anonymous
  • anonymous
need help (x^3+y^2)(x^3-y^2) can someone help me figure this out and explain?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
you have to do four multiplications
anonymous
  • anonymous
I need to multiply
anonymous
  • anonymous
\[x^3\times x^3\] is the first one

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anonymous
  • anonymous
x^6? do I add the exponents?
anonymous
  • anonymous
then \(x^3\times (-y^2)\)
anonymous
  • anonymous
yes
anonymous
  • anonymous
then \(x^3\times y^2\)
anonymous
  • anonymous
and finally \(y^2\times (-y)^2\)
anonymous
  • anonymous
x^6-y^4?
anonymous
  • anonymous
you will notice that middle terms add up to zero, so you are only left with \[x^6-y^4\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
it is always true that \((a+b)(a-b)=a^2-b^2\)
anonymous
  • anonymous
see, all I need is confidence and your great guidings! :) I might make it after all. LOL
jim_thompson5910
  • jim_thompson5910
you can also use a table to help you multiply |dw:1362367881354:dw|
anonymous
  • anonymous
Hi Jim. thank you, that helps as well. :)
jim_thompson5910
  • jim_thompson5910
each cell is the result of multiplying the corresponding row and column headers so the upper left cell is the result of multiplying the two x^3 terms the upper right cell is the result of multiplying x^3 and y^2 etc etc giving you this |dw:1362367968379:dw|
jim_thompson5910
  • jim_thompson5910
then you add up all the terms and combine like terms if you can to get x^6 - y^4 (the x^3y^2 terms will cancel as you got before)
anonymous
  • anonymous
great explanation. I need all the tips I can get. thank you! :)
jim_thompson5910
  • jim_thompson5910
you're welcome

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