Question

If x^2-bx+c=(x+p)(x-q) then fond the factorization of x^2-bxy+cy^2

Answer 1

Given,

x² - bx+c = (x+p)(x - q)

⇒ x² - bx + c = x² + x(p – q) – pq

Comparing both the sides ,

(p – q)  =  –b

c = – pq

∴ b = (q – p)

c = – pq

Now, x² – bxy+cy²

Put b = (q – p) and c = – pq

x²-bxy+cy² will be x² – (p – q)xy – pqy²

= x² – pxy + qxy – pqy²

= x(x – py) + qy(x – py)

= (x + qy)(x – py)