Question

If x^2-bx+c=(x+p)(x-q), then factorize x^2-bxy+cy^2. And please give your guide pdf if you have 1 with this question

Answer 1

Step-by-step explanation:

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)

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Answer 2

Answer:

here is the answer down below

Step-by-step explanation:

Given : x2 – bx + c = (x + p) (x – q)

= x2 + px - qx - pq

= x2 + (p - q)x - pq

⇒ b = q - p , c = -pq

Thus, x2 – bxy + cy2 = x2 + (p - q)xy - pqy2

= x2 + pxy - qxy - pqy2

= x(x + py) - qy(x + py)

= (x + py)(x - qy)