Dividing polynomial by binomial using remainder theorem An A level text book claims that one can find the quotient by first:
1.) Setting up an identity, $f(x)≡ Q(x)(divisor) + remainder$
2.) Finding the coefficients
However, another A level text book says, "Note. This theorem gives a (simple) method for evaluating the remainder only. If the quotient is required, long division must be used."
The question is: Divide $x^3 + x^2 - 7$ by $x-3$ using the remainder theorem. 
In this example,
1.) They set up the identity: $x^3 + x^2 - 7 ≡ (Ax^2 + Bx + C)(x-3) + D$
2.) They let $x=3$ to find coefficient $D$ 
3.) They let $x=0$ to find coefficient $C$
4.) To find coefficients $A$ and $B$, the text book then goes on to "comparing the coefficients". No more detail is given as to how "comparing the coefficients" to find $A$ and $B$ is achieved. 
Can you find coefficients $A$ and $B$ using this method ONLY? If so, how?
 A: Assuming the equation was meant to be:
$$x^3+x^2-7 = (Ax^2+Bx+C)(x-3)+D$$
Setting $x=3$, we sse that $3^3+3^2-7 = 29 = D$.
Setting $x=0$, we see that $-7 = D-3C=29-3C$, so $C=12$.
Setting $x=1$, we see that $-5=(A+B+12)(-2)+29$, or $A+B+12 = 17$.
Setting $x=2$, we see that $5 = (4A+2B+12)(-1)+29 = 4A+2B+12 = 24$.
So now you have two linear equations for $A,B$, and you can solve those.
(Definitely redo my arithemetic, it could be in error.)
A: We write
$$x^3 + x^2 - 7 ≡ (Ax^2 + Bx + C)(x-3) + D.\tag{1}$$
Put $x=3$. That conveniently kills all but $D$ on the right. Substituting on the left, we get that $D=3^3+3^2-7=29$.  We can continue making other simple substitutions, and get linear equations that determine our coefficients.
But you asked about comparing coefficients. That is done as follows. Expand the right-hand side of (1). So in particular, multiply out $(Ax^2 + Bx + C)(x-3)$. We get after a while that the right-hand side is equal to 
$$Ax^3 +(-3A+B)x^2 +(-3B+C)x -3C+D.$$
This has to be the same polynomial as $x^3+x^2-7$.
So the coefficients of $X^3$ must be the same. That means $A=1$.
The coefficients of $x^2$ must be the same. That means $-3A+B=1$.
The coefficents of $x$ must be the same. That means $-3B+C=0$.
And the constant terms must be the same. That means $-3C+D=-7$.
Now solve. We already know $D=27$ a "quickie" way. The rest of the coefficients are now easy to pick up. 
A: (I am not completely sure that this is what you are looking for.)
Let $$p(x) = x^3 + x^2 - 7 = (Ax^2 + Bx + C)(x-3) + D.$$
You want to find $A,B,C,$ and $D$. Note first that you are dividing by $x-3$, so the remainder of the division will be a constant.
That is we have
$$\begin{align}
{\color{blue} 1}x^3 + {\color{red} 1}x^2  + {\color{gray} 0}x + {\color{green}{-7}} &= {\color{blue}A}x^3 + {\color{red}B}x^2 + {\color{gray}C}x + {\color{red}{-3A}}x^2 + ({\color{gray}{-3B}})x + ({\color{green}{- 3C + D}})\\
&= {\color{blue}A}x^3 + ({\color{red}{B - 3A}})x^2 + {\color{gray}{(C - 3B)}}x + ({\color{green}{- 3C + D}})
\end{align}
$$
Now, you know that two polynomials are equal if and only if the coefficients match up. That is, we must have
$$\begin{align}
1 &= A\\
1 &= B - 3A\\
0 &= C - 3B \\
-7 &= -3C + D.
\end{align}
$$
You already have $A$, so from the second equation you can find $B$. That gives $C$ from the third equation. And so from the last equation you have $D$.
