# 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?

• Hint: $x-1$ divides both moduli so evaluating both equations at $x = 1$ yields $\, -3 = P(1) = -5 + a.$ $\ \ \$ – Bill Dubuque Apr 17 at 18:16

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 $$Q(4)$$.

$$Q(x)=h(x)(x^2-5x+4)+7x-8$$ $$Q(4)=h(4)(4^2-5\cdot 4+4)+7\cdot 4-8=h(4)\cdot 0+28-8=20$$

You need only $$P(1)$$, since

• $$x^2-4x+3 = (x-1)(x-3)$$ and
• $$x^2+6x-7 = (x-1)(x+7)$$

Hence,

• $$P(1) = 65-68 = -3$$
• $$P(1) = -5+a \Rightarrow a=2$$

We notied that: $$x^2-4x+3=(x-1)(x-3)$$ and $$x^2+6x-7=(x-1)(x+7)$$.

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

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. $$a=2$$.

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.