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kdarnell08

  • 2 years ago

How about 4x^2 + 81- 36x

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  1. funinabox
    • 2 years ago
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    what exactly is the question

  2. kdarnell08
    • 2 years ago
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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

  3. Hero
    • 2 years ago
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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

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

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

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

  7. Hero
    • 2 years ago
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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

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

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

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

  11. Hero
    • 2 years ago
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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?".

  12. Hero
    • 2 years ago
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    In this case, the two numbers are the same.

  13. Hero
    • 2 years ago
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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.

  14. kdarnell08
    • 2 years ago
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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.

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

  16. funinabox
    • 2 years ago
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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

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

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

  19. funinabox
    • 2 years ago
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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

  20. kdarnell08
    • 2 years ago
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    ok. i think i got it.

  21. kdarnell08
    • 2 years ago
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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

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

  23. Hero
    • 2 years ago
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    Good job

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

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

  26. kdarnell08
    • 2 years ago
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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

  27. Hero
    • 2 years ago
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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.

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

  29. kdarnell08
    • 2 years ago
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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

  30. Hero
    • 2 years ago
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    I'm sorry to hear that.

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

  32. Hero
    • 2 years ago
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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.

  33. kdarnell08
    • 2 years ago
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    yeah. thats what i thought.

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

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