Problem with decomposition into partial fractions: (x^2 + 13)/(x^3 + 4x + 5)

[Citybird]

Got a problem with this.

I suppose there has to be a variable X somewhere in the numerator A and B. Can’t get it though by solving the set of equations. Suggestion?

 

[stapel]

Got a problem with this.

Decompose the following

[imath]\qquad \dfrac{x^2 + 13}{x^3 + 4x + 5}[/imath]

I started with:

[imath]\qquad \dfrac{A}{x + 1} + \dfrac{B}{x^2 - x + 5}[/imath]

I suppose there has to be a variable X somewhere in the numerator A and B. Can’t get it though by solving the set of equations. Suggestion?

You factored the denominator correctly, but then you made an error in setting up the partial fractions. To account for the fact that, in the second fraction, the denominator is quadratic, your fraction for that denominator must have a linear numerator. That is, your initial set-up should be:

[imath]\qquad \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 - x + 5} = \dfrac{x^2 + 13}{x^3 + 4x + 5}[/imath]

Multiplying through by the denominator, you would then get:

[imath]\qquad Ax^2 - Ax + 5A + Bx^2 + Bx + Cx + C = x^2 + 13[/imath]

[imath]\qquad (A + B)x^2 + (-A + B + C)x + (5A + C)1 = x^2 + 0x + 13[/imath]

Equate coefficients, and solve the system. (I get [imath]A = 2[/imath], [imath]B = -1[/imath], and [imath]C = 3[/imath].)

P.S. Thank you for showing all your work so nicely!

 

[Citybird]

Got it! Thanks a lot.