Question

9. factorise x^2 -XZ+XY-yz
O (x-7)(x+2)
0 (x-y)(x-z)
O (x+y)(x-z)
O (x-)(z-x)
Required​

Answer 1

$\implies \sf{ {x}^{2} - xz + xy - yz }$

$\implies\sf{x(x - z) + y(x - z)}$

$\implies\sf{(x + y)(x - z)} \green \bigstar$

So,the correct option is 3rd one (x + y) (x - z)

___________...

• We have to make sure that the variable is common or not.

• If the variable is common then we can write it as result.

• Like this (x + y) (x - z).

Answer 2

We have to factorise the expression (x² - xz + xy - yz).

(x² - xz + xy - yz)

= x² - xz + xy - yz

= x(x - z) + y(x - z)

[Taking (x) and (y) as common]

= (x - z)(x + y)

Therefore, the factorised form of (x² - xz + xy - yz) is [(x - z)(x + y)]

Identities:-

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• a² - b² = (a + b)(a - b)
• x² + (a + b)x + ab = (x + a)(x + b)
• (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
• (a + b)³ = a³ + b³ + 3ab(a + b)
• (a - b)³ = a³ - b³ - 3ab(a - b)
• a³ - b³ (Factor) = (a - b)(a² + ab + b²)
• a³ - b³ (Value fetching) = (a - b)³ + 3ab(a - b)
• a³ + b³ (Factor) = (a + b)(a² - ab + b²)
• a³ + b³ (Value Finding) = (a + b)³ - 3ab(a + b)
• a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
• 4ab = (a + b)² - (a - b)²
• 2(a² + b²) = (a + b)² + (a + b)²