Is the remainder theorem always true for all x ∈ R ? If its not true for all values of x then for which values is it true? According to the remainder theorem, if we divide $p(x)$ by $(ax+b)$, the remainder will be $p(-b/a)$
Let $p(x) = 3x^3+ 5x^2+ 8x +5$ and let the divider be $2x + 7.$
So, $a = 2$ and $b =7$. Thus, the remainder should be $p(-7/2)$
$ = 3×(-7/2)^3 + 5×(-7/2)^2 + 8×(-7/2) + 5$
$= -90.375$
If the remainder theorem is true for all values of $x$, then the remainder should always be $-90.375$.
However, for $x=1$ , $p(x)$ = $p(1)$ = 21 and $2x+7 = 9$. So, the remainder is not $-90.375$. Again, for $x =2,3,4$ its not $-90.375$.
Why is it equal to the estimated value from the remainder theorem? Did I make any mistakes?
Any help will be appreciated.
 A: It is not true that the remainer is polynomial division must equal the remainder in the associated integer divisions (when we fix a value for the variable).
Here is a simpler example.  Let $p(x)=x^2+10$ and consider division by $x$.  Clearly we have $$p(x)=x\times x+10$$ so the remainder is $r(x)=10$, a constant as it should be.  We note that $10=p(0)$, which is again consistent with the remainder theorem.
And yet, if we pick specific values, $10$ is not the remainder for the associated integer division.  For instance, taking $x=2$ we get $p(2)=14$ which is divisible by $2$, so the remainder in that case is $0$.
Worth remarking:  For large enough arguments, there certainly is some connection.  For instance, sticking with $x^2+10$ as an example, if $x>10$ then $10$ is indeed the numerical remainder as well as the polynomial remainder.  A similar claim holds generally though you have to take care of the sign (the numerical remainder must be non-negative, though the polynomial remainder can certainly be negative).
