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anonymous

  • one year ago

Can someone explain Polynomial Long Division to me?

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  1. anonymous
    • one year ago
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    I do very well with the division of a polynomial by a monomial, but once I have to divide by a binomial, I get very confused.

  2. UsukiDoll
    • one year ago
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    let me find an example somewhere brb

  3. UsukiDoll
    • one year ago
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    ah suppose we are given this problem and we need to use long division\[\LARGE \frac{x^2-9x-10}{x+1}\]

  4. UsukiDoll
    • one year ago
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    so we need to rewrite this starting with the denominator outside and the numerator inside|dw:1437820660785:dw|

  5. UsukiDoll
    • one year ago
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    so now we have to start finding what variables we need to multiply x+1 with

  6. UsukiDoll
    • one year ago
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    so how do we produce that x^2 that we need in the numerator? well we need an x so \[\LARGE x(x+1) \rightarrow x^2+x\] after we obtain this result, we need to switch the signs so when we add this to the\[\LARGE x^2-9x\] our \[\LARGE x^2 \] should be gone

  7. UsukiDoll
    • one year ago
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    so \[\LARGE x^2+x\] becomes \[\LARGE -x^2-x \]

  8. UsukiDoll
    • one year ago
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    |dw:1437820910409:dw|

  9. UsukiDoll
    • one year ago
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    we have used all of our variables up, so the only thing remaining is the number

  10. UsukiDoll
    • one year ago
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    so based on what we are seeing we have a -10x-10 so there are negatives. If I choose 10 then we will have 10(x+1) = 10x+10 but we're going to have a problem. Our goal is to cancel -10x-10 and not have a remainder if I switch the sign of 10x+10 I will have -10x-10 and that's something we don't want.

  11. UsukiDoll
    • one year ago
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    so I need to choose -10 so that -10 is distributed all over the x+1 -10(x+1) = -10x-10 switching those signs gives mean 10x+10 and that's what I need so I can cancel the -10x-10 in the division.

  12. UsukiDoll
    • one year ago
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    |dw:1437821086041:dw|

  13. UsukiDoll
    • one year ago
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    so our final answer is (x+1)(x-10) and 0 remainders expanding (x+1)(x-10) with FOIL gives back \[\LARGE x^2-10x+x-10 \rightarrow x^2-9x-10\]

  14. UsukiDoll
    • one year ago
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    so we have done this correctly

  15. UsukiDoll
    • one year ago
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    I'm going to give another example where we do have a remainder.

  16. UsukiDoll
    • one year ago
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    now suppose we are given \[\LARGE \frac{2x^3-9x^2+15}{2x-5}\]

  17. UsukiDoll
    • one year ago
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    whoa oh... we don't have an x... we just put a placeholder like 0x

  18. UsukiDoll
    • one year ago
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    |dw:1437821546765:dw|

  19. UsukiDoll
    • one year ago
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    our x's in the numerator must increase to decrease and it needs to be present.. in this example the regular x was missing so a placeholder 0x is inserted

  20. UsukiDoll
    • one year ago
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    so starting with variables only . I notice that there is a 2x^3 and we have 2x-5 if I multiply \[\LARGE x^2(2x-5) =2x^3-5x^2\] switching the signs gives me this \[\LARGE -(2x^3-5x^2) => -2x^3+5x^2\]

  21. UsukiDoll
    • one year ago
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    |dw:1437821703230:dw|

  22. UsukiDoll
    • one year ago
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    our new goal is to get rid of the -4x^2 we have already used x^2 so we can get rid of the 2x^3 so now we need an x and a number

  23. UsukiDoll
    • one year ago
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    so suppose we choose 2x then \[2x(2x-5) => 4x^2\]... whoops if I change the sign I'll have a negative -4x^2-4x^2 = -8x^2 and that's a big no no ok we need -2x \[\LARGE -2x(2x-5) => -4x^2+10x\] switching signs we have \[\LARGE -(-4x^2+10x) =4x^2-10x \]

  24. UsukiDoll
    • one year ago
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    |dw:1437821921260:dw|

  25. UsukiDoll
    • one year ago
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    so now that we have used up all the variables we are down to numbers our new goal is to get rid of -10x+15 if we have a negative we need to multiply 2x-5 with -5 , so that when the signs are switched we have a 10x. \[\LARGE -5(2x-5) = -10x+25\] switching signs \[\LARGE -(-10x+25) = 10x-25 \]

  26. UsukiDoll
    • one year ago
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    |dw:1437822060614:dw|

  27. UsukiDoll
    • one year ago
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    |dw:1437822081374:dw|

  28. UsukiDoll
    • one year ago
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    we have a remainder of -10... we need to write our answer a little bit differently

  29. UsukiDoll
    • one year ago
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    so we write the \[\LARGE x^2-2x-5\] first the remainder \[\LARGE-10\] will be on the numerator and \[\LARGE 2x-5 \] will be on the denominator so the answer will look like \[\LARGE x^2-2x-5+\frac{-10}{2x-5}\]

  30. UsukiDoll
    • one year ago
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    so do my examples make this a bit easier to understand? do you have questions?

  31. anonymous
    • one year ago
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    I understand more clearly now in terms of dividing a trinomial by a binomial, but once I get to larger polynomials, I run into trouble.

  32. anonymous
    • one year ago
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    Does the process differ at all once you get to the larger polynomials? @UsukiDoll

  33. UsukiDoll
    • one year ago
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    hmmmm... the only types of division (for polynomials) is dividing a trinomial with a monomial and a trinomial with a binomial wait brb I have to do something that needs my attention immediately

  34. anonymous
    • one year ago
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    Okay, no problem.

  35. UsukiDoll
    • one year ago
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    alright I'm back.. so is there any specific example that we can go over together?

  36. UsukiDoll
    • one year ago
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    like a polynomial divided by a trinomial? the process should be the same... just takes longer.

  37. UsukiDoll
    • one year ago
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    ah .. here's an example So suppose we have a longer polynomial divided by a trinomial like \[\LARGE \frac{x^4-3x^3+2x^2-x+1}{x^2-x+1}\]

  38. UsukiDoll
    • one year ago
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    a big heads up... this is going to take more than one drawing to do .

  39. UsukiDoll
    • one year ago
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    |dw:1437823307750:dw|

  40. UsukiDoll
    • one year ago
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    so now our goal to get rid of the x^4 so we need x^2 \[\LARGE x^2(x^2-x+1) =x^4-x^3+x^2\] switching signs \[\LARGE -(x^4-x^3+x^2) = -x^4+x^3-x^2\]

  41. UsukiDoll
    • one year ago
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    |dw:1437823443863:dw|

  42. UsukiDoll
    • one year ago
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    so now our new goal is to get rid of the -2x^3 so we need to multiply -2x with x^2-x+1 \[\LARGE -2x(x^2-x+1) = -2x^3+2x^2-2x\] switching signs we have \[\LARGE 2x^3-2x^2+2x\]

  43. UsukiDoll
    • one year ago
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    |dw:1437823607833:dw|

  44. anonymous
    • one year ago
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    I see, so it's still the same process for the most part. However, I can see how it all aligns when you're dividing polynomials with a dividend and divisor only 1 term in difference, but some of my class problems are 2, sometimes even 3 terms difference.

  45. UsukiDoll
    • one year ago
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    hmm I would do this one at a time though to avoid errors

  46. UsukiDoll
    • one year ago
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    anyway we have used up all of our variables and we are left with numbers. since we have a -x^2, we need to multiply -1 with x^2-x+1 and switch signs \[\LARGE -1(x^2-x+1) = -x^2+x-1\] switching signs \[\LARGE x^2-x+1 \]

  47. UsukiDoll
    • one year ago
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    |dw:1437823911327:dw|

  48. UsukiDoll
    • one year ago
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    |dw:1437823935681:dw|

  49. UsukiDoll
    • one year ago
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    so our remainder is 2

  50. UsukiDoll
    • one year ago
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    \[\LARGE x^2-x+1+\frac{2}{x^2-2x-1}\]

  51. UsukiDoll
    • one year ago
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    sorry got those switched

  52. UsukiDoll
    • one year ago
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    \[\LARGE x^2-2x-1+\frac{2}{x^2-x+1}\]

  53. UsukiDoll
    • one year ago
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    the process is the same. It's just that the bigger the polynomial, the more time consuming and tedious it gets

  54. alekos
    • one year ago
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    good work, this type of maths is not easy

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