Question

Suppose $p(x)$ is a polynomial with integer coefficients. The remainder when $p(x)$ is divided by $x$ -1 is 1 and the remainder when $p(x)$ is divided by $x-4$ is 10 . If $r(x)$ is the remainder when $p(x)$ is divided by $(x-1)(x-4)$, find the value of $r(2006)$.

Answer 1

Answer: 6016

Dividing $p(x)$ by the quadratic $(x-1)(x-4)$ leaves a linear remainder $r(x)=ax+b$
$\frac{p(x)}{(x-1)(x-4)}=q(x)+r(x)\Rightarrow p(x)=(x-1)(x-4)q(x)+(ax+b)$
$p(l)=1\Rightarrow$ where $q(x)\text{ is quotient and }r(x)\text{ is remainder })$
$p(4)=10\Rightarrow 1=a+b$
$10-1=4a+b-a-b\Rightarrow b=4a+b$
$r(x)=3x-2\Rightarrow g=3a\Rightarrow a=3\Rightarrow b=-2$
$r(x)=3x-2\Rightarrow r(2006)=6018-2=6016$
Altematively:
$P(x)=Q_1(x-1)+1$
$P(x)=Q_2(x-4)+10$
$P(x)=Q_3(x-1)(x-4)+ax+b$
$P(1)=1$ and $p(4)=10$
$a+b=1$
$4a+b=10$
$a=3;b=-2$
$\therefore r(x)=3x-2$
$r(2006)=6018-2=6016$