Question

factorise (p-q)-6(p-q)-16

Answer 1

Correct Question :

> (p-q)² - 6(p-q) - 16

Solution :

Let us assume that p + q is equal to some variable , k .

So, this equation reduces to :

> k^2 - 6k - 16

> k^2 - 8k + 2k - 16

> k(k-8) + 2(k-8)

> (k+2)(k-8)

Substituting the original value of k ;

> (p+q+2)(p+q-8)

This is the required answer.

_________________________________________________________

Additional Information :

(a + b)² = a² + 2ab + b²  

(a + b)² = (a - b)² + 4ab  

(a - b)² = a² - 2ab + b²  

(a - b)² = (a + b)² - 4ab  

a² + b² = (a + b)² - 2ab  

a² + b² = (a - b)² + 2ab  

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²  

ab = {(a + b)/2}² - {(a-b)/2}²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)  

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)  

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )