FIND FACTORS OF POLYNOMIALS USING ALGEBRAIC IDENTITIES

Factorize each polynomial using algebraic identity.

Problem 1 :

49p² - q²

Solution :

49p² - q² = (7p)² - q²

a² - b² = (a + b) (a - b)

(7p)² - q² = (7p + q) (7p - q)

Problem 2 :

64 - d²

Solution :

64 - d² = 8² - d²

a² - b² = (a + b) (a - b)

8² - d² = (8 + d) (8 - d)

Problem 3 :

u² - 16v

Solution :

u² - 16v is not factorable.

Problem 4 :

27u³ + 343v³

Solution :

27u³ + 343v³ = (3u)³ + (7v)³

a³ + b³ = (a + b) (a² - ab + b²)

(3u)³ + (7v)³ = (3u + 7v) [(3u)² - (3u)(7v) + (7v)²]

= (3u + 7v) (9u² - 21uv + 49v²)

Problem 5 :

c³ + 216

Solution :

c³ + 216 = c³ + 6³

a³ + b³ = (a + b) (a² - ab + b²)

c³ + 6³ = (c + 6) [(c²) - (6c) + (6)²]

= (c + 6) (c² - 6c + 36)

Problem 6 :

512z³ - 1

Solution :

512z³ - 1 = (8z)³ - 1³

a³ - b³ = (a - b) (a² + ab + b²)

(8z)³ - 1³ = (8z - 1) [(8z)² + (8z)(1) + 1²]

= (8z - 1) (64z² + 8z + 1)

Problem 7 :

16p² + 56p + 49

Solution :

(a + b)² = a² + 2ab + b²

16p² + 56p + 49 = (4p)² + 2(4p)(7) + 7²

= (4p + 7)²

= (4p + 7) (4p + 7)

Problem 8 :

s² + 16st + 64t²

Solution :

(a + b)² = a² + 2ab + b²

s² + 16st + 64t² = s² + 2(s)(8t) + (8t)²

= (s + 8t)²

= (s + 8t) (s + 8t)

Problem 9 :

k² - 18k + 81

Solution :

(a - b)² = a² - 2ab + b²

k² - 18k + 81 = k² - 2(k)(9) + 9²

= (k - 9)²

= (k - 9) (k - 9)

Problem 10 :

m² - 6mn + 9n²

Solution :

(a - b)² = a² - 2ab + b²

m² - 6mn + 9n² = m² - 2(m)(3n) + (3n)²

= (m - 3n)²

= (m - 3n) (m - 3n)

Problem 11 :

2a⁷ - 128a

Solution :

= 2a⁷ - 128a

= 2a(a⁶ - 64)

= 2a[(a³)² - 8²]

= 2a (a³ + 8)(a³ - 8)

= 2a (a³ + 8)(a³ - 2³)

= 2a (a³ + 2³)(a - 2)(a² + a(2) + 2²)

= 2a(a + 2)(a² - 2a + 4)(a - 2)(a² + 2a + 4)

= 2a(a + 2)(a - 2)(a² - 2a + 4)(a² + 2a + 4)

Problem 12 :

a³ - 27

Solution :

= a³ - 27

= a³ - 3³

= (a - 3)(a² + a(3) + 3²)

= (a - 3)(a² + 3a + 9)

Problem 13 :

12(3x - 2y)² - 3x + 2y - 1

Solution :

= 12(3x - 2y)² - 3x + 2y - 1

Let 3x - 2y = a

= 12a² - 3x + 2y - 1

= 12a² - (3x - 2y) - 1

= 12a² - a - 1

= 12a² - 4a + 3a - 1

= 4a(3a - 1) + 1(3a - 1)

= (4a + 1)(3a - 1)

Applying the value of a, we get

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

= (12x - 8y + 1)(9x - 6y - 1)

Problem 14 :

4(2x - 3y)² - 8x + 12y - 3

Solution :

= 4(2x - 3y)² - 8x + 12y - 3

Let 2x - 3y = a

= 4a² - 4(2x - 3y) - 3

= 4a² - 4a - 3

= 4a² - 6a + 2a - 3

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

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

Applying the value of a, we get

= [2(2x - 3y) + 1] [2(2x - 3y) - 3]

= [4x - 6y + 1] [4x - 6y - 3]

Problem 15 :

3 - 5x + 5y - 12(x - y)²

Solution :

= 3 - 5x + 5y - 12(x - y)²

= 3 - 5(x - y) - 12(x - y)²

Let x - y = a

= 3 - 5a - 12a²

= - (12a² + 5a - 3)

= - (12a² + 9a - 4a - 3)

= - [3a(4a + 3) - 1(4a + 3)]

= - (3a - 1) (4a + 3)

Applying the value of a, we get

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

= - [3x - 3y - 1] [4x - 4y + 3]