :

a³  - a², 2a² - 2a, a² - 1

Solution :

a³  - a²

Factor a², we get

= a²(a - 1) ------(1)

2a² - 2a

Factoring 2a, we get

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

a² - 1 = (a + 1)(a - 1) ------(3)

Comparing (1), (2) and (3), we get the common facor.

= (a - 1)

So, the greatest common factor is a - 1.

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