As the author presents it, it seems to me difficult to discover the idea of "partially clearing denomninators"; but notice that we can still clear all denominat

As the author presents it, it seems to me difficult to discover the idea of "partially clearing denomninators"; but notice that we can still clear all denominators—we just have to be lazy about what we do next. (This is one of the things that I have trouble convincing my calculus students to do; I'll frequently see them encounter a product of two binomials, reflexively multiply them together, and then spend time factoring them to find the roots of the product.) Anyway, if we clear denominators in the author's example, then we get: 7x + 2 = A(x - 1)(x + 2) + B(x - 1) + C(x + 2)^2. As long as we don't multiply and re-group as instinct may suggest, we can plug in `x = 1` just as easily as the author describes: 9 = 0 + 0 + 9C. As he suggests, we can also plug in `x = -2` (and now it's maybe less mysterious why this is still OK) to get -12 = -3B ; and then we can proceed as the author does, subtracting off the B term and factoring out `(x + 2)` from what's left, or (which is substantially the same procedure) we can differentiate: 7 = A(x - 1) + B + 2C(x + 2) => 7 = -3A + B. (The differentiation would look more complicated if we had more terms, but notice that we don't actually need to know that the derivative of `C(x + 2)^2` is `2C(x + 2)`, only that the derivative of any polynomial divisible by `(x + 2)^2` is divisible by `(x + 2)`, and so vanishes at `x = -2`.)