FACTORIZATION USING ALGEBRAIC IDENTIES

Factorize each polynomial using algebraic identity.

Problem 1 :

w² - 4

Solution :

w² - 4 = w² - 2²

a² - b² = (a + b) (a - b)

w² - 2² = (w + 2) (w - 2)

Problem 2 :

a² - b²

Solution :

a² - b² = (a + b) (a - b)

Problem 3 :

9 - 16t²

Solution :

9 - 16t² = 3² - (4t)²

a² - b² = (a + b) (a - b)

3² - (4t)² = (3 + 4t) (3 - 4t)

Problem 4 :

8a³ - 27b³

Solution :

8a³ - 27b³ = (2a)³ - (3b)³

a³ - b³ = (a - b) (a² + ab + b²)

= (2a - 3b) [(2a)² + (2a)(3b) + (3b)²]

= (2a - 3b) (4a² + 6ab + 9b²)

Problem 5 :

m³ + n³

Solution :

a³ + b³ = (a + b) (a² - ab + b²)

m³ + n³ = (m + n) (m² - mn + n²)

Problem 6 :

125x³ - 64

Solution :

125x³ - 64 = (5x)³ - 4³

a³ - b³ = (a - b) (a² + ab + b²)

= (5x - 4) [(5x)² + (5x)(4) + (4)²]

= (5x - 4) (25x² + 20x + 16)

Problem 7 :

x² + 10x + 25

Solution :

(a + b)² = a² + 2ab + b²

x² + 10x + 25 = (x)² + 2(x)(5) + (5)²

= (x + 5)²

= (x + 5) (x + 5)

Problem 8 :

36u² - 12uv + v²

Solution :

(a - b)² = a² - 2ab + b²

36u² - 12uv + v² = (6u)² - 2(6u)(v) + (v)²

= (6u - v)²

= (6u - v) (6u - v)

Problem 9 :      

4a² - 4a + 1

Solution :

(a - b)² = a² - 2ab + b²

4a² - 4a + 1 = (2a)² - 2(2a)(1) + 1²

= (2a - 1)²

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

Problem 10 :

25x² - 36

Solution :

25x² - 36 = (5x)² - (6)²

a² - b² = (a + b) (a - b)

(5x)² - (6)² = (5x + 6) (5x - 6)