vault backup: 2024-02-14 09:20:34
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		| @@ -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. | ||||
|   | ||||
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