Question

12. If x2 + bx + c = (x + p) (x + q), then factorise x2 + bxy + cy2

Answer 1

SOLUTION : -

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)

HOPE IT HELPS U!

Answer 2

Hey Mate ✌✌

Here is your answer.....

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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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