Question 1205608
<pre>
You should have grouped the factors in the denominator

 f(x) = (x^2 + 7x - 6)/[(x-1)(x-2)(x+1)]

{{{f(x) =(x^2 + 7x - 6^"")/((x-1)(x-2)(x+1)^"")}}}{{{"=="}}}{{{A/(x-1)+B/(x-2)+C/(x+1)}}}

Multiply through by (x-1)(x-2)(x+1)

{{{f(x) =x^2+7x-6}}}{{{"=="}}}{{{A(x-2)(x+1)+B(x-1)(x+1)+C(x-1)(x-2)}}}

Substituting x=2 will cause two of the terms on the right to become 0, 
then you can solve for B = -2

Substituting x=-1 will also cause two of the terms on the right to become 0, 
then you can solve for C = 4

Substituting x=1 will also cause two of the terms on the right to become 0, 
then you can solve for A = -1

{{{f(x) =-expr(1/(x-1))-expr(2/(x+1))+4/(x-2)}}}  

Those terms look similar to the sum of an infinite geometric series:

{{{f(x) =a/(1-r)}}}{{{""=""}}}{{{a+ar+ar^2+ar^3+""*""*""*""}}}{{{""=""}}}{{{a(1+r+r^2+r^3+""*""*""*"")}}}

We can get the first term in that form:

{{{-1/(x-1)=1/(-x+1)=1/(1-x)}}}

We can get the second term in that form:

{{{-2/(x+1)=-2/(1+x)=-2/(1-(-x))}}}

The third term needs to have first term 1 in the denominator:

{{{4/(x-2)=-4/(2-x)=-4/(2(1-"x/2"))=-2/(1-"x/2")}}}

So we have:

{{{f(x) =1/(1-x)-2/(1-(-x))-2/(1-"x/2")}}}

Using the above equation for {{{a/(1-r)}}} on each term:

{{{f(x) =1(1+x+x^2+x^3+""*""*""*"")-2(1-x+x^2-x^3+""*""*""*"")-2(1+x/2+(x/2)^2+(x/2)^3+""*""*""*"")}}}

Simplifying that gives

{{{f(x) =1+x+x^2+x^3-2+2x-2x^2+2x^3-2-x-2(x/2)^2-2(x/2)^3+""*""*""*"")}}}

{{{f(x) =1+x+x^2+x^3-2+2x-2x^2+2x^3-2-x-2(x^2/4^"")-2(x^3/8^"")+""*""*""*""}}}

{{{f(x) =1+x+x^2+x^3-2+2x-2x^2+2x^3-2-x-x^2/2^""-x^3/4^""+""*""*""*""}}}

{{{f(x) =(1-2-2)+(1+2-1)x+(1-2-1/2)x^2+(1+2-1/4)x^3+""*""*""*""}}}

Simplifying and omitting terms in x<sup>4</sup> and higher we have:

{{{f(x) =-3 + 2x - expr(3/2) x^2 + expr(11/4)x^3}}}

Edwin</pre>