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anonymous

  • one year ago

How to factor these two functions?

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  1. anonymous
    • one year ago
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    f(x)=-(x-3)^2+2 f(x)=-4x^2+16-15

  2. anonymous
    • one year ago
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    I don't really need the answer. I just need a quick review of how to factor.

  3. anonymous
    • one year ago
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    I know the first one can't be factored.

  4. jim_thompson5910
    • one year ago
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    is the second one supposed to be f(x)=-4x^2+16x-15 I put an x after the 16

  5. anonymous
    • one year ago
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    Yes.

  6. anonymous
    • one year ago
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    I apologize, that was careless of me.

  7. jim_thompson5910
    • one year ago
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    that's fine

  8. jim_thompson5910
    • one year ago
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    have you learned the quadratic formula?

  9. anonymous
    • one year ago
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    I know the formula.

  10. jim_thompson5910
    • one year ago
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    you can use the formula to solve -4x^2+16x-15 = 0 for x you'll get two roots x = p and x = q you can use those two solutions to find the factorization in the form k*(x-p)*(x-q) = 0 the k is some fixed number that scales the graph and determines whether the graph opens up or down

  11. anonymous
    • one year ago
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    Oh my gosh. Thank you!

  12. jim_thompson5910
    • one year ago
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    let me know what you get

  13. anonymous
    • one year ago
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    x= 1.5 x=2.5

  14. anonymous
    • one year ago
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    is this correct?

  15. jim_thompson5910
    • one year ago
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    that's the same as x = 3/2 and x = 5/2

  16. anonymous
    • one year ago
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    Oh!

  17. jim_thompson5910
    • one year ago
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    what we can do is multiply both sides of each equation by 2 x = 3/2 ----> 3x = 2 x = 5/2 ----> 5x = 2

  18. jim_thompson5910
    • one year ago
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    then move the 2's over through subtraction 3x = 2 ----> 3x-2 = 0 5x = 2 ----> 5x-2 = 0

  19. jim_thompson5910
    • one year ago
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    what would come next?

  20. anonymous
    • one year ago
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    I'm not sure. I've never seen this done before.

  21. jim_thompson5910
    • one year ago
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    oh wow I made a big typo I'm just noticing now

  22. jim_thompson5910
    • one year ago
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    x = 3/2 should turn into 2x = 3

  23. jim_thompson5910
    • one year ago
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    and x = 5/2 should turn into 2x = 5

  24. anonymous
    • one year ago
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    Oh. Would you multiply them from there to get the original function?

  25. jim_thompson5910
    • one year ago
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    ok so through the quadratic formula, we get these 2 solutions x = 3/2, x = 5/2 multiply both sides by 2 2x = 3, 2x = 5 then move everything to one side 2x-3=0, 2x-5=0

  26. anonymous
    • one year ago
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    Oh.

  27. jim_thompson5910
    • one year ago
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    we have 2x-3=0, 2x-5=0 they would turn into (2x-3)*(2x-5) = 0 I think that's what you had in mind?

  28. anonymous
    • one year ago
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    Yup.

  29. jim_thompson5910
    • one year ago
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    what would (2x-3)*(2x-5) expand out into?

  30. anonymous
    • one year ago
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    4x^2+16x-15

  31. jim_thompson5910
    • one year ago
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    that 16 should be negative try again

  32. anonymous
    • one year ago
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    Oops. 4x^2-16x+15

  33. jim_thompson5910
    • one year ago
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    now compare 4x^2-16x+15 (what you just got when you expanded) with -4x^2+16x-15 (the original function). Are they the same? If not, what can we do to make them the same?

  34. anonymous
    • one year ago
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    I don't think they are the same. I'm not sure. :[

  35. anonymous
    • one year ago
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    Would you need a negative one in front of the one I got?

  36. jim_thompson5910
    • one year ago
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    they aren't the same notice how the 4x^2 is positive in 4x^2-16x+15 then we have -4x^2 in -4x^2+16x-15

  37. jim_thompson5910
    • one year ago
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    same for the 16x terms and the 15 terms too

  38. anonymous
    • one year ago
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    Right. So...what did I do wrong?

  39. anonymous
    • one year ago
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    Why aren't they the same?

  40. jim_thompson5910
    • one year ago
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    well to easily fix this, we can stick a -1 in front of the factorization

  41. jim_thompson5910
    • one year ago
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    since that will make +4x^2 turn into -4x^2 the -16x turn into +16x and the +15 turn into -15

  42. jim_thompson5910
    • one year ago
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    so k = -1 is that constant I was talking about

  43. anonymous
    • one year ago
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    Oh! I think I get it. I understand evrything except the constant. Why is that -1?

  44. anonymous
    • one year ago
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    Where did the -1 come from?

  45. jim_thompson5910
    • one year ago
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    if we expanded out (2x-3)*(2x-5) we end up with 4x^2 as one of the terms we want -4x^2 so we can make (2x-3)*(2x-5) negative and say -1*(2x-3)*(2x-5) instead to fix that issue

  46. jim_thompson5910
    • one year ago
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    expanding out (2x-3)*(2x-5) will have -16x but we want +16x that -1 out front fixes the issue

  47. anonymous
    • one year ago
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    Understood. Thank you so much, Jim!

  48. jim_thompson5910
    • one year ago
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    you're welcome

  49. anonymous
    • one year ago
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    You're the best! Have a good night!

  50. jim_thompson5910
    • one year ago
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    you have a good night as well

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