vault backup: 2024-02-14 09:20:34
This commit is contained in:
parent
62ce58eb8b
commit
d942c00606
@ -1,5 +1,6 @@
|
||||
Partial fraction decomposition is when you break a polynomial fraction down into smaller fractions that add together.
|
||||
## Degree where the numerator is less
|
||||
This is the "main" method of solving, and the next two headings both focus on getting to this point, at which point you solve using the below steps.
|
||||
1. Factor the bottom.
|
||||
2. Create two fractions, $\frac{a}{p1}$, and $\frac{b}{p2}$, where p1 and p2 are the polynomials you just factored out, and a/b are arbitrary variables
|
||||
3. Multiply a by p2, and b by p1., giving you: $$\frac{a*p2}{p1} + \frac{b*p1}{p2}$$
|
||||
@ -9,7 +10,8 @@ Partial fraction decomposition is when you break a polynomial fraction down into
|
||||
$$ \frac{2x+1}{(x+1)(x+2)} $$
|
||||
Given the above fraction, the denominator is already factored, so we can move onto the next step, where two fractions are made with $a$ and $b$ in the numerator, and each of the denominator components in the denominator:
|
||||
$$ \frac{a}{x+1} + \frac{b}{x+2} $$
|
||||
Next, find a common denominator so that you can add the two fractions together. In this case, it's $(x+1)(x+2)$.
|
||||
Next, find a common denominator so that you can add the two fractions together. In this case, it's $(x+1)(x+2)$.
|
||||
Then, to make the denominators equal, you're going to multiply the numerator and the denominator by the component that the denominator is missing. In this example, the denominator for $\frac{a}{x+1}$ is missing $x+2$, so you're going to multiply by $\frac
|
||||
|
||||
## Degree of the numerator is equal
|
||||
1. First perform polynomial division.
|
||||
|
Loading…
Reference in New Issue
Block a user