Question

if x^2-bx+c=(x+p)(x-q), then factorize x^2-bxy+cy^2

Answer 1

Given that x2 - bx+c = (x+p)(x - q)
⇒ x2 - bx + c = x2 + x(p – q) – pq
Comparing both the sides we get (p – q)  =  –b and c = – pq
∴ b = (q – p) and c = – pq
Consider, x2 – bxy+cy2
Put b = (q – p) and c = – pq
x2-bxy+cy2 becomes x2 – (p – q)xy – pqy2
= x2 – pxy + qxy – pqy2
= x(x – py) + qy(x – py)
= (x + qy)(x – py)

Answer 2

Given x2 _ bx+c = (x+p)(x - q)
⇒ x2 _ bx + c = x2 + x(p – q) – pq
Comparing both (p – q)  =  –b and c = – pq
b = (q – p) and c = – pq
Consider, x2 – bxy+cy2
 b = (q – p) and c = – pq
x2-bxy+cy2 =>  x2 – (p – q)xy – pqy2
= x2 – pxy + qxy – pqy2
= x(x – py) + qy(x – py)
= (x + qy)(x – py)