Question

Please help with these partial fractions!!!

Answer 1

a. Factorize the denominator:

`(x+14)/(x^2-2x-8)=(x+14)/((x-4)(x+2))`

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

`(x+14)/(x^2-2x-8)=\frac a{x-4}+\frac 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

`(x+14)/(x^2-2x-8)=\frac3{x-4}-\frac2{x+2}`

as required.

b. Same setup as in (a):

`(-3x^2+5x+6)/(x^3+x^2)=(-3x^2+5x+6)/(x^2(x+1))`

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

`(-3x^2+5x+6)/(x^2(x+1))=\frac ax+\frac b{x^2}+\frac 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

`(-3x^2+5x+6)/(x^2(x+1))=-\frac1x+\frac6{x^2}-\frac2{x+1}`

as required.