Question

Factor x2 – 3xy + 3y2 + 9x – 9y + 18 = 0.

Answer 1

The factorized expression is $x(x-3y+9)+3y(y-3)+18$

Therefore $x^2-3xy+3y^2+9x-9y+18=x(x-3y+9)+3y(y-3)+18$

Step-by-step explanation:

• Given equation is $x^2-3xy+3y^2+9x-9y+18=0$
• To factorize the given equation as below :
• $x^2-3xy+3y^2+9x-9y+18$
• $=x(x-3y+9)+3y(y-3)+18$
• Therefore the factorized expression is $x(x-3y+9)+3y(y-3)+18$
• Therefore $x^2-3xy+3y^2+9x-9y+18=x(x-3y+9)+3y(y-3)+18$