This can be solvec by using partial fractions.
We know that (2x+11)/(x^2+x-6)=a/(x-2)+b/(x+3)2x+11x2+x−6=ax−2+bx+3
We can add these two fractions to get:
a/(x-2)+b/(x+3)=(a(x+3))/((x-2)(x+3))+(b(x-2))/((x-2)(x+3))ax−2+bx+3=a(x+3)(x−2)(x+3)+b(x−2)(x−2)(x+3)
=(a(x+3)+b(x-2))/((x+3)(x-2))=a(x+3)+b(x−2)(x+3)(x−2)
By multiplying by (x-3)(x+2)(x−3)(x+2), we get:
2x+11=a(x+3)+b(x-2)2x+11=a(x+3)+b(x−2)
First, we will use x=2x=2 as this will cancel out bb and allow us to work out aa.
2(2)+11=a(2+3)+b(2-2)2(2)+11=a(2+3)+b(2−2)
4+11=a(5)+b(0)4+11=a(5)+b(0)
15=5a15=5a
a=15/5=3a=155=3
Now, we do the same for x=-3x=−3
2(-3)+11=a(-3+3)+b(-3-2)2(−3)+11=a(−3+3)+b(−3−2)
-6+11=a(0)+b(-5)−6+11=a(0)+b(−5)
5=-5b5=−5b
b=-5/5=-1b=−55=−1
a=3,b=-1a=3,b=−1
(2x+11)/(x^2+x-6)=3/(x-2)+(-1)/(x+3)2x+11x2+x−6=3x−2+−1x+3
=3/(x-2)-1/(x+3)=3x−2−1x+3
