Splitting fractions with a linear denominator: $\frac{2x-1}{x+2}$ How can $$\frac{2x-1}{x+2}$$ be split to give $$A-\frac{B}{x+2}$$
where $A$ and $B$ are integers?
The solution is $$2-\frac{5}{x+2}.$$
 A: Observe that if you set 
$$y=x+2,\quad x=y-2,$$ then you get
$$
\frac{2x-1}{x+2}=\frac{2(y-2)-1}{y}=\frac{2y-5}{y}=2-\frac{5}{y}=2-\frac{5}{x+2}
$$ as wanted.
A: $$\frac{2x-1}{x+2}=\frac{2(x+2)-5}{x+2}=\frac{2(x+2)}{x+2}-\frac{5}{x+2}=2-\frac{5}{x+2}$$
A: Imagine that you divide polynomials as usual numbers.
For example, let's divide $2x-1$ by $x+2$. Firstly, find the number such that if you multiply the divisor by it and substract from a dividend, you vanish the first term. Here $a=2$, because $2x-1 - 2(x+2) = -5$, which doesn't contain a term with $x$. -5 has the degree less than $x+2$, so we must stop. Then just take the result, which is 2, and write the expression: $2x-1=2(x+2) - 5$. Divide each part by $x+2$ and you'll get what you are looking for.
Try on the harder example:
$\frac{4x^2 + 3x + 1}{x-1}$.
Find such monome that $4x^2+3x+1 - a(x-1)$ doesn't contain $x^2$. It's $4x$: 
$4x^2+3x+1 - 4x(x-1) = 7x + 1$.
Then find the next term such that $7x+1 - a(x-1)$ doesn't contain $x$. It is $7$:
$7x+1 - 7(x-1) = 8$.
The degree of $8$ is less than the degree of $x-1$, so stop. 
Now write the full expression:
$4x^2+3x+1 = (x - 1) (4x+7) + 8$
And divide it by $x-1$:
$\frac{4x^2 + 3x + 1}{x-1} = 4x+7 + \frac{8}{x-1}$.
Everything works just the same way as the usual division of the numbers (https://en.wikipedia.org/wiki/Long_division), because numbers are polynomials with $x=10$.You can write this division the same way as you write numbers division.
This algorithm is extremely useful. For example, it is a standart way to find the integrals from functions like $\frac{f(x)}{g(x)}$, where $f(x)$ and $g(x)$ are polynomials.
A: $\dfrac{2x-1}{x+2}:$
\begin{array}{ccccccc}
    &&&& 2 \\
    &&&& -- & -- & --\\
    x & + & 2 & | & 2x & - & 1 \\
    &&&&2x & + & 4 \\
    &&&& -- & -- & --\\
    &&&&&&-5
\end{array}
So $\dfrac{2x-1}{x+2} = 2 + \dfrac{-5}{x+2} = 2 - \dfrac{5}{x+2}$
