Question

Factorise (x+3y)3z^2-(2x+6y)z

Answer 1

GIVEN :

Factorise $(x+3y)3z^2-(2x+6y)z$

TO FIND :

The factors of the given polynomial.

SOLUTION :

Given polynomial is $(x+3y)3z^2-(2x+6y)z$

Now we have to factorise the given polynomial $(x+3y)3z^2-(2x+6y)z$ as below :

$(x+3y)3z^2-(2x+6y)z$

By using the distributive property :

a(x+y)=ax+ay

$=x(3z^2)+3y(3z^2)-(2x)z-(6y)z$

$=3xz^2+9yz^2-2xz-6yz$

$=3xz^2-2xz+9yz^2-6yz$

By taking the common term outside,

= xz(3z-2)+3yz(3z-2)

= (xz+3yz)(3z-2)

By taking the common factor outside,

= z(x+3y)(3z-2)

⇒ $(x+3y)3z^2-(2x+6y)z= z(x+3y)(3z-2)$

∴ the given polynomial $(x+3y)3z^2-(2x+6y)z$ is factorised into z(x+3y)(3z-2).