Question

(x+3y)3z^2-(2x+6y)z factorise this​

Answer 1

GIVEN :

$(x+3y)3z^2-(2x+6y)z$ factorise this​

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 as below :

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

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

By using the distributive property :

a(x+y)=ax+ay

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

By taking the common term outside,

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

By taking the common factor outside,

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

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

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

⇒  [tex](x+3y)3z^2-(2x+6y)z = z(3z-2) (x+3y)[/tex]

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