Answer
The denominator factors as
-2x2 + 7x + 15 = (2x + 3)(5 – x)
or as
-2x2 + 7x + 15 = (-2x – 3)(x – 5).
We'll go with
-2x2 + 7x + 15 = (2x + 3)(5 – x)
first. Then the decomposition will look like
![](https://media1.shmoop.com/images/calculus/calc_indefint_partialfrac_latek_97.png)
Add the partial fractions and set the resulting numerator equal to the original numerator:
-18x – 1 = A(5 – x) + B(2x + 3).
Take x = 5 and solve for B:
-18x – 1 = A(5 – x) + B(2x + 3)
-18(5) – 1 = A(5 – 5) + B(2(5) + 3)
-91 = 13B
-7 = B
Take x = 0 and solve for A:
-18x – 1 = A(5 – x) + B(2x + 3)
-18(0) – 1 = A(5 – 0) + (-7)(2(0) + 3)
-1 = 5A – 21
4 = A
The decomposition is
![](https://media1.shmoop.com/images/calculus/calc_indefint_partialfrac_latek_100.png)
If instead we had started by factoring the denominator as
-2x2 + 7x + 15 = (-2x – 3)(x – 5)
then the decomposition would look like
![](https://media1.shmoop.com/images/calculus/calc_indefint_partialfrac_latek_101.png)
Add the partial fractions and set the resulting numerator equal to the original numerator:
-18x – 1 = A(x – 5) + B(-2x – 3).
Set x = 5 and solve for B:
-18x – 1 = A(x – 5) + B(-2x – 3)
-18(5) – 1 = A(5 – 5) + B(-2(5) – 3)
-91 = -13B
7 = B
Set x = 0 and solve for A:
-18x – 1 = A(x – 5) + B(-2x – 3)
-18(0) – 1 = A(0 – 5) + 7(-2(0) – 3)
-1 = -5A – 21
20 = -5A
-4 = A
Then the decomposition looks like this:
![](https://media1.shmoop.com/images/calculus/calc_indefint_partialfrac_latek_104.png)
While these two answers may look different, they are in fact the same. If we multiply each term of our first answer by 1, we get
![](https://media1.shmoop.com/images/calculus/calc_indefint_partialfrac_latek_105.png)
which is the same as our second answer.