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What formula do I use when i have to find the partial fraction of

$$\frac{10x^2+11x+19 }{ (x-0.5)(2x^2+6x+10)}$$

Is it $A(2x^2+6x+10) + (Bx+C)(x-0.5)$ ?

Or do I have to factorise $(2x^2+6x+10)$ into $2(x^2+6x+10)$ and then use another formula?

Vladimir Vargas
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  • You should format your question with LaTeX. As is, your question is pretty ambiguous. For example, are you trying to use partial fraction decomposition on the function $$\frac{10x^2+11x+19}{ (x-0.5)(2x^2+6x+10)}$$ or $$10x^2+11x+\frac{19}{ (x-0.5)(2x^2+6x+10)}$$ – graydad Nov 19 '14 at 22:20

2 Answers2

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For partial fractions, the numerator of each fraction should always be of lower degree than the denominator. Also, you can move a constant factor between the two denominator components to make some of the numbers look nicer. So you'd have:

$\begin{eqnarray} \frac{10x^2+11x+19}{(x-0.5)(2x^2+6x+10)} & = & \frac{10x^2+11x+19}{(2x-1)(x^2+3x+5)} \\ & \equiv & \frac{A}{2x-1} + \frac{Bx+C}{x^2+3x+5}\end{eqnarray}$

and then you can start mucking around with things to find A, B and C.

ConMan
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Since $2x^2+6x+10$ cannot be factored further (check $b^2-4ac<0$), you write $$\frac{10x^2+11x+19}{(x-0.5)(2x^2+6x+10)} = \frac{A}{x-0.5} + \frac{Bx+C}{2x^2+6x+10}$$

In general:

If you have a simple linear factor $(x-r)$ in the denominator, you include a term $\frac{A}{x-r}$.

If you have a repeated root $(x-r)^k$ in the denominator, you include terms $\frac{A_n}{(x-r)^n}$ for $n=1,2,\dots,k$.

If you have an irreducible quadratic $(ax^2+bx+c)$ in the denominator, you include a term $\frac{Bx+C}{ax^2+bx+c}$.

RHP
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