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Sorry I do not know how to use the formatting will try my best.

Q. Find the binomial expansion up to $x^2$ of:

$$\frac{3+2x^2}{(2x+1)(x-3)^2}$$

For the partial fraction I get:

$$\frac{2}{7}\frac{1}{2x+1} + \frac{6}{7}\frac{1}{x-3} + \frac{3}{(x-3)^2}$$

Then I did the following:

$$(2x+1)^{-1} = 1 - 2x + 4x^2$$

$$(x-3)^{-1}= \frac{1}{3} + \frac{x}{9} + \frac{x^2}{27}$$

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

When I add them I get completely the wrong answer:

Correct answer is $$\frac{1}{3} + \frac{4x}{9} + \frac{11x^2}{9}.$$

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  • $\begingroup$ Do you polynomial division by increasing powers? $\endgroup$ – Bernard Nov 21 '17 at 15:32
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You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator: $$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$ We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):

\begin{alignat}{6} & & \color{red}{\dfrac13 -\dfrac49x}&\color{red}{{}+\dfrac{11}{9}x^2} & \color{red}{{}-\dfrac{182}{81}x^3}&+\dotsm\\% 9 + 12x-11x^2+ 2x^3 & \enspace\biggl(\enspace & 3+2x^2 \\[-18mu] & & -3 - 4x& +\dfrac{11}{3}x^2 &{} -\dfrac23 x^3 \\ & & -4x&+ \dfrac{17}3x^2&{} -\dfrac{2}{3}x^3 \\ &&4x&+\dfrac{16}3x^2&{} -\dfrac{44}{9}x^3&+\dotsm\\ &&&\hskip30mu 11x^2&{}-\dfrac{50}{9}x^3&+\dotsm\\ &&&\hskip12mu - 11x^2&{}-\dfrac{132}{9}x^3&+\dotsm\\ &&&&-\dfrac{182}{9}x^3&+\dotsm \end{alignat}

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Hint:

How about finding $a,b,c$ in

$$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$

by comparing the coefficients of different powers of $x$

For example, $(-3)^2a=3$

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  • $\begingroup$ The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong. $\endgroup$ – boi Shift Nov 21 '17 at 15:45
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$$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^m\left(1-\dfrac{mx}3+\dfrac{m(m-1)}2\left(-\dfrac x3\right)^2+\cdots\right)$$ for $|—x/3|<1$

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