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anonymous

  • one year ago

how the heck do I do this?!?!?! (2xy-5x)^2

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  1. anonymous
    • one year ago
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    what are you trying to do?

  2. anonymous
    • one year ago
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    Im trying to foil this expression, but I dont know what to do with the xy and the x

  3. anonymous
    • one year ago
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    oooh okay so what you want to do is write that same equation out twice so it will be (2xy-5x)(2xy-5x)

  4. anonymous
    • one year ago
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    and what we basically want to do is kind distribute

  5. anonymous
    • one year ago
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    so 2xy times 2xy = 4x^2y^2 so now do that to the -5

  6. anonymous
    • one year ago
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    so 2xy(-5) which is -10xy so so far we have 4x^2y^2-10xy

  7. geerky42
    • one year ago
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    Well, can you do this? \((a-b)^2\) If so, then it's just same process. You can let \(a = 2xy\) and \(b = 5x\).

  8. anonymous
    • one year ago
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    now we want to do the -5 (2xy) which gives -10xy so we now have 4x^2y^2-10xy-10xy

  9. anonymous
    • one year ago
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    how do I multiply xy by x?

  10. anonymous
    • one year ago
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    what do you mean?

  11. anonymous
    • one year ago
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    after foiling 2xy to the other 2xy, I have to multiply the 2xy by the -5x, and thats where I get stuck.

  12. anonymous
    • one year ago
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    oh you just do it dont stop for nothing 2xy(-5x) you do -5 times 2 which gives you -10 how many x's are there? 2 so 10x^2 and now how many y's? 1 so it becomes 10x^2y

  13. anonymous
    • one year ago
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    so when I foil xy and x, I have to combine the x's and leave the y?

  14. geerky42
    • one year ago
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    If by "leaving" y, you mean do nothing, then yes. You have \(xy\cdot x ~~~=~~~ x\cdot x\cdot y = x^2y\)

  15. anonymous
    • one year ago
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    yes and the same would go if it was y y y x z it would be y^3xz....your basically counting the variables

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