Altering the order by method of excess reagent

I don't understand what the last paragraph of the following means (which I came across on the textbook) PLSSS EXPLAINNN


"Consider a reaction: A+2B-->C+D
The rate equarion is of the form rate=k[A]^p[B]^q

However if the initial concentration of B is made at least 10 times that of A, the change in [B] during the reaction will be negligible. This means that [B] is constant within experimental error, so the rate equation becomes rate=constant×[A]^p where the constant=k×the approximately constant value of [B]^q

The value of p in order with respect to A can be found by the usual methods of initial rate or half life.

If the experiment is repeated with the same initial concentration of A but 20 times as much B(doubling the concentration of B from the first experiment) the way in which initial rate alters will depend on the order with respect to B. If the rate doubles, the reaction is 1st order with respect to B. If the rate quadruples, the reaction is 2nd order in B."

What it is saying is that if you are trying to find the order of reaction for A (by half lives) then you need to have B in very large excess so that as the reaction proceeds the concentration of A decreases but B pretty much remains constant. This allows you to determine the order of reaction for A by plotting a concentration - time graph and working out successive half lives without the concentration of B changing and affecting the results.
You now know the order for A and need to find the order for B.

So now you repeat the experiment keeping the concentration of A the same but double the concentration of B and use the initial rates method. If the Initial rate in the second experiment is double the first one then the order of B is 1, whereas if the initial rate is quadrupled then the order of B is 2.