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

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 }) $
$p (4)=10 \Rightarrow 1= a + b$
$10-1=4 a + b - a - b \Rightarrow b =4 a + b$
$r ( x )=3 x -2 \Rightarrow g =3 a \Rightarrow a =3 \Rightarrow b =-2$
$r(x)=3 x-2 \Rightarrow r(2006)=6018-2=6016$
Altematively:
$P(x)=Q_1(x-1)+1$
$P(x)=Q_2(x-4)+10$
$P(x)=Q_3(x-1)(x-4)+a x+b$
$P (1)=1$ and $p (4)=10$
$a+b=1$
$4 a+b=10$
$a =3 ; b =-2$
$\therefore r ( x )=3 x -2$
$r(2006)=6018-2=6016$