How do you write the partial fraction decomposition of the rational expression (-3x^2-5x+18)/((x+1)(x^2+9))?

Apr 1, 2016

Left this solution is left in place. Whilst the values work, the solution format does not comply with the standardised form. Tony B
$\textcolor{g r e e n}{- \frac{3}{x + 1} - \frac{5}{{x}^{2} + 9} + \frac{50}{\left(x + 1\right) \left({x}^{2} + 9\right)} \textcolor{b l u e}{= \frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}}}$

Explanation:

For partial fractions we have to break down the denominator into its factorised components and then attempt to find the appropriate numerators that would take us back to the original expression. When I tried this on paper I had problems until I realised that I should go back to the first principles. Given that the right hand side is true and that the left hand side does not match the right, then the left needs adjusting until it does match. This is done using the process of mathematics.

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Given:$\text{ } \textcolor{b l u e}{\frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}}$

Write this as:

$\frac{A}{x + 1} + \frac{B}{{x}^{2} + 9} = \frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}$

Multiply throughout by $\left(x + 1\right) \left({x}^{2} + 9\right)$

$A \left({x}^{2} + 9\right) + B \left(x + 1\right) \text{ "->" } - 3 {x}^{2} - 5 x + 18$

I have used the $\to$ instead of $=$ as I know in advance that the left is not yet of the same intrinsic value as the right. An adjustment will be made later.

Multiply out the brackets giving:

$A {x}^{2} + 9 A + B x + B \text{ "->" } - 3 {x}^{2} - 5 x + 18$

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$\textcolor{b l u e}{\text{Comparing the "x^2" terms}}$

$\implies A {x}^{2} = - 3 {x}^{2} \implies \textcolor{b l u e}{A = - 3}$

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$\textcolor{b l u e}{\text{Comparing the "x" terms}}$

$\implies B x = - 5 x \implies \textcolor{b l u e}{B = - 5}$

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$\textcolor{b l u e}{\text{Comparing the constant terms}}$

$\implies 9 A + B \to + 18$

$\implies 9 \left(- 3\right) + \left(- 5\right) \to + 18$

$\implies - 32 \to + 18 \text{ this is where the problem is.}$

Let $k$ be the correction factor. Then what we really have is

$- 32 + k = + 18$

$k = + 50$

$\textcolor{b r o w n}{\text{So we need to find a way of adding 50 so that the system works, and that is:}}$

$\text{ } \textcolor{b l u e}{\frac{50}{\left(x + 1\right) \left({x}^{2} + 9\right)}}$
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$\textcolor{b l u e}{\text{Putting it all together}}$

$\textcolor{g r e e n}{- \frac{3}{x + 1} - \frac{5}{{x}^{2} + 9} + \frac{50}{\left(x + 1\right) \left({x}^{2} + 9\right)} = \frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}}$

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Check the LHS

$\frac{- 3 \left({x}^{2} + 9\right) - 5 \left(x + 1\right) + 50}{\left(x + 1\right) \left({x}^{2} + 9\right)}$

$\frac{- 3 {x}^{2} - 27 - 5 x - 5 + 50}{\left(x + 1\right) \left({x}^{2} + 9\right)}$

$\frac{- 3 {x}^{2} - 5 x - 32 + 50}{\left(x + 1\right) \left({x}^{2} + 9\right)}$

$\frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}$

$L H S = R H S$
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Apr 1, 2016

$\frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)} = \frac{2}{x + 1} - \frac{5 x}{{x}^{2} + 9}$

Explanation:

The denominator is already factored as far as possible with Real coefficients, since ${x}^{2} + 9 \ge 9 > 0$ for all $x \in \mathbb{R}$.

Assuming we want to stay with Real coefficients we are looking for a partial fraction decomposition of the form:

$\frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}$

$= \frac{A}{x + 1} + \frac{B x + C}{{x}^{2} + 9}$

$= \frac{A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x + 1\right)}{\left(x + 1\right) \left({x}^{2} + 9\right)}$

$= \frac{\left(A + B\right) {x}^{2} + \left(B + C\right) x + \left(9 A + C\right)}{\left(x + 1\right) \left({x}^{2} + 9\right)}$

Equating coefficients we get the following system of equations:

$\left\{\begin{matrix}A + B = - 3 \\ B + C = - 5 \\ 9 A + C = 18\end{matrix}\right.$

Hence we find:

$\left\{\begin{matrix}A = 2 \\ B = - 5 \\ C = 0\end{matrix}\right.$

So:

$\frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)} = \frac{2}{x + 1} - \frac{5 x}{{x}^{2} + 9}$