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anonymous

  • one year ago

Help for a medal and fan please?

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  1. anonymous
    • one year ago
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    Factor completely 3x3 + 6x2 - 24x.

  2. anonymous
    • one year ago
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    @johnweldon1993 can you help?

  3. johnweldon1993
    • one year ago
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    Hey :) Wel first off...notice anything about this expression? Common things I mean?

  4. anonymous
    • one year ago
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    Well I noticed there is a cubed and a squared this time

  5. anonymous
    • one year ago
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    Like most of my questions didn't have a cubed exponent

  6. johnweldon1993
    • one year ago
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    Well yes that's true lol, but also dont let that deter you...this will get easy very quick! \[\large 3x^3 + 6x^2 - 24x\] Well first thing I notice...is that each term has at least 1 'x' right?

  7. johnweldon1993
    • one year ago
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    And another thing I notice is that all these numbers are divisible by 3 right?

  8. anonymous
    • one year ago
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    Oh, haha thats what you meant XD

  9. johnweldon1993
    • one year ago
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    Lol yeah but ehh what can ya do :P

  10. johnweldon1993
    • one year ago
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    So! if each of these numbers is divisible by 3...we can factor it out \[\large 3x^2 + 6x^2 - 24x\] will then become \[\large 3(x^2 + 2x^2 - 8x)\] still with me?

  11. anonymous
    • one year ago
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    Yes I am!

  12. johnweldon1993
    • one year ago
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    Sorry typo... \[\large 3(x^3 + 2x^2 - 8x)\]

  13. anonymous
    • one year ago
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    Its alright

  14. johnweldon1993
    • one year ago
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    But okay now...since each term ALSO has at least 1 'x' ...we can ALSO factor out an 'x' \[\large 3(x^3 + 2x^2 - 8x)\] will then become \[\large 3x(x^2 + 2x - 8)\] And NOW we have something that can be factored very easily right :)

  15. anonymous
    • one year ago
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    Okay, right.

  16. johnweldon1993
    • one year ago
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    So after we factor what is inside those parenthesis...we get...?

  17. anonymous
    • one year ago
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    Stop! When you factor, do you multiply or divide? I always get confused by this

  18. johnweldon1993
    • one year ago
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    Well I use factor loosely here I admit Before what we were doing...we were factoring out a common term (With can be thought of as dividing each term we have by that term) Here when I say factor \(\large x^2 + 2x - 8\) I mean tell me what that can be broken down into

  19. anonymous
    • one year ago
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    Oooh, okay. Yeah sorry I was often told different definitions of the word factor so I got confused. So, the -8 can be broken down into -2? Is that what you are looking for?

  20. johnweldon1993
    • one year ago
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    Not quite...well as you can see we have a quadratic, so lets write it as such... \[\large x^2 + 2x - 8 = 0\] Now, we can go through the whole quadratic formula and get our roots...OR we can factor since this one is each So... \[\large x^2 + 2x - 8 = (x + ?) (x - ?)\] Does that look familiar?

  21. anonymous
    • one year ago
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    Haha, sorry but I'm still confused by what factoring means XD Sorry for being a pain D:

  22. johnweldon1993
    • one year ago
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    No that's okay...here lets take a tangent for a second If I give you a quadratic expression \[\large ax^2 + bx + c = 0\] How would you solve that?

  23. anonymous
    • one year ago
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    Well first you have to break them down, right?

  24. johnweldon1993
    • one year ago
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    How so?

  25. anonymous
    • one year ago
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    By dividing a certain number by a certain number. Sorry I'm not good at explaining so I'll give an example. Earlier, you broke down 3x^3+6x^2−24x into 3(x3+2x2−8x) by dividing them all by 3 since 3 went into all of them.

  26. anonymous
    • one year ago
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    Ugh forgot to put the ^ symbol. 3(x^3+2x^2-8x)

  27. johnweldon1993
    • one year ago
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    Okay I see what you were going for Not quite That process was a way of simplifying our "hard" expression into an easier one...what we are doing now is the next step now that we have an easy expression

  28. johnweldon1993
    • one year ago
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    SO, what I gonna do is a quick little refresher about factoring :)

  29. anonymous
    • one year ago
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    Wow, I am more clueless about this then I thought I was :O

  30. johnweldon1993
    • one year ago
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    If we have a quadratic expression \[\large ax^2 + bx+ c = 0\] We can solve this in 3 different ways...but I will focus on the basic one first The quadratic equation \[\large \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] does that look familiar?

  31. anonymous
    • one year ago
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    Oooh, yeah I remember this. I can actually finish it from here, I remember what to do from here. Thank you for your help and time!

  32. johnweldon1993
    • one year ago
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    There ya go :) yeah forget the word "factor" for now...it'll come to you a little later on :) Lemme know what you get for an answer!

  33. anonymous
    • one year ago
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    Alright, I'll also let you know what I get on my test!

  34. anonymous
    • one year ago
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    Well, not test, but assignment

  35. johnweldon1993
    • one year ago
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    Lol either way :P

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