How do you write the partial fraction decomposition of the rational expression #x^2/ (x^2+x+4)#?

1 Answer
Jan 21, 2017

Real solution:

#x^2/(x^2+x+4) = 1-(x+4)/(x^2+x+4)#

Complex solution:

#x^2/(x^2+x+4) = 1-(1/2-7/30sqrt(15)i)/(x+1/2-sqrt(15)/2i)-(1/2+7/30sqrt(15)i)/(x+1/2+sqrt(15)/2i)#

Explanation:

Note that the discriminant of #x^2+x+4# is negative:

#Delta = 1^2-4(1)(4) = 1-16 = -15 < 0#

So this quadratic is irreducible over the real numbers.

So (assuming we want real coefficients), the best we can do is split the given rational expression into a polynomial part and an expression of the form #(Ax+B)/(x^2+x+4)#, which we can do as follows:

#x^2/(x^2+x+4) = ((x^2+x+4)-(x+4))/(x^2+x+4) = 1-(x+4)/(x^2+x+4)#

#color(white)()#
Complex solution

If we want a partial fraction decomposition with Complex coefficients, then we can proceed as follows:

#(x+4)/(x^2+x+4) = (x+4)/((x+1/2-sqrt(15)/2i)(x+1/2+sqrt(15)/2i))#

#color(white)((x+4)/(x^2+x+4)) = A/(x+1/2-sqrt(15)/2i)+B/(x+1/2+sqrt(15)/2i)#

#color(white)((x+4)/(x^2+x+4)) = (A(x+1/2+sqrt(15)/2i)+B(x+1/2-sqrt(15)/2i))/(x^2+x+4)#

Equating coefficients, we find:

#{ (A+B = 1), (1/2(A+B)+sqrt(15)/2i(A-B) = 4) :}#

Subtracting #1/2# of the first equation from the second we get:

#sqrt(15)/2i(A-B) = 7/2#

Multiplying both sides by #-2/sqrt(15)i# we get:

#A-B = -7/15sqrt(15)i#

Adding or subtracting the first equation and dividing by #2# we find:

#A = 1/2(1-7/15sqrt(15)i) = 1/2-7/30sqrt(15)i#

#B = 1/2(1+7/15sqrt(15)i) = 1/2+7/30sqrt(15)i#

So:

#(x+4)/(x^2+x+4) = (1/2-7/30sqrt(15)i)/(x+1/2-sqrt(15)/2i)+(1/2+7/30sqrt(15)i)/(x+1/2+sqrt(15)/2i)#

And:

#x^2/(x^2+x+4) = 1-(1/2-7/30sqrt(15)i)/(x+1/2-sqrt(15)/2i)-(1/2+7/30sqrt(15)i)/(x+1/2+sqrt(15)/2i)#