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
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 $ax + bx + c + d$.
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:
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 $\frac{b}{x+2}$:
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:
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.