a. Factorize the denominator:
[tex]\dfrac{x+14}{x^2-2x-8}=\dfrac{x+14}{(x-4)(x+2)}[/tex]
Then we're looking for [tex]a,b[/tex] such that
[tex]\dfrac{x+14}{x^2-2x-8}=\dfrac a{x-4}+\dfrac b{x+2}[/tex]
[tex]\implies x+14=a(x+2)+b(x-4)[/tex]
If [tex]x=4[/tex], then [tex]18=6a\implies a=3[/tex]; if [tex]x=-2[/tex], then [tex]12=-6b\implies b=-2[/tex]. So we have
[tex]\dfrac{x+14}{x^2-2x-8}=\dfrac3{x-4}-\dfrac2{x+2}[/tex]
as required.
b. Same setup as in (a):
[tex]\dfrac{-3x^2+5x+6}{x^3+x^2}=\dfrac{-3x^2+5x+6}{x^2(x+1)}[/tex]
We want to find [tex]a,b,c[/tex] such that
[tex]\dfrac{-3x^2+5x+6}{x^2(x+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac c{x+1}[/tex]
Quick aside: for the second term, since the denominator has degree 2, we should be looking for another constant [tex]b'[/tex] such that the numerator of the second term is [tex]b'x+b[/tex]. We always want the polynomial in the numerator to have degree 1 less than the degree of the denominator. But we would end up determining [tex]b'=0[/tex] anyway.
[tex]\implies-3x^2+5x+6=ax(x+1)+b(x+1)+cx^2[/tex]
If [tex]x=0[/tex], then [tex]b=6[/tex]; if [tex]x=-1[/tex], then [tex]c=-2[/tex]. Expanding everything on the right then gives
[tex]-3x^2+5x+6=ax^2+ax+bx+b+cx^2=(a-2)x^2+(a+6)x+6[/tex]
which tells us [tex]a-2=-3[/tex] and [tex]a+6=5[/tex]; in both cases, we get [tex]a=-1[/tex]. Then
[tex]\dfrac{-3x^2+5x+6}{x^2(x+1)}=-\dfrac1x+\dfrac6{x^2}-\dfrac2{x+1}[/tex]
as required.
