The partial fraction decomposition suggests that the function can be broken down into the sum of two other functions, or;
5x−1(x−2)(x+1)=Ax−2+Bx+1
Where we need to solve for A and B. We can cross multiply to combine the terms on the right hand side over a common denominator. We get;
5x−1(x−2)(x+1)=A(x+1)+B(x−2)(x−2)(x+1)
We can now cancel the denominator on each side, leaving;
5x−1=A(x+1)+B(x−2)
Now we can solve for A and B. We can make one of the terms cancel out by choosing the right value for x. Lets try x=~1.
5(~1)−1=A(~1+1)+B(~1−2)
The A term goes away since it is multiplied by zero, leaving;
~6=~3B
Solving for B;
B=2
We can substitute B and solve for A, but it would be easier to do the same trick that we used to solve for B. Let x=2.
5(2)−1=A(2+1)+B(2−2)
This time, the B term goes away;
9=3A
A=3
Now that we have our values for A and B we can plug into our first function and get;
5x−1(x−2)(x+1)=3x−2+2x+1
