Rational Expression equivalent form EDIT: I know how to find the answer, but does anyone know why plugging in numbers for x does not work?
The Question: If the rational expression $\frac {3x^2}{3x-1}$ is rewritten in the equivalent form $\frac {\frac 13}{3x-1}+A$, what must expression A be in terms of x?
The four answer choices:  A) $x+ \frac 13$  B) $x+1$  C) $x-1$  D) $x-3$
The Answer:

 A) $x+ \frac 13$

I don't really know the way this question is "supposed" to be solved, so I just tried plugging in some numbers. 


*

*Plugging in 0 for x, I would get $0=\frac {-1}3+ A$

*Plugging in 1 for x, I would get $\frac 32=\frac 23 + A$.

*Plugging in 2 for x, I would get $\frac {12}{5}=\frac 53 +A$.


The equation for $x=0$ makes it seem like $A$ really is the correct answer, but plugging in 1 and 2, I wasn't sure which one, if any were correct. Pretty sure I'm just missing something really obvious here, but asking anyway. Answer Key contains the answer listed above.
 A: HINT From $\frac {3x^2}{3x-1} = \frac {\frac 13}{3x-1}+A $ you get $A = \frac {3x^2}{3x-1} - \frac {\frac 13}{3x-1}$.
A: You can solve the equation for A and get 
$A = \frac {3x^2-\frac 13}{3x-1}$. 
In the numerator 3 can be factored out.
$A = \frac {3(x^2-\frac 19)}{3x-1}$
$x^2-\frac 19$ is equivalent to the third binomial formula $a^2-b^2=(a-b)\cdot (a+b)$.
Therfore $ x^2-\frac 19=(x-\frac 13)\cdot (x+\frac 13)$
$A= \frac {3\cdot (x-\frac 13)\cdot (x+\frac 13)}{3x-1}=\frac { (3x-1)\cdot (x+\frac 13)}{3x-1}$

Numerical example: $x=2$
$\frac {3x^2}{3x-1}=\frac {3\cdot 2^2}{3\cdot 2-1}=\frac{12}{5}$
This result has to be the same like
$\frac{\frac13}{3x-1}+x+\frac 13=\frac{1}{9x-3}+x+\frac13$
$=\frac{1}{15}+2+\frac 13=\frac{1}{15}+\frac{30}{15}+\frac{5}{15}=\frac{36}{15}=\frac{12}{5}\quad \checkmark$
A: You have
$$\frac{3x^2}{3x-1}=\frac{1/3}{3x-1}+A$$
so
$$3x^2=\frac{1}{3}+A(3x-1)$$
$$A=\frac{3x^2-1/3}{3x-1}=\frac{1}{3}\cdot\frac{9x^2-1}{3x-1}=\frac {1}{3}\frac{(3x-1)(3x+1)}{3x-1}$$
etc.
A: $\frac{3x^2}{3x-1} = \frac{3x^2-x+x}{3x-1} = \frac{x(3x-1)}{3x-1}+\frac{x}{3x-1} = x+\frac{x}{3x-1} = x+\frac{x-\frac{1}{3}+\frac{1}{3}}{3x-1}$
$=x+\frac{\frac{1}{3}(3x-1)}{3x-1} + \frac{\frac{1}{3}}{3x-1} = x+\frac{1}{3}+\frac{\frac{1}{3}}{3x-1}$
At each step, I generally did one of two things: I added and subtracted the same amount (equivalent to adding zero which is always allowed) to the same place, or I factored a common factor, or I separated a fraction into multiple pieces.
A: One may write
$$
\frac {3x^2}{3x-1}=\frac {\frac13\left(9x^2-1\right)+\frac13}{3x-1}=\frac {\frac13}{3x-1}+\frac {\frac13\left(3x-1\right)(3x+1)}{3x-1}=\frac {\frac13}{3x-1}+x+\frac13.
$$
A: Although there are various ways to solve this problem, especially since you're given the expression $\frac{1/3}{3x-1}+A$ for $\frac{3x^2}{3x-1}$ and so methods like solving for $A$ by subtraction are apparent, it seems most natural to me to interpret this as an exercise in dividing polynomials. The "best" approach is always a rather subjective assessment, but long division gives you a systematic way to obtain expressions like your $\frac{1/3}{3x-1} + A$ without needing to have a part given.
You may already be familiar with this, but just as you perform long division of integers, there is a similar notion of long division of polynomials. When dividing integers, you can divide integers $n$ by $d \neq 0$ to obtain two integers: a quotient $q$ and a remainder $r$, satisfying
$$\frac{n}{d} = q + \frac{r}{d}.$$
Equivalently, this can be written in terms of integers as
$$n = qd + r,$$
where $n$ is expressed as an integer multiple of $d$ plus a remainder.
(For example, when you divide $34$ by $6$ you get the quotient $5$ and remainder $4$, which means that $34 = 5 \cdot 6 + 4$, or $\frac{34}{6} = 5 + \frac{4}{6}$, which in lowest terms is $5\frac{2}{3}$.)
With the customary requirement that the remainder be smaller than the divisor ($0 \leq r < |q|$, which is simply $0 \leq r < q$ for positive $q$), the quotient $q$ and remainder $r$ are unique.
Similarly, we can divide a polynomial $p(x)$ by another, nonconstant polynomial $d(x)$ to obtain a quotient $q(x)$ and a remainder $r(x)$ such that
$$p(x) = q(x)d(x) + r(x),$$
or in rational expressions
$$\frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)}.$$
Here, we impose a similar requirement that the remainder be smaller than the divisor, which is that the degree of $r(x)$ be strictly less than that of $d(x)$. That is, the highest power of $x$ which appears in $r(x)$ is less than the highest power of $x$ in $d(x)$. (We require $d(x)$ to be nonconstant above because dividing by $p(x)$ by a nonzero constant $d(x) = d$ would leave a zero remainder, and $0$ conventionally does not have a well-defined degree. But it's not as if there's any practical difficulty in dividing a polynomial by a nonzero constant.)
So, how do we actually perform this division? Write the long division similarly to that of integers, but instead of organizing by place value, have columns organized by degree. At each step, use the leading terms (those with highest powers of $x$) as your guide.
First, we have
$$
\require{enclose}
\begin{array}{r}
3x-1 \enclose{longdiv}{3x^2} \\[-3pt]
\end{array}
$$
and note that $3x^2 = (3x) x$, or if you prefer $\frac{3x^2}{3x} = x$, so begin by placing $x$ into the quotient we're forming. (As noted above, look at the leading terms.) Now, subtract from the dividend $3x^2$ the product of the divisor $3x-1$ and $x$, giving
$$3x^2 - (3x-1)x = 3x^2 - (3x^2 - x) = x,$$
which amounts to the long division below.
$$
\require{enclose}
\begin{array}{r}
                    x  \\[-3pt]
3x-1 \enclose{longdiv}{3x^2\phantom{-1x}} \\[-3pt]
         \underline{3x^2-x} \\[-3pt]
                    x  \\[-3pt]
\end{array}
$$
Now, the degree of $x$ is $1$, just like the degree of $3x-1$, so we're not finished yet. (The remainder must have strictly smaller degree.) Next, we divide this remaining $x$ by $3x-1$, watching the leading terms. Since $x = \frac{1}{3}(3x)$, or in other words $\frac{x}{3x} = \frac{1}{3}$, now we add that $\frac{1}{3}$ to the quotient, obtaining
$$
\require{enclose}
\begin{array}{r}
                    x+\frac{1}{3}  \\[-3pt]
3x-1 \enclose{longdiv}{3x^2\phantom{-x-1\frac{1}{3}}} \\[-3pt]
         \underline{3x^2-x}\phantom{-1\frac{1}{3}} \\[-3pt]
                    x\phantom{-1\frac{1}{3}}  \\[-3pt]
         \underline{x-\frac{1}{3}} \\[-3pt]
                    \frac{1}{3}
\end{array}
% (Sorry, that 1 in $-1\frac{1}{3}$ in the phantom bits is a rather hacky thing I did because of having trouble getting the spacing right. It's certainly not the Right Thing, but looked close enough for a casual glance.)
$$
and finally, since $\frac{1}{3}$ is a constant, with degree $0$, it is the true remainder. So what does this result mean? Put everything into
$\frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)}$
to obtain the desired
$$\frac{3x^2}{3x-1} = x+\frac{1}{3} + \frac{1/3}{3x-1}.$$
P.S. I suspect this topic was presented to you to prepare you for its use in integration later. When we want to antidifferentiate a rational expression, a typical first step is division, followed by applying the method of partial fractions. While you may not need to worry about partial fractions now, it's worth remembering how to divide fractions to prepare for it.
A: $$\frac{3x^2}{3x-1} = \frac{0+3x^2}{3x-1}=\frac{(\frac{1}{3}-\frac{1}{3})+3x^2}{3x-1}=\frac{\frac{1}{3}+(-\frac{1}{3}+3x^2)}{3x-1}=\dots$$
Noting that
$$x = \frac{3x}{3}=\frac{3x+0}{3}=\frac{3x+(-1+1)}{3}=\frac{(3x-1)+1}{3}$$
Then we have
$$A = \frac{3x^2-\frac{1}{3}}{3x-1} = \frac{3\bigg(\frac{(3x-1)+1}{3}\bigg)^2-\frac{1}{3}}{3x-1}=\frac{\frac{(3x-1)^2}{3}+\frac{2(3x-1)}{3}+\frac{1}{3}-\frac{1}{3}}{3x-1}=x+\frac{1}{3}$$
