I seem to have serious problem understanding entry points of Partial fractions. I would like to decompose the following: $$\dfrac{x^4-8}{x^2+2x}$$.

My workings. Please help me judge if I am getting the concept or completely lost: I first simply the denominator ${x^2+2x}$ to become $x(x+2)$. Using $x(x+2)$ as LCD I create my partial fractions in form: $$\dfrac{x^4-8}{x^2+2x}=\dfrac{A}{x}+\dfrac{B}{x+2}$$

Multiply both sides by LCD, I get: $$x^4-8=Ax+2A+Bx$$ Collecting like terms: $x^4-8 =Ax+Bx+2A$. Simplified with coeff. to: $$(x^3)x-8=(A+B)x+2A$$ When I match with coefficients, I get $$ A+B = x^3$$ and $$2A=-8 \therefore A=-4 $$ substituting to get value of B: $B=(x^3-(-4))$ i.e. $B=(x^3+4)$

My Solution: $$\dfrac{x^4-8}{x^2+2x}=\dfrac{-4}{x}+\dfrac{x^3+4}{x+2}$$

  • $\begingroup$ first the fraction is improper, make it proper by making the degree of the numerator less than the denominator. Then apply partial fractions after. $\endgroup$ – WhizKid Oct 1 '13 at 13:14
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    $\begingroup$ Before doing partial fractions, if the numerator has degree $\ge$ the degree of the denominator, divide. $\endgroup$ – André Nicolas Oct 1 '13 at 13:15
  • $\begingroup$ @AndréNicolas, that probably is the concept I need to understand first. How are you detecting that the degree of numerator is lesser? Is it from $x^4$ or from $x^2$? $\endgroup$ – Sylvester Oct 1 '13 at 13:19
  • $\begingroup$ @Sylvester, that's correct; $4\ge 2$ so you need to divide. $\endgroup$ – vadim123 Oct 1 '13 at 16:54

You missed the first step: long division on polynomials to make the numerator of smaller degree than the denominator.

After long division, you should get $$x^2-2x+4+\frac{-8x-8}{x^2+2x}$$ and then you proceed as in your initial attempt, i.e. $$\frac{-8x-8}{x^2+2x}=\frac{A}{x}+\frac{B}{x+2}$$


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