## anonymous 5 years ago can someone help me with partial fraction ? 4/x(x^2+4)

1. anonymous

i might

2. anonymous

so the function is 4/( x * ( x^2 + 4) )?

3. anonymous

yes thats right

4. anonymous

yes thats right

5. anonymous

cool, so what's the first step? separate the function based upon the denominator, you have two functions mutliplied by eachother on the bottom: x * (x^2+4)

6. anonymous

so you could create a sum of two fractions: one with x as the denonm. and one with (x^2+4) as the denom. can you post those please?

7. anonymous

i don't quite get it.. ><

8. anonymous

np, ok, the goal is to express the function as the sum of other functions, which are actually "fractions"...where the denominators of these functions that you want to sum are the parts of the denominator of the larger function....clear?

9. anonymous

so basically, here you have 4 ?( x * (x^2 + 4) )...the larger function on the denom is (x * (x^2 + 4) )....so can you separate this into two functions?

10. anonymous

4 / x + 1/(x^2+4) im not very sure.. i suck at maths

11. anonymous

right now, ignore the numerator so you're right, but ignore 4...the setup is now: \[4/(x * (x^2 + 4) ) = A/(x) + B/(x^2+4)\]

12. anonymous

does that look familiar, like something in your text?

13. anonymous

but shouldn't it be A / x + Bx+c / (x^2+4)?

14. anonymous

ok, why do you think that you use "x" twice?

15. anonymous

because of the demo has a (x^2)?

16. anonymous

sorry, you are right about that equation not being quite right, because there is x^2+4...so you actually need to adjust the numerator of the fraction with (x^2+4)

17. anonymous

good catch! the equation becomes: \[4/(x∗(x2+4))=A/(x)+(Bx+C)/(x2+4) \]

18. anonymous

yeap.. that about it.. the only thing is i don't know how to solve for Bx+C

19. anonymous

ok, now do you think you can make these fractions so that they have a common denominator? hint: multiply by "1"

20. anonymous

i did a cross multipy which end up to be 4= A(x^2+4) + X(Bx+c)?

21. anonymous

ok, however you are adding these fractions...so you cannot cross multiply...you need to "give" the fractions the same denominator...and easy way would be multiply each fraction by 1, where 1 is actually a fraction with the numerator and denominator equal, in this case you would want to multiply A/x by (x^2+4)/(x^2+4)...see how this multiplying by 1 is OK?

22. anonymous

so the fraction on becomes: \[( A * (x^2 + 4))/(x * (x^2 +4))\]

23. anonymous

cool?

24. anonymous

ohh icic

25. anonymous

nice, now can you apply the concept to the other fraction? then post the new equation?

26. anonymous

A(x^2+4) / x(x^2+4) + Bx+C / x(x^2+4) am i right?

27. anonymous

almost, what about the numerator on the second fraction?

28. anonymous

the second fraction demo is x(x^2+4) isnt it same as the 1st fraction 's demo?

29. anonymous

yes, that's right, but the numerator is missing something...

30. anonymous

\[(A * (x^2 + 4) ) /( x * x^2 +4 )) + ((Bx + C)*(????))/(x * (x^2 + 4))\]

31. anonymous

is it 1 ?

32. anonymous

why 1?

33. anonymous

right now you've only multiplied the second fraction by (1/x)...so you didn't multiply by "1" like you did for the first with (x^2+4)/(X^2+4)

34. anonymous

then is it Bx+C * x(x^2+4)?

35. anonymous

so you're close, right?

36. anonymous

so i got to muplity Bx+C with X(x^2+4)?

37. anonymous

not quite, right it is multiplied by 1/x...see how you put x in the denominator? multiply the numerator by x too...so that you have multiplied by "1"..right?

38. anonymous

Bx+c * 1/x / x(x^2+4) ?? im confused

39. anonymous

no, you were very close with (Bx + c) /x * (x^2 + 4)...how did you get here from (Bx+c)/(x^2 + 4)?

40. anonymous

oh cancel out the x from the x(9x^2+4) but then the 1st fraction's demo isnt the same as 2nd one?

41. anonymous

typo error x(x^2+4)

42. anonymous

er, no canceling now

43. anonymous

actually do the opposite....so (Bx+c)/(x^2+4) becomes ((Bx+c)*x)/((x^2+4)*x)...see how you multiplied numerator and denominator by x because you mutliplied the fraction by "1" where "1" = x/x

44. anonymous

so, what are the two fractions now, now that they have common denominators of x*(x^2+4)...rewrite the equation

45. anonymous

A( x^2+4) / x(x^2+4) + Bx+c * (x) / x (x^2+4)

46. anonymous

right on

47. anonymous

so the entire equation looks like this: \[4/(x * (x^2 + 4)) = ((A * (x^2 + 4)) + ((Bx + c) *(x)))/(x * (x^2 + 4))\]

48. anonymous

yeap..

49. anonymous

nice, now you can ignore denominators for the time being...so we actually want to look at the equality between the numerators

50. anonymous

\[4 = (A *(x^2 + 4)) + (Bx+c)*x \]

51. anonymous

this eqn isnt it same as what just now i had cross multiply them together?

52. anonymous

i'm not sure what you mean by cross multiply...

53. anonymous

as stated just now A/x + Bx+c/ x^2+4 i cross mutiply them.. meaning. A * (x^2+4) and x* (bx+c)

54. anonymous

in the original equation you're saying that you multiplied everything by the denominator?

55. anonymous

yea.

56. anonymous

cross multiply refers to when you have two fractions: like a/b = c/d , cross multiplying would yield a*d = c*b...

57. anonymous

that is what cross-multiplying technically is...it is best to the procedure as above so as to not move too hastily and miss a step...

58. anonymous

ohh..

59. anonymous

so anyway, we are close to solving

60. anonymous

4=(A∗(x2+4))+(Bx+c)∗x

61. anonymous

\[4 = (A * (x^2 + 4)) +(Bx + c)*(x)\] expand this

62. anonymous

4 = Ax^2 + 4A + Bx^2 + cx?

63. anonymous

\[4 = (A * (x^2 + 4)) + (Bx + C) * x = Ax^2 + 4A + Bx^2 + Cx\]

64. anonymous

so, now, let's consider this, both equations are assumed to be equal for all values of x...right, because it's an equation

65. anonymous

so you see how only one term does not have x in it?

66. anonymous

the 4A

67. anonymous

right on! so now we need to figure out what each letter should equal

68. anonymous

therefore A = 1

69. anonymous

dude, nice!

70. anonymous

yeah, the simplest case was to consider that x = 0, now we have that A =1...so let's find C

71. anonymous

there are no x terms on the left side, so, what do you think C equals?

72. anonymous

but as from just now eqn 4 = A(x^2+4) + x(Bx+c) can i sub X = 0 for Bx+c(x) so that 4=A(0^2+4)? which turn out to be 4 = 4A?

73. anonymous

Bx^2+Cx...that is the idea

74. anonymous

Ax^2+4A+Bx^2+Cx, ok we should group these with like terms --> (A+B)*x^2 +(C)*x + 4A

75. anonymous

(A+B)*x^2 +(C)*x + 4A = 4

76. anonymous

you figured out that A = 1 by looking at the case when x=0

77. anonymous

yea

78. anonymous

and that where i struck finding b and c

79. anonymous

now, on the right side here there are no terms with x^2 nor x...so what do you think C would be equal to?

80. anonymous

C = 0?

81. anonymous

right on!

82. anonymous

now, you know A=1, C=0, now...to find B....so we look at where we have B, only ing (A+B)*x^2

83. anonymous

which (1 + B) *x^2 because A=1

84. anonymous

again, there are no x^2 terms on the other side..so B must equal what?

85. anonymous

3?

86. anonymous

87. anonymous

why 3?

88. anonymous

the idea is that on the other side the term is 4 + 0*x^2, so we have (1+B)*x^2 = (0)*x^2

89. anonymous

ohh.. its -1

90. anonymous

yes, so what are A, B and C

91. anonymous

A= 1 B= -1 C= 0

92. anonymous

one you have them all, substitute them back in the fractions

93. anonymous

oicic

94. anonymous

where they were separted

95. anonymous

in here: 4/(x*(x^2+4))=A/(x)+B/(x^2+4)

96. anonymous

ok got it.. thx a lot man~ sry to wate such a long time.. hope u don't mind

97. anonymous

no problem, now do you think you can solve another partial fraction problem?

98. anonymous

not really.. cos' theres a lot of type of partial fraction..

99. anonymous

well the premise is the same....separate the denominator into a product of functions...then create the letter fractions with these smaller functions...solve for the letters...

100. anonymous

just use this conversation as notes....try another and good luck!

101. anonymous

ok thanks a lot

102. anonymous

np