p(x) is a polynomial of degree>2. When p(x) is divided by x-2, it leaves remainder 1, and when divided by x-3, leaves rem. 3. Find the remainder when p(x) is divided by (x-2)(x-3)
Hello,
p(x)=(x-2)Q(x)+1
p(x)=(x-3)R(x)+3
thus we know that p(2)=1 and p(3)=3
Let s(x) be the remainder of p when divided by (x-2)(x-3)
p(x)=(x-2)(x-3)T(x)+s(x)
Since (x-2)(x-3) is of degree 2, s(x) has to be of degree 0 or 1.
Substituting x=2 and x=3, this gives the system :
$\displaystyle \left\{\begin{array}{ll} 1=p(2)=0+s(2) \\ 3=p(3)=0+s(3) \end{array} \right.$
So we're looking for $\displaystyle s(x)=ax+b$ and such that $\displaystyle s(2)=1$ and $\displaystyle s(3)=3$
Substitute :
$\displaystyle \left\{\begin{array}{ll} s(2)=1=2a+b \\ s(3)=3=3a+b \end{array} \right.$
You can finish it eh ?