Partial Fraction Decompostion of (x^4 - 1) / (x^3 + x^2 + x)

In summary: Really sorry about that. Is there any particular textbook that you recommend I study from to review this sort of stuff as well as Calc I, II and III.I would recommend "Calculus" by James Stewart for reviewing Calculus I, II, and III. As for partial fractions, "Calculus: Early Transcendentals" by James Stewart also covers it in Chapter 7.
  • #1
WK95
139
1

Homework Statement


Find Partial Fraction Decomposition of
##(x^4 - 1) / (x^3 + x^2 + x)##

Homework Equations



The Attempt at a Solution


##(x^4 - 1) / (x^3 + x^2 + x) = (x^4 - 1) / (x)(x^2 + x + 1) = A/x + B/(x^2 + x + 1)##
##(x^4 - 1) = A(x^2 + x + 1) + B(x)##
##(x^4 - 1) = Ax^2 + Ax + A + Bx##
##(x^2 - 1)(x^2 + 1) = Ax^2 + A + Ax + Bx##
##(x^2 - 1)(x^2 + 1) = A(x^2 + 1) + x(A + B)##
 
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  • #2
WK95 said:

Homework Statement


Find Partial Fraction Decomposition of
##(x^4 - 1) / (x^3 + x^2 + x)##

Homework Equations



The Attempt at a Solution


##(x^4 - 1) / (x^3 + x^2 + x) = (x^4 - 1) / (x)(x^2 + x + 1) = A/x + B/(x^2 + x + 1)##
##(x^4 - 1) = A(x^2 + x + 1) + B(x)##
##(x^4 - 1) = Ax^2 + Ax + A + Bx##
##(x^2 - 1)(x^2 + 1) = Ax^2 + A + Ax + Bx##
##(x^2 - 1)(x^2 + 1) = A(x^2 + 1) + x(A + B)##

Looks pretty messy. You could try doing a polynomial long division which would get it all over with in one go rather than trying to equate all those coefficients.
 
  • #3
Zondrina said:
Looks pretty messy. You could try doing a polynomial long division which would get it all over with in one go rather than trying to equate all those coefficients.
That would be preferred but I specifically need to approach the problem in this manner as part of teh assignment.

Through long division, I got
##(x-1) + (x-1)/(x^3 + x^2 + x)##
 
  • #4
WK95 said:
That would be preferred but I specifically need to approach the problem in this manner as part of teh assignment.

Alright then. You made a small error jumping to your second equal sign. Remember your numerator is going to look different if you have a quadratic in the denominator.
 
  • #5
Zondrina said:
Alright then. You made a small error jumping to your second equal sign. Remember your numerator is going to look different if you have a quadratic in the denominator.

I'm not seeing what you mean.
 
  • #6
WK95 said:
I'm not seeing what you mean.

If your quotient, let's call it ##Q(x)##, contains factors which are irreducible over the given field, then the numerator ##N(x)## of each partial fraction with such a factor ##F(x)## in the denominator must be found as a polynomial with ##deg(N) < deg(F)##, rather than as a constant.

The meat of that basically states :

##\frac{x^4-1}{x(x^2 + x + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + x + 1}##

In the second partial fraction notice that since ##x^2 + x + 1## is irreducible over ##ℝ##, we must find ##N(x)## such that ##deg(N) < deg(x^2 + x + 1) = 2##.

In this case that amounts to finding an arbitrary linear polynomial since ##deg(Bx + C) = 1 < 2##.
 
  • #7
What does deg( mean?
 
  • #8
WK95 said:
What does deg( mean?

##deg(f(x))## is used to denote the degree of the polynomial ##f(x)##.

##deg(x) = 1##
##deg(x^2) = 2##
 
  • #9
Partial fraction decomposition is recommended when the degree of numerator is less than the degree of the denominator. Using WK95's approach, you would first divide the original polynomials and then use partial fractions on the remainder.
 
  • #10
SteamKing said:
Partial fraction decomposition is recommended when the degree of numerator is less than the degree of the denominator. Using WK95's approach, you would first divide the original polynomials and then use partial fractions on the remainder.

##(x-1) + (x-1)/(x^3 + x^2 + x)##

##(x-1)/(x^3 + x^2 + x)=(x-1)/(x)(x^2 + x + 1)##
##(x-1)/(x)(x^2 + x + 1) = A/x + B/(x^2 + x + 1)##
##(x-1) = A(x^2 + x + 1) + B(x)##

##x=0##
##(x-1) = A(x^2 + x + 1) + (Bx + C)(x)##
##A = -1##

And then I'm stuck with finding B since I can't find x for (x^2 + x + 1) = 0
 
Last edited:
  • #11
WK95 said:
##(x-1) + (x-1)/(x^3 + x^2 + x)##

##(x-1)/(x^3 + x^2 + x)=(x-1)/(x)(x^2 + x + 1)##
##(x-1)/(x)(x^2 + x + 1) = A/x + B/(x^2 + x + 1)##
You really need to review "partial fractions". You need
##\frac{x- 1}{x(x^2+ x+ 1)}= \frac{A}{x}+ \frac{Bx+ C}{x^2+ x+ 1}##
##(x-1) = A(x^2 + x + 1) + B(x)##

##x=0##
##(x-1) = A(x^2 + x + 1) + B(x)##
##A = -1##

And then I'm stuck with finding B since I can't find x for (x^2 + x + 1) = 0
That's because there is NO such x!

After multiplying through by ##x(x^2+ x+ 1)## you have
##x- 1= A(x^2+ x+ 1)+ (Bx+ C)x= Ax^2+ Ax+ A+ Bx^2+ Cx= (A+ B)x^2+ (A+ C)x+ A##
For those to be true for all x, you must have A+ B= 0, A+ C= 1, A= -1.
 
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  • #12
HallsofIvy said:
You really need to review "partial fractions".
Really sorry about that. Is there any particular textbook that you recommend I study from to review this sort of stuff as well as Calc I, II and III.

HallsofIvy said:
After multiplying through by ##x(x^2+ x+ 1)## you have
##x- 1= A(x^2+ x+ 1)+ (Bx+ C)x= Ax^2+ Ax+ A+ Bx^2+ Cx= (A+ B)x^2+ (A+ C)x+ A##
For those to be true for all x, you must have A+ B= 0, A+ C= 1, A= -1.

Thanks a lot. I get it now
 

Related to Partial Fraction Decompostion of (x^4 - 1) / (x^3 + x^2 + x)

What is partial fraction decomposition?

Partial fraction decomposition is a method used in mathematics to simplify a rational expression into simpler forms. It is particularly useful in integration and solving differential equations.

Why is partial fraction decomposition important?

Partial fraction decomposition allows us to break down complex rational expressions into simpler forms, making them easier to work with and solve. It also helps in finding the roots of polynomials and in evaluating indefinite integrals.

How do you perform partial fraction decomposition?

To perform partial fraction decomposition, we first factor the denominator of the rational expression into linear or irreducible quadratic factors. Then, we write the expression in the form of A/(x-a) + B/(x-b) + ... , where A, B, etc. are constants. We solve for these constants by equating coefficients and then substitute them back into the original expression.

What is the purpose of partial fraction decomposition in the expression (x^4 - 1) / (x^3 + x^2 + x)?

In this expression, partial fraction decomposition allows us to simplify the denominator and break it down into linear factors, making it easier to integrate or solve for the roots. It also helps us identify any potential singularities or poles in the expression.

Can partial fraction decomposition be applied to all rational expressions?

Yes, partial fraction decomposition can be applied to any rational expression, as long as the denominator can be factored into linear or irreducible quadratic factors.

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