<s>[INST] 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?
 [/INST] 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. 
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