Page 37 - Calculus for the Clueless, Calc II
P. 37
We now multiply out the top on the right and set the left numerator equal to the right numerator.
These three equations in three unknowns are not bad but not particularly nice to solve. So we can use the other
2
2
technique. Going back to the original top on the right, we have (Ax + B)(x- 2) + C(x + 5) = 9x -5x + 19. There
is only one linear factor, x - 2, but it is enough. Substituting x = 2 in this equation, we get [A(2) + B](2 - 2) +
2
2
C(2 + 5) = 9(2) - 5(2) + 19. From this we get 9C = 45 or C = 5. Substituting C = 5 into Eq. (1), we get A = 4.
Substituting A = 4 into Eq. (2), we get B = 3.
Splitting the first fraction on the right,
Example 23—
We have two linear factors, and 1 is to the second power—soooo .....
Multiply and group; we get:
