Verified Answer :

3y^2z*(2y^4yz-y+z)

=6y^(4z)+12y^(1+2z)z-3y^(1+2z)+3zy^(2z)

Answer: 48(yz)^2 - 6zy^2 - 6yz^2

Step-by-step explanation: We want to find the product:

3y2z(2y2z + 4yz - y - z)

First, distribute the product:

3y2z2y2z + 3y2z4yz -3y2zy - 3y2zz

now, let's simplify the equation:

(3222)(yz)^2 + (324)(yz)^2 - (32)zy^2 - (32)yz^2

24(yz)^2 + 24(yz)^2 - 6zy^2 - 6yz^2

48(yz)^2 - 6zy^2 - 6yz^2

If the equation you want to distribute is:

(3y^2z)(2y^2z + 4yz - y - z)

the distribution is:

(3y^2z)ยด(2y^2z) + (3y^2z)4yz -(3y^2z)y - (3y^2z)z

6y^4z + 12y^(2z + 1)z - 3y^(3z + 1) - (3y^2z)*z