How to compute $\frac{t^2}{t+1}$ to the form $\frac{1}{t+1} +t -1$ One of my attempts would look like below.
$\frac{t^2}{t+1}$ = $\frac{t \times t+1-1}{t+1}$ = 1+ $\frac{t-1}{t+1}$ = $\frac{t-1+1-1}{t+1} + 1$ = $\frac{-2}{t+1} +2$
Also, I put into t an arbitrary number in both equations and have gotten the same answer. The first number was 18785 to which both contained the answer $\frac{352876225}{18786}$. 
 A: we have $$\frac{t^2}{t+1}=\frac{t^2-1+1}{t+1}=\frac{(t-1)(t+1)+1}{t+1}$$
A: Several responses have you make the following factorization:
$$\frac{t^2}{t+1} = \frac{(t+1)(t-1) + 1}{t+1}$$
But in general, you can take a slightly more iterative approach:
$$\frac{t^2}{t+1} = \frac{t^2 + t - t}{t+1} = \frac{t\cdot (t+1)}{t+1} - \frac{t}{t+1} = t - \big(\frac{t+1-1}{t+1}\big) = t - 1 + \frac1{t+1}$$
You just need to consider which terms are required to turn the largest term in the numerator into a multiple of the denominator.
A: This is not the most effective way to do this, but it could be a useful strategy if you're completely lost on how to proceed for a similar problem, or lost on how to think about problems like this. 
The key idea for polynomial division is to remember turning fractions like $\frac{25}{7}$ into mixed numbers (i.e. $3\frac{4}{7}$)--we want to look for something in the denominator to divide something in the numerator. In the integer case, we look for multiples of the denominator nearby the number in the numerator (i.e. 21=3*7 is nearby 25).
$t+1$ does not divide $t^2$. This next part is sort of arbitrary, but you might say $t^2+t$ is "near" $t^2$ but more importantly, it's divisible by $t+1$:
$t^2+t=t(t+1)$
To proceed, re-express the numerator in terms of this "nearby" multiple (think: 25=3*7+4)
$\frac{t^2}{t+1}=\frac{t^2+t-t}{t+1}=\frac{t(t+1)-t}{t+1}$
Now pull out the part divisible by the denominator (think: $\frac{3*7+4}{7}=3+\frac{4}{7}$):
$\frac{t^2}{t+1}=t-\frac{t}{t+1}$.
Turns out we guessed poorly, but we can repeat the procedure to get the answer you wanted--what's a multiple of $t+1$ close to $t$? $1*(t+1)$ is close to $t$:
$\frac{t}{t+1}=\frac{(t+1)-1}{t+1}=1-\frac{1}{t+1}$
Adding this to our first step:
$\frac{t^2}{t+1}=t-\frac{t}{t+1}=t-(1-\frac{1}{t+1})=\frac{1}{t+1}+t-1$.
A: Do you know long-division method:
$$\rm dividend(x)=quotient(x)+\frac{remainder(x)}{divisor(x)}$$
$$\begin{array}{r|l|l}\hline
t+1&t^2&t\\
&t^2+t\\\hline
&-t&-1\\
&-t-1\\\hline
&1
\end{array}$$
So:
$$\frac{t^2}{t+1}=(t-1)+\frac1{t+1}$$
A: I'm not sure what you did in your computation, but here's an idea and a hint: Write it as
$$\frac{t^2}{t+1}=\frac{(t^2-1)+1}{t+1} = \frac{(t+1)(t-1)+1}{t+1}.$$
Can you see what happened and where to go from there?
