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calculusxy

  • one year ago

factoring quadratics..

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  1. calculusxy
    • one year ago
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    \[5a^2 - 180\]

  2. Zale101
    • one year ago
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    Hi!

  3. calculusxy
    • one year ago
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    Hello!

  4. Zale101
    • one year ago
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    Do you have a question? :3

  5. Zale101
    • one year ago
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    Lol

  6. Zale101
    • one year ago
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    nvm

  7. calculusxy
    • one year ago
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    Yeah i just posted it above but here it is anyway: \[5a^2 - 180\]

  8. Zale101
    • one year ago
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    It didnt appear for a moment. Alright, let us begin!

  9. calculusxy
    • one year ago
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    okay!

  10. Zale101
    • one year ago
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    Let us first make the equation look clear to us. Let's factor out the 5.

  11. calculusxy
    • one year ago
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    5(a^2 - 36)

  12. Zale101
    • one year ago
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    Awesome!

  13. Zale101
    • one year ago
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    What can you say about x^2-36? Well, we know that there are 2 x's multiplied together. (x )(x ) We need to find the missing numbers. To find the missing numbers, we need to know the factors of 36 that when added together is zero and multiplied together is 36.

  14. Zale101
    • one year ago
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    \(a^2-36=a^2+0a+36\) Correct?

  15. Zale101
    • one year ago
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    Those two equations are technically the same.

  16. calculusxy
    • one year ago
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    yes

  17. Zale101
    • one year ago
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    0a means that there's two numbers that got added and were the same Those two numbers got multiplied and equaled to 36

  18. calculusxy
    • one year ago
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    wait.. i still don't understand that part about 0a

  19. Zale101
    • one year ago
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    Okay, here's a different way of explaining it a^2 means x*x 0a means number + number =0 36 means number * number =36 But, those numbers that got added and multiplied together have to be the same two numbers that got added and multiplied.

  20. calculusxy
    • one year ago
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    but what two numbers would lead to that 0

  21. calculusxy
    • one year ago
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    but when multiplied will give me 36

  22. Zale101
    • one year ago
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    Correct!

  23. calculusxy
    • one year ago
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    ok so what's the next step?

  24. Zale101
    • one year ago
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    Member, those numbers must equal 36 when multiplied, so it has to be a factor of 36.

  25. Zale101
    • one year ago
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    Getting those two numbers is the last step

  26. calculusxy
    • one year ago
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    ok...

  27. calculusxy
    • one year ago
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    1, 2, 3, 4, 9, 12, 18, 36

  28. calculusxy
    • one year ago
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    factors of 36

  29. Zale101
    • one year ago
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    |dw:1444099970567:dw|

  30. calculusxy
    • one year ago
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    6 oh!

  31. calculusxy
    • one year ago
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    okay so the answer would be : 5(a - 6)(a + 6)

  32. calculusxy
    • one year ago
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    would that be correct?

  33. Zale101
    • one year ago
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    Yes :)

  34. calculusxy
    • one year ago
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    thank you! i have another question.

  35. Zale101
    • one year ago
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    Does it make sense?

  36. calculusxy
    • one year ago
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    \[5n^2 + 5n - 360\]

  37. calculusxy
    • one year ago
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    and yes

  38. Astrophysics
    • one year ago
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    Nice one @Zale101 There is a trick to avoid all of this for the future, it's called complete the square (of course if you understand it that is) \[\huge a^2-b^2 = (a-b)(a+b)\]

  39. calculusxy
    • one year ago
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    5(n^2 + n - 75) factors of 75: 1,3,5,15,25,75 i can't find a number whose sum will be n or 1

  40. Zale101
    • one year ago
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    Nice one astro! Someone aleady sent me to try this, but i didn't want to change what i was gearing towards lol

  41. Astrophysics
    • one year ago
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    Haha, I see

  42. Zale101
    • one year ago
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    Dont look at the factors, sorry if i included this term. look at the 2 numbers that can be multiplied to get the last term but that's of course after factoring. Always look for the difference in squares. When we had 5(x^2-36) We can always use the rule astro given to us \((a^2−b^2)=(a−b)(a+b)\)

  43. calculusxy
    • one year ago
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    i didn't understand

  44. calculusxy
    • one year ago
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    @Zale101 Can you please hurry up a bit? I need to go to bed soon..

  45. Zale101
    • one year ago
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    Are you sure you factored it right?

  46. calculusxy
    • one year ago
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    i am not sure.. i desperately need help with this

  47. Zale101
    • one year ago
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    Try factoring it again.

  48. Zale101
    • one year ago
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    there was a mistake when you factored 5 out

  49. calculusxy
    • one year ago
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    5(n^2 + 1n - 72)

  50. Zale101
    • one year ago
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    Much better

  51. Zale101
    • one year ago
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    Now it should not be confusing :)

  52. Zale101
    • one year ago
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    Can you work that out?

  53. calculusxy
    • one year ago
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    5(n+9)(n-8)

  54. Zale101
    • one year ago
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    Perfect.

  55. calculusxy
    • one year ago
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    sorry i just need help with another question

  56. calculusxy
    • one year ago
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    9v^2 - 12v + 4

  57. Zale101
    • one year ago
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    Show me your attempt on ths.

  58. calculusxy
    • one year ago
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    i don't know how to factor the 9 when i have the 12 or 4 (which can't be fully divisible by 9)

  59. calculusxy
    • one year ago
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    i don't know what to do

  60. Zale101
    • one year ago
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    That's because all three terms dont have a common factor. This needs a slightly different approach.

  61. Zale101
    • one year ago
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    You have your two parenthesis prepared. Are you familiar with group factoring?

  62. calculusxy
    • one year ago
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    not that much

  63. calculusxy
    • one year ago
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    can you plz help me quickly?

  64. Zale101
    • one year ago
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    So, before i group factor i'll try to use a different approach and this approach is examining what got added that made 12. Okay, -6v-6v=-12 v we can choose to replace -12v with -6v-6v \(9v^2 - 12v + 4\) \(9v^2 -6v-6v + 4\)

  65. Zale101
    • one year ago
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    @calculusxy how much do you know about group factoring?

  66. Zale101
    • one year ago
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    Learning how to group factor will save you when factoring these types of problems.

  67. Zale101
    • one year ago
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    This will help http://www.mesacc.edu/~scotz47781/mat120/notes/factoring/grouping/grouping.html

  68. calculusxy
    • one year ago
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    http://www.purplemath.com/modules/factquad2.htm i was looking at this . i don't know if this can be useful

  69. Zale101
    • one year ago
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    You can try the bar model, it helps too!

  70. calculusxy
    • one year ago
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    okay thanks! :)

  71. Zale101
    • one year ago
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    But after i arranged the equation with -6v-6v replacing -12v, then group factoring would be easier from there.

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