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teenandtoddlermom

  • 2 years ago

need help (x^3+y^2)(x^3-y^2) can someone help me figure this out and explain?

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  1. satellite73
    • 2 years ago
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    you have to do four multiplications

  2. teenandtoddlermom
    • 2 years ago
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    I need to multiply

  3. satellite73
    • 2 years ago
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    \[x^3\times x^3\] is the first one

  4. teenandtoddlermom
    • 2 years ago
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    x^6? do I add the exponents?

  5. satellite73
    • 2 years ago
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    then \(x^3\times (-y^2)\)

  6. satellite73
    • 2 years ago
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    yes

  7. satellite73
    • 2 years ago
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    then \(x^3\times y^2\)

  8. satellite73
    • 2 years ago
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    and finally \(y^2\times (-y)^2\)

  9. teenandtoddlermom
    • 2 years ago
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    x^6-y^4?

  10. satellite73
    • 2 years ago
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    you will notice that middle terms add up to zero, so you are only left with \[x^6-y^4\]

  11. satellite73
    • 2 years ago
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    yes

  12. satellite73
    • 2 years ago
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    it is always true that \((a+b)(a-b)=a^2-b^2\)

  13. teenandtoddlermom
    • 2 years ago
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    see, all I need is confidence and your great guidings! :) I might make it after all. LOL

  14. jim_thompson5910
    • 2 years ago
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    you can also use a table to help you multiply |dw:1362367881354:dw|

  15. teenandtoddlermom
    • 2 years ago
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    Hi Jim. thank you, that helps as well. :)

  16. jim_thompson5910
    • 2 years ago
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    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|

  17. jim_thompson5910
    • 2 years ago
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    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)

  18. teenandtoddlermom
    • 2 years ago
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    great explanation. I need all the tips I can get. thank you! :)

  19. jim_thompson5910
    • 2 years ago
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    you're welcome

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