Remainder theorem for polynomials (JUEE 1990) 
Suppose the polynomial $P(x)$ with integer coefficients satisfies the following conditions:
  (A) If $P(x)$ is divided by $x^2 − 4x + 3$, the remainder is $65x − 68$.
  (B) If $P(x)$ is divided by $x^2 + 6x − 7$, the remainder is $−5x + a$.
  Then we know that $a =$?

I am struggling with this first question from the 1990 Japanese University Entrance Examination. Comments from the linked paper mention that this is a basic application of the "remainder theorem".  I'm only familiar with the polynomial remainder theorem but I don't think that applies here since the remainders are polynomials. Do they mean the Chinese remainder theorem, applied to polynomials?
So for some $g(x)$ and $h(x)$ we have:
$$P(x) = g(x)(x^2-4x+3) + (65x-68),\\
P(x) = h(x)(x^2+6x-7) + (-5x+a),$$
which looks to have more unknowns than equations. How should I proceed from here?
 A: You need only $P(1)$, since


*

*$x^2-4x+3 = (x-1)(x-3)$ and

*$x^2+6x-7 = (x-1)(x+7)$
Hence,


*

*$P(1) = 65-68 = -3$

*$P(1) = -5+a \Rightarrow a=2$
A: Hint:
If you know that the remainder of the division of some polynomial $Q$ by, say, $x^2-5x+4$ is $7x-8$ you can find some values of $Q$ by substituting $x$ by the zeros of the divisor.
Indeed, the zeros of $x^2-5x+4$ are $x=1$ and $x=4$. So you can find what is $Q(4)$.
$$Q(x)=h(x)(x^2-5x+4)+7x-8$$
$$Q(4)=h(4)(4^2-5\cdot 4+4)+7\cdot 4-8=h(4)\cdot 0+28-8=20$$
A: We notied that: $x^2-4x+3=(x-1)(x-3)$ and $x^2+6x-7=(x-1)(x+7)$.

So, we have: $P(1)=65*1-68=-5*(1)+a$.
  $\implies -3=-5+a\iff a=2$.

A: Since $P(x)=g(x)(x^2-4x+3)+(65x-68)$, for $x=1$ you get $P(1)=-3$. This imply that
$P(1)=h(1)\cdot 0+(-5+a)$, i.e. $a=2$.
A: You could use Polynomial Remainder Theorem here, it's just impractical. It states P(x) mod (x-b) is congruent to P(b). It never says: let b, be a number.  $x^2-4x+3=x-(-x^2+5x-3)$.
Easier to note the second divisor is $10x-10$ more than the first so $(10x-10)y-5x+a= 65x-68$ so y = 7 produces $70x-70-5x+a=65x-(70-a)$ so $a=2$ works. 
