Question 558330
first off its multiplication and z^0=1 since 1 is being multiplied and not added it isnt even needed and can be erased from problem because 1 times y = y
{{{(-15y^2z)(-2y^2)}}}
{{{30y^(2+2)z}}} multiplying 2 negatives makes a possitive, and then need to add the like exponents
{{{30y^4z}}}