Intermediate Algebra, Chapter 5, 5.1, Section 5.1, Problem 104

Simplify the expression $3a^2(-5a^{-6})(-2a)^0$ so that no negative exponents appear in the final result. Assume that the variables represent nonzero real numbers.

Remove the negative exponent in the numerator by rewriting $-5a^{-6}$ as $\dfrac{-5}{a^6}$. A negative exponent follows the rule: $a^{−n}= \dfrac{1}{a^n}$.

$3a^2 \left(\dfrac{-5}{a^6} \right)(-2a)^0$


Anything raised to the th power is $1$.

$(3a^2 \cdot \left(\dfrac{-5}{a^6} \right) \cdot (1))$


Multiply all the factors in $3a^2 \cdot \left(-\dfrac{5}{a^6} \right) \cdot (1)$.

$(3a^2 \left(-\dfrac{5}{a^6} \right)(1))$


Multiply $-\dfrac{5}{a^6}$ by $1$ to get $-\dfrac{5}{a^6}$.

$(3a^2 \left(-\dfrac{5}{a^6} \right))$


Multiply $3a^2$ by each term inside the parentheses.

$-\dfrac{15}{a^4}$