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

Guest Jun 22, 2014

Best Answer 

 #5
avatar+26406 
+10

Or you could lay it out like this:

boxmultiply

Alan  Jun 22, 2014
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6+0 Answers

 #1
avatar+3450 
+10

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

                    2n3+3n -2

x                        3n2 -5

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

             -10n3       -15n + 10

+    6n5 +9n3 -6n2

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

6n5 -n3 -6n2 -15n +10

 

There you go! I hope this helps

NinjaDevo  Jun 22, 2014
 #2
avatar+91479 
+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 3n2 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
avatar+91479 
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
avatar+81070 
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
 #5
avatar+26406 
+10
Best Answer

Or you could lay it out like this:

boxmultiply

Alan  Jun 22, 2014
 #6
avatar+91479 
0

Please not that this is an expression, not an equation.  An equations has an equal sign.   ♬

Melody  Jun 22, 2014

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