Question

Please help with these partial fractions!!!

Answer 1

a. Factorize the denominator:

$\dfrac{x+14}{x^2-2x-8}=\dfrac{x+14}{(x-4)(x+2)}$

Then we're looking for $a,b$ such that

$\dfrac{x+14}{x^2-2x-8}=\dfrac a{x-4}+\dfrac b{x+2}$

$\implies x+14=a(x+2)+b(x-4)$

If $x=4$, then $18=6a\implies a=3$; if $x=-2$, then $12=-6b\implies b=-2$. So we have

$\dfrac{x+14}{x^2-2x-8}=\dfrac3{x-4}-\dfrac2{x+2}$

as required.

b. Same setup as in (a):

$\dfrac{-3x^2+5x+6}{x^3+x^2}=\dfrac{-3x^2+5x+6}{x^2(x+1)}$

We want to find $a,b,c$ such that

$\dfrac{-3x^2+5x+6}{x^2(x+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac c{x+1}$

Quick aside: for the second term, since the denominator has degree 2, we should be looking for another constant $b'$ such that the numerator of the second term is $b'x+b$. 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 $b'=0$ anyway.

$\implies-3x^2+5x+6=ax(x+1)+b(x+1)+cx^2$

If $x=0$, then $b=6$; if $x=-1$, then $c=-2$. Expanding everything on the right then gives

$-3x^2+5x+6=ax^2+ax+bx+b+cx^2=(a-2)x^2+(a+6)x+6$

which tells us $a-2=-3$ and $a+6=5$; in both cases, we get $a=-1$. Then

$\dfrac{-3x^2+5x+6}{x^2(x+1)}=-\dfrac1x+\dfrac6{x^2}-\dfrac2{x+1}$

as required.