Write (9x^2+26x+20)/((1+x)(2+x)^2) as partial fraction, then expand as binomials up to and including x^3. Show the expansion is roughly 5-(7x)/2+Bx^2+Cx^3?

For partial fractions I got: $\frac{3}{\left(1 + x\right)} + \frac{2}{\left(2 + x\right)} + \frac{4}{2 + x} ^ 2$ For the expansion, I got: $5 - \frac{9 x}{2} + 4 {x}^{2} - \frac{27 {x}^{3}}{8}$, I'm not sure why I'm getting $- \frac{9 x}{2}$ and not $- \frac{7 x}{2}$

Dec 13, 2017

partial fraction decomposition is:

$\frac{9 {x}^{2} + 26 x + 20}{\left(1 + x\right) {\left(2 + x\right)}^{2}} \equiv \frac{3}{1 + x} + \frac{6}{2 + x} - \frac{4}{2 + x} ^ 2$

The binomial expansion is:

$\frac{9 {x}^{2} + 26 x + 20}{\left(1 + x\right) {\left(2 + x\right)}^{2}} = 5 - \frac{7}{2} x + 3 {x}^{2} - \frac{23}{8} {x}^{3} + \ldots$

Leading to $B = 3$ and $C = - \frac{23}{8}$

Explanation:

The partial fraction decomposition will be of the form:

$\frac{9 {x}^{2} + 26 x + 20}{\left(1 + x\right) {\left(2 + x\right)}^{2}} \equiv \frac{A}{1 + x} + \frac{B}{2 + x} + \frac{C}{2 + x} ^ 2$
$\text{ } = \frac{A {\left(2 + x\right)}^{2} + B \left(2 + x\right) \left(2 + x\right) + C \left(1 + x\right)}{\left(1 + x\right) {\left(2 + x\right)}^{2}}$

$9 {x}^{2} + 26 x + 20 \equiv A {\left(2 + x\right)}^{2} + B \left(1 + x\right) \left(2 + x\right) + C \left(1 + x\right)$

Where $A , B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put $x = - 1 \implies 9 - 26 + 20 = A \implies A = 3$
Put $x = - 2 \implies 36 - 52 + 20 = - C \implies C = - 4$
Coeff$\left({x}^{0}\right) \implies 20 = 4 A + 2 B + C \implies B = 6$

So we can now write:

$\frac{9 {x}^{2} + 26 x + 20}{\left(1 + x\right) {\left(2 + x\right)}^{2}} \equiv \frac{3}{1 + x} + \frac{6}{2 + x} - \frac{4}{2 + x} ^ 2$

Now we can write the as

$S = \frac{3}{1 + x} + \frac{6}{2 + x} - \frac{4}{2 + x} ^ 2$
$\setminus \setminus = \frac{3}{1 + x} + \frac{6}{2 \left(1 + \frac{x}{2}\right)} - \frac{4}{2 \left(1 + \frac{x}{2}\right)} ^ 2$
$\setminus \setminus = 3 {\left(1 + x\right)}^{- 1} + 3 {\left(1 + \frac{x}{2}\right)}^{- 1} - {\left(1 + \frac{x}{2}\right)}^{- 2}$

The Binomial Series tells us that:

 (1+x)^n = 1+nx + (n(n-1))/(2!)x^2 (n(n-1)(n-2))/(3!)x^3 + ...

${S}_{1} = 3 {\left(1 + x\right)}^{- 1}$
$\setminus \setminus \setminus \setminus = 3 \left\{1 + \left(- 1\right) x + \frac{\left(- 1\right) \left(- 2\right)}{2} {x}^{2} + \frac{\left(- 1\right) \left(- 2\right) \left(- 3\right)}{6} {x}^{3} + \ldots\right\}$
$\setminus \setminus \setminus \setminus = 3 \left\{1 - x + {x}^{2} - {x}^{3} + \ldots\right\}$
$\setminus \setminus \setminus \setminus = 3 - 3 + 3 {x}^{2} - 3 {x}^{3} + \ldots$

${S}_{2} = 3 {\left(1 + \frac{x}{2}\right)}^{- 1}$
$\setminus \setminus \setminus \setminus = 3 \left\{1 + \left(- 1\right) \left(\frac{x}{2}\right) + \frac{\left(- 1\right) \left(- 2\right)}{2} {\left(\frac{x}{2}\right)}^{2} + \frac{\left(- 1\right) \left(- 2\right) \left(- 3\right)}{6} {\left(\frac{x}{2}\right)}^{3} + \ldots\right\}$
$\setminus \setminus \setminus \setminus = 3 \left\{1 - \frac{1}{2} x + \frac{1}{4} {x}^{2} - \frac{1}{8} {x}^{3} + \ldots\right\}$
$\setminus \setminus \setminus \setminus = 3 - \frac{3}{2} x + \frac{3}{4} {x}^{2} - \frac{3}{8} {x}^{3} + \ldots$

${S}_{3} = {\left(1 + \frac{x}{2}\right)}^{- 2}$
$\setminus \setminus \setminus \setminus = 1 + \left(- 2\right) \left(\frac{x}{2}\right) + \frac{\left(- 2\right) \left(- 3\right)}{2} {\left(\frac{x}{2}\right)}^{2} + \frac{\left(- 2\right) \left(- 3\right) \left(- 4\right)}{6} {\left(\frac{x}{2}\right)}^{3} + \ldots$
$\setminus \setminus \setminus \setminus = 1 - x + \frac{3}{4} {x}^{2} - \frac{1}{2} {x}^{3} + \ldots$

Combining these results:

 S= {3-3x+3x^2 -3x^3 } + {3-3/2x+3/4x^2 -3/8x^3} - { 1-x+3/4x^2-1/2x^3}+O(x^4)

$\setminus \setminus = \left(3 + 3 - 1\right) + \left(- 3 - \frac{3}{2} + 1\right) x + \left(3 + \frac{3}{4} - \frac{3}{4}\right) {x}^{2} + \left(- 3 - \frac{3}{8} + \frac{1}{2}\right) {x}^{3}$

$\setminus \setminus = 5 - \frac{7}{2} x + 3 {x}^{2} - \frac{23}{8} {x}^{3} + \ldots$