Question 715801
<pre>
A = {{{1/2}}}(6h²-4h) is correct but why not multiply it all the way out?

A = {{{1/2}}}h[(h-4)+(5h)] = {{{1/2}}}h(h-4+5h) = {{{1/2}}}h(6h-4) = 3h²-2h

or you could factor out h and write it as h(3h-2), but don't leave it with
a fraction factor when it is not necessary.

Edwin</pre>