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anonymous

  • 5 years ago

x^3-2x^2+5x-4=0 solve for x

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  1. amistre64
    • 5 years ago
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    +- 1,2,4 x=2? 2 | 1 -2 5 -4 0 2 0 10 ------------ 1 0 5 6 nope, but since its all+ it aint bigger than 2

  2. anonymous
    • 5 years ago
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    wait how'd you get 2 and 4?

  3. amistre64
    • 5 years ago
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    x=1? 1 | 1 -2 5 -4 0 1 -1 4 ----------- 1 -1 4 0 goody (x+1)(x^2 -x +4)

  4. amistre64
    • 5 years ago
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    the rule is: factors of last# ------------- factors of first#

  5. amistre64
    • 5 years ago
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    typo (x-1)(x^2 -x +4)

  6. amistre64
    • 5 years ago
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    1,2,4 ----- leads to our options to test 1

  7. anonymous
    • 5 years ago
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    i dont understand that

  8. amistre64
    • 5 years ago
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    4 factors to what? 1.4 ; 2.2 all the factors of 4 are: 1,2,4

  9. amistre64
    • 5 years ago
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    1 factors to 1 ....

  10. anonymous
    • 5 years ago
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    ?????

  11. amistre64
    • 5 years ago
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    what is our last number in the equation?

  12. amistre64
    • 5 years ago
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    the last number is always a product of 2 other numbers ....

  13. amistre64
    • 5 years ago
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    1(4) = 4 2(2) = 4 thats all we get right?

  14. anonymous
    • 5 years ago
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    sorry i dont understand that..... im only in algebra 2..... all i know is you graph it and you get x=1, then i don't know what to do from there

  15. amistre64
    • 5 years ago
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    well, if x=1; then that means that (x-1) is a factor and you divide it out of the original equation to get whats left

  16. anonymous
    • 5 years ago
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    yea....

  17. anonymous
    • 5 years ago
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    but i get a remainder and i dont know what to do with it

  18. amistre64
    • 5 years ago
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    you dont get a remainder ....

  19. anonymous
    • 5 years ago
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    i did

  20. anonymous
    • 5 years ago
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    when i synthetically divide i got -2 as a remainder

  21. anonymous
    • 5 years ago
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    ooooo. i see i forgot the negative sign okay so i got a remainder of 0

  22. anonymous
    • 5 years ago
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    then i get (x-1)(x^2-x+4) now what

  23. amistre64
    • 5 years ago
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    x^2 -x +4 ------------------ x-1 | x^3 -2x^2 +5x -4 +x^2 ------------- -x^2 +5x - x ------ 4x -4 +4 ----- 0

  24. anonymous
    • 5 years ago
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    then do i use the quadratic formula?

  25. amistre64
    • 5 years ago
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    you can try; but your gonna end up with a sqrt(-) whih means it wont factor over the reals

  26. anonymous
    • 5 years ago
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    i dont know what that is..... but do i get an imaganary number?

  27. amistre64
    • 5 years ago
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    you get imaginary numbers yes

  28. anonymous
    • 5 years ago
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    okay... can you help me with this? ((2)/(x+2))-((x)/(2-x))=((x^2+4)/(x^2-4))

  29. amistre64
    • 5 years ago
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    2 x x^2+4 ---- - ---- = ------ x+2 2-x x^2-4 ^^^ -(x-2) -2 x x^2+4 ---- - ---- = ------ 2-x 2-x x^2-4 -2(2-x) x(2-x) x^2+4 ------ - ------- = ------ 4 -x^2 x^2-4

  30. anonymous
    • 5 years ago
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    yea and then i got 4/(x+2) but i dont know how to solve for x from there

  31. amistre64
    • 5 years ago
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    -4 +2x -2x +x^2 x^2+4 ---------------- = ------ -(x^2-4) x^2-4 4-x^2 = x^2+4 -x^2 = x^2 when x=0

  32. amistre64
    • 5 years ago
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    that almost works :) but 1 doesnt= -1

  33. anonymous
    • 5 years ago
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    hold on let me check my work

  34. anonymous
    • 5 years ago
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    wait what would the answer be then?

  35. amistre64
    • 5 years ago
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    let me do it on paper to keep better track of it

  36. anonymous
    • 5 years ago
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    wait how'd you get -(x^2-4) in the denom. i only got (x^2-4)

  37. anonymous
    • 5 years ago
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    when i simplify i get 4/(x+2) but i dont know what to do from there

  38. amistre64
    • 5 years ago
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    2(2-x) -x(2+x) = x^2 +4 4 -2x -2x -x^2 = x^2+4 0 = x^2+x^2 +4x +4-4 0 = 2x^2 +4x 0 = 2x(x+2) x = 0, -2

  39. anonymous
    • 5 years ago
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    how did you get rid of the denom.?

  40. amistre64
    • 5 years ago
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    i multiplied evething by x^2-4; but i got an error still

  41. anonymous
    • 5 years ago
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    .............

  42. anonymous
    • 5 years ago
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    ok well can you just tell me how to solve 4/(x+2)

  43. amistre64
    • 5 years ago
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    there is no solution; you end up with 0=8

  44. anonymous
    • 5 years ago
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    how?

  45. amistre64
    • 5 years ago
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    \[\frac{2}{x+2}-\frac{x}{2-x}=\frac{x^2 +4}{(x-2)(x+2)}\]

  46. anonymous
    • 5 years ago
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    ok

  47. amistre64
    • 5 years ago
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    \[\frac{2(x+2)(x-2)}{x+2} + \frac{x(x+2)(x-2)}{x+2}=x^2+4\]

  48. anonymous
    • 5 years ago
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    shouldnt the right hand side have a denom.?

  49. amistre64
    • 5 years ago
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    no denom, it canceled: \[frac{(x^2+4)(x-2)(x+2)}{(x-2)(x+2)}\] cancels \[2(x-2)+x(x-2)=x^2+4\] \[2x-4+x^2-2x=x^2+4\] \[x^2 -4 = x^2+4\] \[x^2-x^2 = 4+4\] \[0=8 \implies \emptyset \]

  50. amistre64
    • 5 years ago
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    \[\frac{(x^2+4)(x−2)(x+2)}{(x−2)(x+2)}\]

  51. anonymous
    • 5 years ago
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    ooooooo...... ok

  52. anonymous
    • 5 years ago
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    but i dont see how it cancels on the left hand side

  53. amistre64
    • 5 years ago
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  54. anonymous
    • 5 years ago
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    omygosh thank you soooo much :)

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