Question

If x²-bx+c = (x+p) (x-q) , Then find the factors of x²-bxy+cy² ?​

Answer 1

Explanation:

Given 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)

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