#1**+10 **

You can set this up just like a basic multiplication problem.

2n^{3}+3n -2

x 3n^{2 }-5

------------------------------------

-10n^{3} -15n + 10

+ 6n^{5} +9n^{3} -6n^{2}

--------------------------------------

6n^{5} -n^{3} -6n^{2} -15n +10

There you go! I hope this helps

NinjaDevo
Jun 22, 2014

#2**+5 **

It is easiest if you swap them around so that you put the one with the fewest terms first.

Now, in the first bracket you have 2 terms, the first is 3n^{2} and the second is -5

what you have to do is take the first term and multiply it with the second bracket and then take the second term and multiply it by the second bracket. If there was a third term you would do that next. so

$$3n^2(2n^3+3n-2)+(-5)(2n^3+3n-2)\\\\

=(3n^2\times2n^3+3n^2\times 3n-3n^2\times2)+(-5\times 2n^3+-5\times 3n--5\times 2)\\\\

=6n^5+9n^3-6n^2+(-10n^3+-15n--10)\\\\

=6n^5+9n^3-6n^2-10n^3-15n+10\\\\

=6n^5-n^3-6n^2-15n+10$$

Melody
Jun 22, 2014

#3**0 **

You beat me to it Ninja, it doesn't pay to get too side tracked! ♬

I have just taken a look at your answer. I have never seen anyone do it like that before.

Melody
Jun 22, 2014

#4**0 **

ND's method is similar to "vertical" multiplication. I actually like this because the terms can be "lined up" nicely and we don't lose track of what we have. Variables with similar powers are "stacked" and then we just add or subtract their coefficients, as the case may be.

Points and "Thumbs" from me, ND....!!! Your method of presentation is uniformly good.........

CPhill
Jun 22, 2014