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unimatix

  • one year ago

Simplify Algebra (in comments)

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  1. unimatix
    • one year ago
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    \[4(-8)(2-8x)^3(x^2-9)^3+3(2x)(2-8x)^4(x^2-9)^2\]

  2. unimatix
    • one year ago
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    The next step shown looks like \[4(2-8x)^3(x^2-9)^2(20x^2-3x-72)\]

  3. unimatix
    • one year ago
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    I'm having trouble understand how that was arrived at.

  4. anonymous
    • one year ago
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    hold on solving them

  5. unimatix
    • one year ago
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    Okay, thanks!

  6. anonymous
    • one year ago
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    alright here it is, use this website to do the rest http://prntscr.com/8nfl0b

  7. anonymous
    • one year ago
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    (www.Mathway.com)

  8. unimatix
    • one year ago
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    wut?

  9. Nnesha
    • one year ago
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    :=)

  10. Nnesha
    • one year ago
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    heya

  11. Nnesha
    • one year ago
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    take out the common factor

  12. unimatix
    • one year ago
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    I just need it simplified to the second step.

  13. unimatix
    • one year ago
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    because then I am solving equal to 0.

  14. unimatix
    • one year ago
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    that's where my confusion is

  15. Nnesha
    • one year ago
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    \[4(-8)\color{Red}{(2-8x)^3}\color{blue}{(x^2-9)^3}+3(2x)\color{Red}{(2-8x)^4}\color{blue}{(x^2-9)^2}\] see there is GCF (greatest common factor

  16. Nnesha
    • one year ago
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    |dw:1443912816495:dw| now what is common in both terms sorry about the writing ....:/\:

  17. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @unimatix The next step shown looks like \[4(2-8x)^3(x^2-9)^2(20x^2-3x-72)\] \(\color{blue}{\text{End of Quote}}\) i think you forgot the negative sign it's `-4`

  18. Nnesha
    • one year ago
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    \[4(-8)\color{Red}{(2-8x)^3}\color{blue}{(x^2-9)^3}+3(2x)\color{Red}{(2-8x)^4}\color{blue}{(x^2-9)^2}\] (x^2-9)^3 can be wriitten as (x^2-9)^2(x^2-9) \[4(-8)\color{Red}{(2-8x)^3}\color{blue}{(x^2-9)^2(x^2-9)}+3(2x)\color{Red}{(2-8x)^4}\color{blue}{(x^2-9)^2}\] so as you can (x^2-9)^2 is common

  19. Nnesha
    • one year ago
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    ohhh well you're not online can't work without u

  20. mathstudent55
    • one year ago
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    \(4(-8)(2-8x)^3(x^2-9)^3+3(2x)(2-8x)^4(x^2-9)^2\) \(=-32(2-8x)^3(x^2-9)^3+3(2x)(2-8x)^4(x^2-9)^2\) \(=(2 - 8x)^3(x^2 - 9)^2[-32(x^2-9)+3(2x)(2-8x)]\) \(=(2 - 8x)^3(x^2 - 9)^2[-32x^2+288+12x - 48x^2]\) \(=(2 - 8x)^3(x^2 - 9)^2(-80x^2+12x + 288)\)

  21. unimatix
    • one year ago
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    3rd line down I'm confused

  22. unimatix
    • one year ago
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    How are you moving the -32?

  23. mathstudent55
    • one year ago
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    \(=(2 - 8x)^3(x^2 - 9)^2(-80x^2+12x + 288)\) \(= -4 (2 - 8x)^3(x^2 - 9)^2(20x^2 -3 x - 72)\)

  24. mathstudent55
    • one year ago
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    Here is the second line. You're ok with it? \(=-32(2-8x)^3(x^2-9)^3+3(2x)(2-8x)^4(x^2-9)^2\)

  25. unimatix
    • one year ago
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    Yeah!

  26. mathstudent55
    • one year ago
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    From the 2nd line to the third line, we factor out two factors: \(=-32\color{red}{(2-8x)^3}\color{purple}{(x^2-9)^3}+3(2x)\color{red}{(2-8x)^4}\color{purple}{(x^2-9)^2}\)

  27. mathstudent55
    • one year ago
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    We have (in red) the factor 2 - 8x. It is to the 3 power in the first part, and to the 4th power in the second part. We can factor out (8 - 2x)^3. Then (in purple) we have the factor x^2 - 9 to the 3rd power and to the 2th power. We can factor out (x^2 - 9)^2. Ok so far?

  28. unimatix
    • one year ago
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    OKay I understand that step

  29. unimatix
    • one year ago
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    good on the 4th line too

  30. unimatix
    • one year ago
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    Okay I got it.

  31. mathstudent55
    • one year ago
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    Here is the 3rd line after factoring out those two factors: \(=\color{red}{(2 - 8x)^3}\color{purple}{(x^2 - 9)^2}[-32\color{purple}{(x^2-9)}+3(2x)\color{red}{(2-8x)}]\)

  32. mathstudent55
    • one year ago
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    Ok.

  33. unimatix
    • one year ago
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    I'm not really used to factoring out those factors I guess.

  34. mathstudent55
    • one year ago
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    Then the 5th line is just simplification inside the square brackets.

  35. unimatix
    • one year ago
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    That all makes sense. It's pretty straightforward. Thank you!

  36. mathstudent55
    • one year ago
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    Finally the last line which is on the next response above, is simply taking out a factor of -4 from inside the parentheses and bringing it out.

  37. mathstudent55
    • one year ago
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    You're welcome.

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