When a polynomial $f(x)$ is divided by $(x-2),$ the remainder is $7$. When $f(x)$ is divided by $(x+1)$ the remainder is $-2$.
(a) If the remainder is $px+q$ when $f(x)$ is divided by $(x-2)(x+1)$, find the values of $p$ and $q$.
(b) Find the remainder when $f(x+3)$ is divided by $(x+1)(x+4)$
I found the values of $p$ and $q$ which are $3$ and $1$ by using remainder theorem.
$f(x)=Q(x)*(x-2)(x+1)+(px+q)$
However I don't know how to get the remainder of part(b).
Let $g(x)=f(x+3)$.
When $g(x)$ is divided by $(x+1)$, the remainder
$=g(-1)$
$=f(-1+3)$
$=f(2)$
When $g(x)$ is divided by $(x+4)$, the remainder
$=g(-4)$
$=f(-4+3)$
$=f(-1)$
At this point, since the two remainders I got are the same when $f(x)$ is divided by $(x-2)$ and $(x+1)$,
I assume $f(x+3)=Q(x)*(x+1)(x+4)+(px+q)$.
But $3x+1$ is not the answer.
What mistakes have I made and how to solve part(b)?
Hint $ $ Simply shift $\,x\to x\!+\!3\,$ in the prior division with remainder, i.e.
Lemma $\,\ f(x\!+\!a)\bmod g(x\!+\!a)\, = \,(f\bmod g)(x\!+\!a)$
Proof $\,\ f = gh + r,\ \deg r < \deg g,\ \ r = (f\bmod g),\ $ by Euclidean Division wtih Remainder
hence $\ f(x\!+\!a) = g(x\!+\!a)h(x\!+\!a) + r(x\!+\!a),\,\ \deg r(x\!+\!a) = \deg r < \deg g =\deg g(x\!+\!a)$
therefore $\ f(x\!+\!a)\bmod g(x\!+\!a) \,=\, r(x\!+\!a)\ $ by the uniqueness of the remainder.
Remark $ $ Your argument doesn't work because - though the remainders are the same - they are at different (shifted) points, i.e. $\,g(-1) = f(2) =7\,$ and $\, g(-4) = f(-1) = -2,\,$ i.e. the line $\, r(x+3)\,$ is a shifted version of the line $\,r(x) = 3x+1\,$ While they take the same values at the corresponding shifted points, they are different lines.