I did the beginning but am not able to expand it this is my work the question is linked to another part I will link that too.
This is how a video extended the exact question but I don't know what is going on.
Thank you very much I appreciate all the help.
This seems to be an A-Level Puremath-3 question, you have to apply the general binomial theorem, assuming your partial fraction decomposition was correct:
First set up for the expansion by taking out factors from each of the terms so that they are in the general form $a^n(1+ax)^n$;
$$f(x)=(1-2x)^{-1}+3*2^{-1}\left(1-\frac{x}{2}\right)^{-1}-2*2^{-2}\left(1-\frac{x}{2}\right)^{-2}$$
$$f(x)=(1-2x)^{-1}+\frac{3}{2}\left(1-\frac{x}{2}\right)^{-1}-\frac{1}{2}\left(1-\frac{x}{2}\right)^{-2}$$
Next noting that the general binomial theorem expansion is of the form $$(1+ax)^n=1+(n)(ax)+\frac{(n)(n-1)(ax)^2}{2!}+\frac{(n)(n-1)(n-2)(ax)^3}{3!}...$$
We can start to expand each term bearing in mind that we are limited up to and including the $x^2$ term
For the first term;
$$1+(-1)(-2x)+\frac{(-1)(-2)(-2x)^2}{2}= 1+2x+4x^2 $$
For the second term; $$\frac{3}{2}\left[1+(-1)\left(-\frac{x}{2} \right) +\frac{(-1)(-2)(-\frac{x}{2})^2}{2} \right]=\frac{3}{2}\left(1+\frac{x}{2}+\frac{x^2}{4} \right)$$
For the third term $$-\frac{1}{2}\left[1 +(-2)\left(-\frac{x}{2}\right)+ \frac{(-2)(-3)\left(-\frac{x}{2}\right)^2}{2} \right]=-\frac{1}{2}\left(1+x+\frac{3x^2}{4}\right)$$
Finally adding all the terms up the final answer is
$$f(x)=2+\frac{9}{4}x+4x^2$$