How can I calculate the uncertainty of two combined uncertainties? If uncertainty can be calculated as half the range, and percentage uncertainty is the uncertainty over a mean all multiplied by 100, how can I find what the uncertainty is of A^2 * B^3 when A = 21.3 and the uncertainty of A is +-0.14, and B = 4 and the uncertainty is +-0.01?

Am I right in thinking you just do +-0.14*2 added to +-0.01*3?
No, just adding the uncertainties is wrong; and it is certainly wrong to ever square or cube uncertainties and use the results as uncertainties themselves.  You can see the latter by imagining a trivial re-scaling of the variables; quantities like an uncertainty cubed scale improperly.

The way to combine uncertainties depends on certain assumptions about the quantities being combined.  The usual assumptions applicable to your case are:


A and B are independent of one another.  
The statistical distributions of the measurements for A and B are sufficiently similar to gaussians that you can use the math for combining gaussian random variates.
The uncertainites themselves are "small", meaning that certain calculations can be done dropping higher order powers of the uncertainties.


In the case of your problem, first we get the uncertainty on $A^2$.  Obviously $A$ and $A$ are not uncorrelated with each other.  The uncertainty with $\sigma_a = 0.14&lt;A&gt;$ is calculated in this dirty but justifiable way:
$$
( &lt;A&gt; \pm 0.14 &lt;A&gt; )( &lt;A&gt; \pm 0.14 &lt;A&gt; ) = &lt;A&gt;^2 \pm 0.28 &lt;A&gt; &lt;A&gt; + O(0.14^2) =\approx
&lt;A&gt;^2 \left( 1 + \pm 0.28 \right)
$$
Similarly, when the fractional error in $B$ is $0.01$,
$$
&lt;B&gt;^3 (1 \pm 0.01)^3 \approx &lt;B&gt;^3 \left( 1 + \pm 0.03 \right)
$$
Then we use the independence of $A$ and $B$ and the Gaussian property of the distributions to add the uncertainties "in quadrature":
$$
&lt;A&gt;^2 \left( 1 + \pm 0.28 \right) &lt;B&gt;^3 \left( 1 + \pm 0.03 \right) \approx
&lt;A&gt;^2 &lt;B&gt;^3 \left( 1 \pm \sqrt{.28^2+.03^2}\right) 
$$
which gives a fractional error on $A^2B^3$ of $0.2816$.
If we assume that $A_\text{min} \leq A \leq A_\text{max}$ and $B_\text{min} \leq B \leq B_\text{max}$, and wish to work out $\min{A^2B^3}$ and $\max{A^2B^3}$, the following is useful. (See Mark's answer for discussions on probability, spread, independence etc). This is meant more to be a "calculate rough bounds for school physics" level answer.

The best way to do these questions is to work out the extremes of values and plug them in - though this might not be how you have been taught.

The lowest value is taking: $A = A_\text{min} = 21.3 - 0.14 = 21.16$, $B = B_\text{min} = 4 - 0.01 = 3.99$, thus min answer $ = A^2 B^3 = 28441.33736$

Similarly, $A_\text{max} = 21.3 + 0.14 = 21.44$, $B_\text{max} = 4.01$, max answer $= 29640.3058$.

Actual calculated answer $ = 21.3^2 \ast 4^3 = 29036.16$

So we have:

\begin{align*}
A^2 B^3 = 29036.16 \begin{cases} + &amp; 604.145796 \\ - &amp; 594.82264 \end{cases}
\end{align*}
(Round as appropriate)

Of course...

You may well have been taught "if you multiply or divide, add the percentage errors". This gives approximately the right answer in most cases, in the above, we get A%Error = 0.6572769953...%, B%Error = 0.25...%, so total percentage error is: $2 \ast 0.6572769953...\% + 3 \ast 0.25\% = 2.064553991...\%$

This gives a rough absolute final error of $\pm 29036.16 * 2.064553991\% = \pm 599.472 \approx \pm 600$.

Note that this is conceptually more confusing, almost more work than our previous working, and we get a less accurate answer. (Here ends my rant about education of calculating errors).