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`?

The partial fraction decomposition will be of the form:

`(9x^2+26x+20)/((1+x)(2+x)^2) -= A/(1+x) + B/(2+x) + C/(2+x)^2`
`" " = ( A(2+x)^2 + B(2+x)(2+x) + C(1+x) ) / ((1+x)(2+x)^2)`

Leading to the identity:

`9x^2+26x+20 -= A(2+x)^2 + B(1+x)(2+x) + C(1+x)`

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 => 9-26+20=A=> A = 3`
Put `x = -2 => 36-52+20=-C=>C=-4`
Coeff`(x^0) => 20=4A+2B+C=> B=6`

So we can now write:

`(9x^2+26x+20)/((1+x)(2+x)^2) -= 3/(1+x) + 6/(2+x) - 4/(2+x)^2`

Now we can write the as

`S = 3/(1+x) + 6/(2+x) - 4/(2+x)^2`
`\ \ = 3/(1+x) + 6/(2(1+x/2)) - 4/(2(1+x/2))^2`
`\ \ = 3(1+x)^(-1) + 3(1+x/2)^(-1) - (1+x/2)^(-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(1+x)^(-1)`
`\ \ \ \ = 3{1+(-1)x+((-1)(-2))/(2)x^2 + ((-1)(-2)(-3))/6x^3 + ... }`
`\ \ \ \ = 3{1-x+x^2 -x^3 + ... }`
`\ \ \ \ = 3-3+3x^2 -3x^3 + ...`

`S_2 = 3(1+x/2)^(-1)`
`\ \ \ \ = 3{1+(-1)(x/2)+((-1)(-2))/(2)(x/2)^2 + ((-1)(-2)(-3))/6(x/2)^3 + ... }`
`\ \ \ \ = 3{1-1/2x+1/4x^2 -1/8x^3+ ... }`
`\ \ \ \ = 3-3/2x+3/4x^2 -3/8x^3+ ...`

`S_3 = (1+x/2)^(-2)`
`\ \ \ \ = 1+(-2)(x/2)+((-2)(-3))/2(x/2)^2+((-2)(-3)(-4))/6(x/2)^3+...`
`\ \ \ \ = 1-x+3/4x^2-1/2x^3+...`

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)`

`\ \ = (3+3-1) + (-3-3/2+1)x +(3+3/4-3/4)x^2 + (-3-3/8+1/2)x^3`

`\ \ = 5 -7/2x+3x^2 -23/8x^3 + ...`