What is this "trick" to solve for the unknowns in partial fractions?

Question

For example, $\frac{4}{(x-2)(x+7)} = \frac{A}{x+7} + \frac{B}{x-2}$

setting the numerator: $4= A(x-2) + B(x+7) $

The method I always use is to expand and compare coefficients, which can be very long if it is a more complicated fraction.

However, to solve this very quickly, we can take $x=2$ and therefore $A = -4/9$

and $x= -7$ and therefore $B=4/9$

What is this trick and when can I use it? I've not learnt this, but I've seen many people do this trick to solve integral partial fractions very quickly. Can I get some examples on when can I use it and how to determine it as I've heard that we cannot use this for every situation.

For partial fractions, I've learnt about

Linear: $Q(x) = ax+b : \frac{A}{ax+b}$

Irreducible/Quadratic: $Q(x) = ax^2 + bx+c : \frac{Ax+B}{ax^2 +bx+c}$

Answer 1

I don't know a name for the trick. The original equation comes with a hidden $x \neq 2,-7$ because of the denominator so you are technically taking the limit as $x \to 2$. You got it backwards in your post-when you substitute in $x=2$ you get the value for $B$ because the $A(x-2)$ goes to zero.

Another way would be to substitute any two other values for $x$. You would get two simultaneous equations for $A,B$. Using $2,-7$ is convenient because the equations decouple.

Answer 2

It looks like you are referring to the Heaviside cover-up method! In your case, you can multiply both sides of the equation by $(x+7)$. Setting $x = -7$ eliminates B and solves directly for A. This works for ${most}$ of the linear terms that you mention. For the irreducible terms, you may have to compare coefficients.

If you had all linear terms and one irreducible term, you could use this method to solve for all the linear terms, then solve the other term algebraically.