## Intermediate Algebra (6th Edition)

We are given the expression $4y^{0}-(4y)^{0}$. $4y^{0}-(4y)^{0}=4(y)^{0}-1=4\times1-1=3$ In general, $a^{0}$ will be 1 when a does not equal 0. In the first term, 4 and y are not in parentheses, so only y (the base of the exponent) is evaluated by the exponent.