0
$\begingroup$

I have this expression

enter image description here

Using distributive property, It became like this

enter image description here

How did it arrived at $-4a^{-1}a^{1/3}$ ?

  • 3
    Do you mean $-4a^{-1}a^{1/3}$ is (part of) the given answer? It is wrong.2017-01-12
  • 0
    @StackTD Not part of the given answer but the previous step before we arrive with the final answer2017-01-13

1 Answers 1

2

by multiplication we get $$2a^{2+1/2}-4a^{-1+1/2}$$ using the power rule we get $$2a^{\frac{5}{2}}-4a^{-\frac{1}{2}}$$

  • 0
    You inadvertently left out the $a$ in the first term. Should be $2a^\frac{5}{2}$2017-01-12