37 lines
2.9 KiB
Markdown
37 lines
2.9 KiB
Markdown
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: $$\frac{a*p2}{p1} + \frac{b*p1}{p2}$$
|
|
4. Now you can distribute $a$ and $b$, giving you $ax + c$ and $bx + d$. Group and factor. Your equation equals the original equation, so the numerator of the first equals $ax + bx + c + d$.
|
|
|
|
### Example
|
|
$$ \frac{2x+1}{(x+1)(x+2)} $$
|
|
1. 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:
|
|
$$ \frac{a}{x+1} + \frac{b}{x+2} $$
|
|
2. Next, find a common denominator so that you can add the two fractions together. In this case, it's $(x+1)(x+2)$
|
|
3. 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 $\frac{x+2}{x+2}$, and vice versa for $\frac{b}{x+2}$:
|
|
$$ \frac{a(x+2)}{(x+1)(x+2)} + \frac{b(x+1)}{(x+2)(x+1)} $$
|
|
4. You can now add the two equations together and distribute $a$ and $b$, giving you:
|
|
$$ \frac{ax+2a + bx+b}{(x+1)(x+2)} $$
|
|
This equals the first equation, so:
|
|
$$ \frac{2x+1}{(x+1)(x+2)} = \frac{ax+2a + bx+b}{(x+1)(x+2)} $$
|
|
5. Notice that the denominator on both sides is equal, meaning you can cancel them out, giving you:
|
|
$$ 2x + 1 = ax + 2a + bx + b $$
|
|
6. Next, group your $x$ values on one side, and your constants on the other side. You'll notice that you can factor $ax + bx$, giving you $x(a+b)$.
|
|
$$ 2x+1 = x(a + b) + (2a + b) $$
|
|
7. With the above equation, each side is in the same form. it's $x$ multiplied by a constant ($2$ on the left, and $(a+b)$ on the right, and with a constant of $1$ on the left and $2a + b$) on the right, letting you find the two equations below:
|
|
$$ 2 = a + b $$
|
|
$$ 1 = 2a + b $$
|
|
8. The above equations can be solved as a system of equations, giving you:
|
|
$$ a=-1,\space b=3 $$
|
|
9. Your answer would be:
|
|
$$ \frac{-1}{x+1} + \frac{3}{x+2} $$
|
|
|
|
## Degree of the numerator is equal
|
|
1. First perform polynomial division.
|
|
2. Then find a partial fraction with the remainder
|
|
|
|
## Degree where the numerator is greater
|
|
1. First perform polynomial division to reach a point where the degree of the numerator is less than the degree of the denominator, then follow those steps for the remainder. Your answer is in the form of $a + r$, where $a$ was your answer for division, and $r$ is your reformatted remainder. |