Find the value of $r$ such that \[\frac{r^2 - 5r + 4}{r^2-8r+7} = \frac{r^2 - 2r -15}{r^2 + r - 12}.\]
1 answer
We set the fractions equal to each other and cross-multiply: \[\begin{aligned} (r^2 - 5r + 4)(r^2 + r - 12) &= (r^2 - 2r - 15)(r^2-8r+7)\\ r^4 + 2r^3 - 8r^2 - 10r^3 - 20r^2 + 80r - 60r - 48 &= r^4 - 8r^3 + 7r^2 -2r^3 +16r^2 -15r^2 - 120r + 112r - 105\\ r^4 - 8r^3 - 3r^2 - 104r + 57 &= 0. \end{aligned}\]If the quadratic $r^2 - 8r - 3$ has any rational roots, they must be divisors of 3. We check 1, -1, 3, and -3, and quickly find that only $\boxed{r = -1}$ and $r = 3$ are solutions.