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YanaSidlinskiy

  • one year ago

Help? Need some checked and answered as well... NO SPAMMING!

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  1. YanaSidlinskiy
    • one year ago
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    \[2x^2+20x+18\]

  2. YanaSidlinskiy
    • one year ago
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    I got: (x+1)(x+9)

  3. jim_thompson5910
    • one year ago
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    let's expand out (x+1)(x+9) to see what we get |dw:1438385037947:dw|

  4. jim_thompson5910
    • one year ago
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    fill out the table by multiplying the headers |dw:1438385069259:dw|

  5. jim_thompson5910
    • one year ago
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    then you add up all the terms x^2 + x + 9x + 9 = x^2 + 10x + 9 that shows us x^2 + 10x + 9 factors to (x+1)(x+9)

  6. YanaSidlinskiy
    • one year ago
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    Yeah, but the original equation is what I've posted first and factored out and came out to as (x+1)(x+9)

  7. YanaSidlinskiy
    • one year ago
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    Wait, so I factored it wrong?

  8. jim_thompson5910
    • one year ago
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    if you have a 2 outside, then it works because 2(x+1)(x+9) = 2(x^2 + 10x + 9) = 2x^2 + 20x + 18

  9. YanaSidlinskiy
    • one year ago
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    Ohh! Ok! I think that's what I was missing..For this problem: \[2x^2+6x-36\] I got: 2(x-3)(x+6) Is that correct?

  10. jim_thompson5910
    • one year ago
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    yes it is

  11. YanaSidlinskiy
    • one year ago
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    Whee! \[s^2-s-56\] MY ANSWER: (s-8)(s+7)

  12. jim_thompson5910
    • one year ago
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    very good

  13. YanaSidlinskiy
    • one year ago
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    \[m^2+13m+12\] My answer: (m+1)(m+12)

  14. anonymous
    • one year ago
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    That's right!

  15. jim_thompson5910
    • one year ago
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    agreed, you are correct

  16. YanaSidlinskiy
    • one year ago
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    \[x^2-x-72\] My answer: (x-9)(x+8)

  17. jim_thompson5910
    • one year ago
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    also correct

  18. YanaSidlinskiy
    • one year ago
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    Anyways.... I need help with this one: \[x^2-20x-13\] I'm guessing it could be prime?

  19. jim_thompson5910
    • one year ago
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    ways to multiply to -13 -1*13 1*(-13) none of those pairs add to -20 -1+13 = 12 1+(-13) = -12 so yes it is prime

  20. anonymous
    • one year ago
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    You can use the quadratic formula.

  21. jim_thompson5910
    • one year ago
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    specifically you can use the discriminant formula D = b^2 - 4ac if D is a positive number and a perfect square, then you can factor

  22. YanaSidlinskiy
    • one year ago
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    I know I can, but I do not want to take it that far alongside, that's not what I'm required to do..

  23. anonymous
    • one year ago
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    Oh okay I thought you were trying to solve for x for all of them

  24. YanaSidlinskiy
    • one year ago
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    \[3x^2-27x+54\] This can be factored out, right? No, all I need to do is factor not solving any further..

  25. anonymous
    • one year ago
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    Yes it can

  26. jim_thompson5910
    • one year ago
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    yes that can be factored. if you're stuck, try factoring out the GCF 3

  27. YanaSidlinskiy
    • one year ago
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    Ok, lemme see what I can get lol. I thought so about 3 lol.. Don't do the problem for me.. I wanna do it haha.

  28. YanaSidlinskiy
    • one year ago
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    Ok, so I got: \[3(x^2-9x+18)\] Does this seem right? If it does, not all of it has been factored out, correct?

  29. jim_thompson5910
    • one year ago
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    so far so good. can you factor the inside stuff at all?

  30. YanaSidlinskiy
    • one year ago
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    Yes, I think so.. A 2?

  31. YanaSidlinskiy
    • one year ago
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    3

  32. jim_thompson5910
    • one year ago
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    try to find two numbers that multiply to 18 and add to -9

  33. anonymous
    • one year ago
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    Yes a 3

  34. YanaSidlinskiy
    • one year ago
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    (x-3)(x-6)?

  35. jim_thompson5910
    • one year ago
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    correct

  36. anonymous
    • one year ago
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    You still need a number before the parentheses

  37. jim_thompson5910
    • one year ago
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    yes don't forget about the 3 out front

  38. anonymous
    • one year ago
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    Since you factored out a 3.

  39. YanaSidlinskiy
    • one year ago
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    Wheee! Thanks guys! Haha! Yes, definitely! I appreciate you guys helping out. I seriously can't thank you enough! Again thanks for the help put in!

  40. jim_thompson5910
    • one year ago
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    you're welcome

  41. anonymous
    • one year ago
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    Your welcome :)

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