# FACTORING POLYNOMIALS USING ALGEBRAIC IDENTITIES

Any one of the Algebraic identities will be useful to find factors of expression.

(a + b)2 = a2 + 2ab + b2

(a - b)= a2 - 2ab + b2

a2 - b2 = (a + b)(a - b)

a3 - b3 = (a - b)(a2 + ab + b2)

a3 + b3 = (a + b)(a2 - ab + b2)

Factorize each polynomial using algebraic identity.

Problem 1 :

36k² - 1

Solution :

36k² - 1 = (6k)² - 1²

a² - b² = (a + b) (a - b)

(6k)² - 1² = (6k + 1) (6k - 1)

Problem 2 :

x² - 9y²

Solution :

x² - 9y² = x² - (3y)²

a² - b² = (a + b) (a - b)

x² - (3y)² = (x + 3y) (x - 3y)

Problem 3 :

25m² - n²

Solution :

25m² - n² = (5m)² - n²

a² - b² = (a + b) (a - b)

(5m)² - n² = (5m + n) (5m - n)

Problem 4 :

8r³ - 729

Solution :

8r³ - 729 = 8r³ - 9³

a³ - b³ = (a - b) (a² + ab + b²)

8r³ - 9³ = (8r - 9) [(8r)² + (8r)(9) + 9²]

= (8r - 9) (64r² + 72r + 81)

Problem 5 :

p³ - 1000q³

Solution :

p³ - 1000q³ = p³ - (10q)³

a³ - b³ = (a - b) (a² + ab + b²)

p³ - (10q)³ = (p - 10q) [(p)² + (p)(10q) + (10q)²]

= (p - 10q) (p² + 10pq + 100q²)

Problem 6 :

27c³ + 125d³

Solution :

27c³ + 125d³ = (3c)³ + (5d)³

a³ + b³ = (a + b) (a² - ab + b²)

(3c)³ + (5d)³ = (3c + 5d) [(3c)² - (3c)(5d) + (5d)²]

= (3c + 5d) (9c² - 15cd + 25d²)

Problem 7 :

64y³- 216

Solution :

64y³- 216 = (4y)³ - 6³

a³ - b³ = (a - b) (a² + ab + b²)

(4y)³ - 6³ = (4y - 6) [(4y)² + (4y)(6) + 6²]

= (4y - 6) (16y² + 24y + 36)

Problem 8 :

36y² + 84y + 49

Solution :

(a + b)² = a² + 2ab + b²

36y² + 84y + 49 = (6y)² + 2(6y)(7) + 7²

= (6y + 7)²

= (6y + 7) (6y + 7)

Problem 9 :

h² + 4h + 4

Solution :

(a + b)² = a² + 2ab + b²

h² + 4h + 4 = h² + 2(h)(2) + 2²

= (h + 2)²

= (h + 2) (h + 2)

Problem 10 :

z² - 8z + 16

Solution :

(a - b)² = a² - 2ab + b²

z² - 8z + 16 = z² - 2(z)(4) + 4²

= (z - 4)²

= (z - 4) (z - 4)

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