Page 36 - Calculus for the Clueless, Calc II

Now match coefficients.

Solve two equations in two unknowns. It is really important for your algebra to be good.

Substitute in either equation.

Method 2

This is true for all values of x. If we substitute x = 3 in both sides and then x = -3 in both sides,
we will get both A and B with no work.

If x = 3, A(3 + 3) + B(3 - 3) = 2(3) + 18; 6A = 24; A = 4. If x = -3, A(-3 + 3) + B(-3 - 3)= 2(-3) + 18; -6B = 12;
B = -2.

This way is so much easier—why don't we always use it? It is only perfect if we have all linear factors to the
first power. Otherwise, it will not totally work. If there are no linear factors, you can't use this method. That is
why both methods are needed. Let us finally finish the problem!

Note how easy the calculus part is. The algebra can be overwhelming.

Example 22—

Notice the degree of the top (2) is less than the degree of the bottom (3), so long division is not necessary. The
bottom is already factored. There is one quadratic factor and one linear factor. The form is