Remainder theorem for polynomials (JUEE 1990)

You need only $P(1)$, since

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

Hence,

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

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 $Q(4)$.

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

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

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