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

[WK95]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

 

[STEMucator]

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.

 

[WK95]

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

 

[STEMucator]

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.

 

[WK95]

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.

 

[STEMucator]

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 $deg(Bx + C) = 1 < 2$.

 

[WK95]

What does deg( mean?

 

[STEMucator]

What does deg( mean?

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

$deg(x) = 1$
$deg(x^2) = 2$

 

[SteamKing]

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.

 

[WK95]

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

 

[HallsofIvy]

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

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

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.

 

[WK95]

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.

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