From cd83cb8b61c8ed13b13dde25d6de2e324a130164 Mon Sep 17 00:00:00 2001 From: zleyyij Date: Wed, 14 Feb 2024 09:25:34 -0700 Subject: [PATCH] vault backup: 2024-02-14 09:25:34 --- education/math/Partial Fractions.md | 7 ++++++- 1 file changed, 6 insertions(+), 1 deletion(-) diff --git a/education/math/Partial Fractions.md b/education/math/Partial Fractions.md index 11fb782..2b28616 100644 --- a/education/math/Partial Fractions.md +++ b/education/math/Partial Fractions.md @@ -11,7 +11,12 @@ $$ \frac{2x+1}{(x+1)(x+2)} $$ 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 $(x+1)(x+2)$. -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 + +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 $\frac{b}{x+2}$: +$$ \frac{a(x+2)}{(x+1)(x+2)} + \frac{b(x+1)}{(x+2(x+1))} $$ +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 ## Degree of the numerator is equal 1. First perform polynomial division.