Partial fractions

  • Thread starter deryk
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Main Question or Discussion Point

(5x^4-6x^3+31x^2-46x-20)/(2x^5-3x^4+10x^3-14x^2+5)

I got
it = 1/(2x+1) + 4.75/(x-1) + -2/(x-1)^2 + 8.75(x^2+5)

My working was several pages so Im not gonna post it. I was wondering if any of you know if that is right? Are there any geniuses on here who can do them in there head?
 

Answers and Replies

VietDao29
Homework Helper
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Hmmm, I don't think it's correct.
You can check your answer by pluging x in and check if they are equal.
x = 0: The first one gives: -4.
While the second gives: 38.
So it's wrong...
Viet Dao,
 
TD
Homework Helper
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Factoring the denominator should give:

[tex]2x^5 - 3x^4 + 10x^3 - 14x^2 + 5 = \left( {x - 1} \right)^2 \cdot \left( {x^2 + 5} \right) \cdot \left( {2x + 1} \right)[/tex]

So propose the partial fractions:

[tex]\frac{A}{{\left( {x - 1} \right)^2 }} + \frac{B}{{x - 1}} + \frac{{Cx + D}}{{x^2 + 5}} + \frac{E}{{2x + 1}}[/tex]

Now, rather than working it out and rearranging in powers of x to get a large 5x5 system, try choosing values of x that simplify the calculations, i.e. x's for which parts become zero (zero's of the denominators)
 
lurflurf
Homework Helper
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deryk said:
(5x^4-6x^3+31x^2-46x-20)/(2x^5-3x^4+10x^3-14x^2+5)

I got
it = 1/(2x+1) + 4.75/(x-1) + -2/(x-1)^2 + 8.75(x^2+5)

My working was several pages so Im not gonna post it. I was wondering if any of you know if that is right? Are there any geniuses on here who can do them in there head?
Looks like you factored the denominator correctly
try the cover up shortcut, each highest order term can be found by substituting the a in to the function except for the factor that is zero. Lower order terms are found by subtracting off higher order terms. For terms of the form x^2+a^2 x^2 may be substituted for to avoid complex numbers, or they may be treated with the others using complex numbers.
factor denominator
(5x^4-6x^3+31x^2-46x-20)/[(2x+1)(x-1)^2(x^2+5)]
x=-1/2
(5a^4-6a^3+31a^2-46a-20)/[(2x+1)(a-1)^2(a^2+5)]|a=-1/2
=1/(2x+1) [you were right]
x=1 (order 2)
(5a^4-6a^3+31a^2-46a-20)/[(2a+1)(x-1)^2(a^2+5)]|a=1
=-2/(x-1)^2 [you were right]
x=1 (order 1)
(5a^3-a^2+30a-16)/[(2a+1)(x-1)(a^2+5)]|a=1
=3/(x-1) [you were wrong]
where the quotient without remainder is used
(5x^4-6x^3+31x^2-46x-20)=(x-1)(5x^3-x^2+30x-16)-36
x^2=-5
here to make things simple we substitute for a^2 instead of a. We multiply numerator and denominator by the conjugate of the denominator to rationalize. The general form that results is (ua+v)/(x^2-a^2) we replace a with x in the answer.
(5a^4-6a^3+31a^2-46a-20)/[(2a+1)(a-1)^2(x^2+5)]
(5a^4-6a^3+31a^2-46a-20)/[(2a^3-3a^2+1)(x^2+5)]|a^2=-5
(125+30a-155-46a-20)/[(-10a+15+1)(x^2+5)]
(-50-16a)(10a+16)/[(-10a+16)(10a+16)((x^2+5)]
(-756a/756)/(x^2+5)|a=x
-x/(x^2+5) [you were wrong]
THUS THE FINAL ANSWER
1/(2x+1)-2/(x-1)^2+3/(x-1)-x/(x^2+5)
 

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