Question

factorise the following (X+3y)3z^2-(2x+6y)z​

Answer 1

Answer:

The factors of the given polynomial.

SOLUTION :

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

2

−(2x+6y)z

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

2

−(2x+6y)z as below :

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

2

)+3y(3z

2

)−(2x)z−(6y)z

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

2

+9yz

2

−2xz−6yz

=3xz^2-2xz+9yz^2-6yz=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)(x+3y)3z

2

−(2x+6y)z=z(x+3y)(3z−2)

∴ the given polynomial (x+3y)3z^2-(2x+6y)z(x+3y)3z

2

−(2x+6y)z is factorised into z(x+3y)(3z-2).