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anonymous

  • one year ago

How to factor? 12x^6-27y^4. I know to take out the three, but I don't know what to do from there. Any help? Thanks.

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  1. anonymous
    • one year ago
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    once you take out the 3, you have the difference of two squares

  2. anonymous
    • one year ago
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    \[3(4x^6-9y^4)\] the inner piece is like \(a^2-b^2=(a+b)(a-b)\) with \(a=2x^3, b=3y^2\)

  3. anonymous
    • one year ago
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    Ah, okay, I didn't know the formula. Thank you.

  4. anonymous
    • one year ago
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    May I ask for the formula if it is a^2+b^2, @satellite73 ?

  5. jim_thompson5910
    • one year ago
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    `a^2+b^2` is a sum of squares and can't be factored (assuming 'a' and 'b' have nothing in common)

  6. anonymous
    • one year ago
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    I'm working on the problem now, just gimme a minute and we'll see if I got it right. :-)

  7. anonymous
    • one year ago
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    I got 2x^5+3y^2)(2x^4-3y^2)

  8. anonymous
    • one year ago
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    what @jim_thompson5910 said you can factor the "difference of two squares" but not the "sum of two squares"

  9. anonymous
    • one year ago
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    hmm no

  10. anonymous
    • one year ago
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    @satellite73 How did you get 2x^3?

  11. anonymous
    • one year ago
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    the exponent for the \(x\) term should be \(3\) in both cases

  12. anonymous
    • one year ago
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    that is a good question

  13. anonymous
    • one year ago
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    i visualized \(4x^6\) as \((2x^3)^2\)

  14. anonymous
    • one year ago
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    I thought you just subtract. Since to the squared is to, you'd minus two from 6.

  15. anonymous
    • one year ago
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    So you divide the exponent?

  16. jim_thompson5910
    • one year ago
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    satellite73 is using the rule \[\LARGE (x^a)^b = x^{a*b}\] notice how the inner exponent 'a' is being multiplied with the outer exponent b

  17. jim_thompson5910
    • one year ago
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    so with 4x^6, satellite73 broke up the 6 into 3*2 and used the rule above

  18. anonymous
    • one year ago
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    Does my understanding of it? You divide?

  19. anonymous
    • one year ago
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    *does it work ,too?

  20. jim_thompson5910
    • one year ago
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    yeah 6/2 = 3

  21. jim_thompson5910
    • one year ago
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    and 4/2 = 2

  22. anonymous
    • one year ago
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    Does that work in general or was I just lucky there?

  23. jim_thompson5910
    • one year ago
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    well again, notice how 6 factors into 3*2 we want the *2 portion so we can ultimately have something squared so that's why we divide by 2 to figure out that other factor

  24. jim_thompson5910
    • one year ago
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    if we had say x^10, we'd break it up like 5*2 how do we find 5? well we can say 10/2 = 5

  25. anonymous
    • one year ago
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    Ah, okay, I am working on another problem right now. It would mean so much if you could stay for a bit to see if I got it right. :-)

  26. anonymous
    • one year ago
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    This may take more for more as me as it involves big numbers. Thanks for your patience.

  27. anonymous
    • one year ago
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    Finally got the factors. 28 and 28.

  28. anonymous
    • one year ago
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    Hold on a bit.

  29. anonymous
    • one year ago
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    (4x^2+7)^3

  30. anonymous
    • one year ago
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    Is that right?

  31. anonymous
    • one year ago
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    @satellite73 ?

  32. anonymous
    • one year ago
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    I just realized I didn't give you the problem. XD

  33. anonymous
    • one year ago
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    16x^4+56x^2+49

  34. anonymous
    • one year ago
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    your answer is correct

  35. anonymous
    • one year ago
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    Actually, according to the calculator, it's squared not cubed. :(

  36. anonymous
    • one year ago
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    oh no it isn't the exponent should be 2

  37. anonymous
    • one year ago
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    \[16x^4+56x^2+49\] is a "perfect square"

  38. anonymous
    • one year ago
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    \[(a+b)^2=a^2+2ab+b^2\] in your case \(a=4x^2,b=7\)

  39. anonymous
    • one year ago
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    Thank you. I'm fine for now, but I may be good. Thanks again for your time! :-)

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