Question

How do you solve `(20r+4)/r + (5r+1)/(r^3)`?

Answer 1

`(20r+4)/r+(5r+1)/r^3 = ((4r^2+1)(5r+1))/r^3`

has zeros: `-1/5` and `+-i/2`

Explanation:

The question asks to "solve":

`(20r+4)/r+(5r+1)/r^3`

We can simplify it, factor it and/or find its zeros.

First note that to add two rational expressions, they need to have identical denominators - just like adding ordinary fractions.

In order to do that in our current example, we can multiply the numerator and denominator of the first expression by `r^2`. Also note that `(20r+4) = 4(5r+1)`. So we find:

`(20r+4)/r+(5r+1)/r^3 = ((20r+4)r^2)/r^3+(5r+1)/r^3`

`color(white)((20r+4)/r+(5r+1)/r^3) = ((20r+4)r^2+(5r+1))/r^3`

`color(white)((20r+4)/r+(5r+1)/r^3) = (4r^2(5r+1)+1(5r+1))/r^3`

`color(white)((20r+4)/r+(5r+1)/r^3) = ((4r^2+1)(5r+1))/r^3`

We could multiply out the numerator to get `20r^3+4r^2+5r+1`, but if we want to find the zeros or any further factors then it is better in the form `(4r^2+1)(5r+1)`.

Note that we can immediately identify `r=-1/5` is a zero, since that makes `(5r+1) = 0`.

The quadratic factor `(4r^2+1)` is always non-zero for real values of `r`. If you allow complex values, then it can be factored as a difference of squares:

`4r^2+1 = (2r)^2-i^2" "color(white)(XXXX)color(grey)("(where "i^2 = -1")")`

`color(white)(4r^2+1) = (2r-i)(2r+i)`

which identifies the two remaining zeros `r = i/2` and `r = -i/2`