Question

Factorise:
pq²+q(p-1) -1=?​

Answer 1

Given

Factorise:

pq²+q(p-1) -1

Solution

➟ pq² + q(p - 1) - 1

➟ pq² + qp - q - 1

➟ pq(q + 1) - 1(q + 1)

➟ (q + 1)(pq - 1)

Some Identities

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• a² - b² = (a + b)(a - b)
• (a + b)³ = a³ + b³ + 3ab(a + b)
• (a - b)³ = a³ - b³ - 3ab(a - b)
• a³ - b³ = (a - b)(a² + ab + b²)
• a³ + b³ = (a + b)(a² - ab + b²)

$\rule{200}3$

Answer 2

Heya!!

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Given,

Factorise,

pq²+q(p-1) -1

Solution:-

➟ pq² + q(p - 1) - 1

➟ pq² + qp - q - 1

➟ pq(q + 1) - 1(q + 1)

➟ (q + 1)(pq - 1)

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Some Identities:-

$\bigstar$(a + b)² = a² + b² + 2ab

$\bigstar$(a - b)² = a² + b² - 2ab

$\bigstar$a² - b² = (a + b)(a - b)

$\bigstar$(a + b)³ = a³ + b³ + 3ab(a + b)

$\bigstar$(a - b)³ = a³ - b³ - 3ab(a - b)

$\bigstar$a³ - b³ = (a - b)(a² + ab + b²)

$\bigstar$a³ + b³ = (a + b)(a² - ab + b²)

Types of factorise:-

• Greatest Common Factor.
• Grouping.
• Difference in Two Squares.
• Sum or Difference in Two Cubes.
• Trinomials.
• General Trinomials.