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kdarnell08

How about 4x^2 + 81- 36x

  • one year ago
  • one year ago

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  1. funinabox
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    what exactly is the question

    • one year ago
  2. kdarnell08
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    i dont know how to do the problem, and i want someone to walk me throught the steps so i can understand the futture questions. 4x^2+81-36x

    • one year ago
  3. Hero
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    When factoring any quadratic, the goal is always the same. Find two numbers (x and y) that add to get b, but multiply to get ac. x + y = b xy = ac

    • one year ago
  4. Hero
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    In this case, we need to find: x + y = -36 xy = 324

    • one year ago
  5. Hero
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    The only two numbers that work is x = -18 y = -18

    • one year ago
  6. Hero
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    Anytime both x and y are the same, it means you will have a perfect square.

    • one year ago
  7. Hero
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    4x^2 - 36x + 81 4x^2 - 18x - 18x + 81 2x(2x - 9) - 9(2x - 9) (2x - 9)(2x - 9) (2x - 9)^2

    • one year ago
  8. kdarnell08
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    alright. so how about 4y^2 +16y+16 How would that work..

    • one year ago
  9. Hero
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    It works the same way: x + y = 16 xy = 64 Find x and y

    • one year ago
  10. Hero
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    I'm pretty sure you can find the two numbers

    • one year ago
  11. Hero
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    It doesn't require doing any actual Algebra. All you do is ask yourself, "What two numbers add to get 16, yet multiply to get 64?".

    • one year ago
  12. Hero
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    In this case, the two numbers are the same.

    • one year ago
  13. Hero
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    I'm not going to your homework for you. I'll show you how to do one problem. But, don't think I'm going to show you how to do every single problem. You need to figure this one out. The steps are exactly the same as the previous problem.

    • one year ago
  14. kdarnell08
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    i dont expect you to do it for me lol. i actually need to learn it. im not in hs like a bunch of the people on there. i am in nursing school and personally do not want to have to repay for the class.

    • one year ago
  15. Hero
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    Okay, so did you find the two numbers yet?

    • one year ago
  16. funinabox
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    if you are having trouble, a simple way of solving this is substitution. x + y = 16 xy = 64 or in other words y = 16 - x x(16 - x) = 64 further simplified... 16x - x^2 = 64 x^2 - 16x + 64 = 0 then you can factor and solve

    • one year ago
  17. funinabox
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    which is a much more simple polynomial to answer than the previous one

    • one year ago
  18. Hero
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    Funny thing is, she would still need to find the same two numbers either way.

    • one year ago
  19. funinabox
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    i know i realized that after i posted it :D but it's still a more simple polynomial to look at, at least in terms of co-efficients

    • one year ago
  20. kdarnell08
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    ok. i think i got it.

    • one year ago
  21. kdarnell08
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    4y^2 + 16y +16 4(y^2 + 4y + 4) 4(y^2+ 2y + 2y+ 2) 4 [y(y+2) + 2(y+2)] =4(y+2)^2

    • one year ago
  22. kdarnell08
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    right? yes no? lol ive always been terrible with numbers :/

    • one year ago
  23. Hero
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    Good job

    • one year ago
  24. kdarnell08
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    sheesh. glad i dont ddeal with this in my profession ha

    • one year ago
  25. Hero
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    You have to think of math as "playing with numbers", rather than "a chore".

    • one year ago
  26. kdarnell08
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    its just too difficult for me to understand. anyting else you can put in front of me and i get it right away, and sure certain parts of math i can too, but when it comes to this i just...ugggg. wanna curl up in a ball and cry

    • one year ago
  27. Hero
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    Most of the time, students that don't get it are usually those who are not very good with seeing the relationships between addition and multiplication. Either that or they don't have their multiplication tables well memorized.

    • one year ago
  28. Hero
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    This concept is definitely more "mental math" compared to other algebra concepts.

    • one year ago
  29. kdarnell08
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    my tables are fine...and maybe i do have a problem with my adding and multiplying and i just dont realize it? cuz ive never gotten the factoring polynomials things. not now, not in hs

    • one year ago
  30. Hero
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    I'm sorry to hear that.

    • one year ago
  31. kdarnell08
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    hey. only gotta deal with this until march so i think i can stick it out

    • one year ago
  32. Hero
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    If you have difficulty with "finding the two numbers that add to get one number, but multiply to get another", then your problem is not with factoring, but with the multiplication/addition relationships.

    • one year ago
  33. kdarnell08
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    yeah. thats what i thought.

    • one year ago
  34. kdarnell08
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    well thank you! i appreciate it very much!! back to other homework. Happy Holidays!

    • one year ago
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