Question

can someone help me with partial fraction ? 4/x(x^2+4)

Answer 1

i might

Answer 2

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

Answer 3

yes thats right

Answer 4

yes thats right

Answer 5

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)

Answer 6

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?

Answer 7

i don't quite get it.. ><

Answer 8

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?

Answer 9

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?

Answer 10

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

Answer 11

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

Answer 12

does that look familiar, like something in your text?

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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)

Answer 17

good catch! the equation becomes:


\[4/(x∗(x2+4))=A/(x)+(Bx+C)/(x2+4) \]

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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?

Answer 22

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

Answer 23

cool?

Answer 24

ohh icic

Answer 25

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

Answer 26

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

Answer 27

almost, what about the numerator on the second fraction?

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

is it 1 ?

Answer 32

why 1?

Answer 33

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)

Answer 34

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

Answer 35

so you're close, right?

Answer 36

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

Answer 37

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?

Answer 38

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

Answer 39

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

Answer 40

oh cancel out the x from the x(9x^2+4)

but then the 1st fraction's demo isnt the same as 2nd one?

Answer 41

typo error x(x^2+4)

Answer 42

er, no canceling now

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

right on

Answer 47

so the entire equation looks like this:

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

Answer 48

yeap..

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

yea.

Answer 56

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

Answer 57

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...

Answer 58

ohh..

Answer 59

so anyway, we are close to solving

Answer 60

4=(A∗(x2+4))+(Bx+c)∗x

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

the 4A

Answer 67

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

Answer 68

therefore A = 1

Answer 69

dude, nice!

Answer 70

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

Answer 71

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

Answer 72

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?

Answer 73

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

Answer 74

Ax^2+4A+Bx^2+Cx, ok we should group these with like terms -->

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

Answer 75

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

Answer 76

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

Answer 77

yea

Answer 78

and that where i struck finding b and c

Answer 79

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?

Answer 80

C = 0?

Answer 81

right on!

Answer 82

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

Answer 83

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

Answer 84

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

Answer 85

3?

Answer 86

why 3?

Answer 87

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

Answer 88

ohh.. its -1

Answer 89

yes, so what are A, B and C

Answer 90

A= 1
B= -1
C= 0

Answer 91

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

Answer 92

oicic

Answer 93

where they were separted

Answer 94

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

Answer 95

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

Answer 96

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

Answer 97

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