Question

Factorise:
$ab^{2} + (a - 1)b - 1$​

Answer 1

$\implies\sf{ab^{2} + (a - 1)b - 1}$

$\implies\sf{ {ab}^{2} + ab - b - 1}$

$\implies\sf{ab(b + 1) - 1(b + 1)}$

$\green \bigstar\boxed{\sf{(ab - 1)(b + 1)}} \ \green\bigstar$

$\sf \underline\purple{How \: to \: Factorise \: ?}$

• At first we have take common terms like (b + 1)

• Then (b + 1) will be common from both sides.

• Then we can write (ab - 1)

• And finally the result will be (ab - 1) (b + 1)

Answer 2

We have to factorise [ab² + (a - b)b - 1].

ab² + (a - 1)b - 1

= ab² + ab - b - 1

= ab(b + 1) - 1(b + 1)

= (b + 1)(ab - 1)

Identities:-

1. (a + b)² = a² + b² + 2ab
2. (a - b)² = a² + b² - 2ab
3. a² - b² = (a + b)(a - b)
4. x² + (a + b)x + ab = (x + a)(x + b)
5. (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
6. (a + b)³ = a³ + b³ + 3ab(a + b)
7. (a - b)³ = a³ - b³ - 3ab(a - b)
8. a³ - b³ (Factor) = (a - b)(a² + ab + b²)
9. a³ - b³ (Value fetching) = (a - b)³ + 3ab(a - b)
10. a³ + b³ (Factor) = (a + b)(a² - ab + b²)
11. a³ + b³ (Value Finding) = (a + b)³ - 3ab(a + b)
12. a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
13. 4ab = (a + b)² - (a - b)²
14. 2(a² + b²) = (a + b)² + (a + b)²