From 4a569448edd8e89c68d8d97cb41877a70cca53e0 Mon Sep 17 00:00:00 2001 From: zleyyij Date: Wed, 14 Feb 2024 09:30:34 -0700 Subject: [PATCH] vault backup: 2024-02-14 09:30:34 --- education/math/Partial Fractions.md | 21 ++++++++++++--------- 1 file changed, 12 insertions(+), 9 deletions(-) diff --git a/education/math/Partial Fractions.md b/education/math/Partial Fractions.md index 2b28616..7e80d17 100644 --- a/education/math/Partial Fractions.md +++ b/education/math/Partial Fractions.md @@ -8,16 +8,19 @@ This is the "main" method of solving, and the next two headings both focus on ge ### Example $$ \frac{2x+1}{(x+1)(x+2)} $$ -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: +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: $$ \frac{a}{x+1} + \frac{b}{x+2} $$ -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 $\frac{x+2}{x+2}$, 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 - +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 $ax +$$ 2x+1 = $$ ## Degree of the numerator is equal 1. First perform polynomial division. 2. Then find a partial fraction with the remainder