## anonymous 5 years ago x^3-2x^2+5x-4=0 solve for x

1. amistre64

+- 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

wait how'd you get 2 and 4?

3. amistre64

x=1? 1 | 1 -2 5 -4 0 1 -1 4 ----------- 1 -1 4 0 goody (x+1)(x^2 -x +4)

4. amistre64

the rule is: factors of last# ------------- factors of first#

5. amistre64

typo (x-1)(x^2 -x +4)

6. amistre64

1,2,4 ----- leads to our options to test 1

7. anonymous

i dont understand that

8. amistre64

4 factors to what? 1.4 ; 2.2 all the factors of 4 are: 1,2,4

9. amistre64

1 factors to 1 ....

10. anonymous

?????

11. amistre64

what is our last number in the equation?

12. amistre64

the last number is always a product of 2 other numbers ....

13. amistre64

1(4) = 4 2(2) = 4 thats all we get right?

14. anonymous

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

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

yea....

17. anonymous

but i get a remainder and i dont know what to do with it

18. amistre64

you dont get a remainder ....

19. anonymous

i did

20. anonymous

when i synthetically divide i got -2 as a remainder

21. anonymous

ooooo. i see i forgot the negative sign okay so i got a remainder of 0

22. anonymous

then i get (x-1)(x^2-x+4) now what

23. amistre64

x^2 -x +4 ------------------ x-1 | x^3 -2x^2 +5x -4 +x^2 ------------- -x^2 +5x - x ------ 4x -4 +4 ----- 0

24. anonymous

then do i use the quadratic formula?

25. amistre64

you can try; but your gonna end up with a sqrt(-) whih means it wont factor over the reals

26. anonymous

i dont know what that is..... but do i get an imaganary number?

27. amistre64

you get imaginary numbers yes

28. anonymous

okay... can you help me with this? ((2)/(x+2))-((x)/(2-x))=((x^2+4)/(x^2-4))

29. amistre64

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

yea and then i got 4/(x+2) but i dont know how to solve for x from there

31. amistre64

-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

that almost works :) but 1 doesnt= -1

33. anonymous

hold on let me check my work

34. anonymous

wait what would the answer be then?

35. amistre64

let me do it on paper to keep better track of it

36. anonymous

wait how'd you get -(x^2-4) in the denom. i only got (x^2-4)

37. anonymous

when i simplify i get 4/(x+2) but i dont know what to do from there

38. amistre64

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

how did you get rid of the denom.?

40. amistre64

i multiplied evething by x^2-4; but i got an error still

41. anonymous

.............

42. anonymous

ok well can you just tell me how to solve 4/(x+2)

43. amistre64

there is no solution; you end up with 0=8

44. anonymous

how?

45. amistre64

$\frac{2}{x+2}-\frac{x}{2-x}=\frac{x^2 +4}{(x-2)(x+2)}$

46. anonymous

ok

47. amistre64

$\frac{2(x+2)(x-2)}{x+2} + \frac{x(x+2)(x-2)}{x+2}=x^2+4$

48. anonymous

shouldnt the right hand side have a denom.?

49. amistre64

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

$\frac{(x^2+4)(x−2)(x+2)}{(x−2)(x+2)}$

51. anonymous

ooooooo...... ok

52. anonymous

but i dont see how it cancels on the left hand side

53. amistre64

54. anonymous

omygosh thank you soooo much :)