26 lines
1.8 KiB
Markdown
26 lines
1.8 KiB
Markdown
Partial fraction decomposition is when you break a polynomial fraction down into smaller fractions that add together.
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## Degree where the numerator is less
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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.
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1. Factor the bottom.
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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
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3. Multiply a by p2, and b by p1., giving you: $$\frac{a*p2}{p1} + \frac{b*p1}{p2}$$
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4. When you split the
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### Example
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$$ \frac{2x+1}{(x+1)(x+2)} $$
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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:
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$$ \frac{a}{x+1} + \frac{b}{x+2} $$
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Next, find a common denominator so that you can add the two fractions together. In this case, it's $(x+1)(x+2)$.
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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 $\frac{b}{x+2}$:
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$$ \frac{a(x+2)}{(x+1)(x+2)} + \frac{b(x+1)}{(x+2(x+1))} $$
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You can now add the two equations together and distribute $a$ and $b$, giving you:
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$$ \frac{ax+2a + bx+b}{(x+1)(x+2)}$$
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This equals the first equation
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## Degree of the numerator is equal
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1. First perform polynomial division.
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2. Then find a partial fraction with the remainder
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## Degree where the numerator is greater
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1. First perform polynomial division to reach a point where the degree of the numerator is less than the degree of the denominator |