find variable from polynomial when divisor and remainder is known I am asked the following question:

When ${x^4 - 3x^3 + px - 5}$ is divided by ${x - 3}$ the remainder is $16$, find $p$.

So my way of solving this was to use synthetic long division to divide ${x^4 - 3x^3 + px - 5}$ by $3$ that would leave me with a remainder expression that would have $p$.
$$
\begin{matrix}3&amp;1&amp;0&amp;-3&amp;p&amp;-5\\
&amp;&amp;3&amp;9&amp;18&amp;54+3p\\
&amp;1&amp;3&amp;6&amp;18+p&amp;49+3p
\end{matrix}
%3| 1 |  0 | -3  |  p      |  -5
% |      3 |  9  |  18     |  54 + 3p
%   
%  | 1 | 3 |  6  |  18 + p |  49 + 3p
$$

I get the remainder to be ${3p + 49}$

I would say then that ${3p + 49 = 16}$ and ${p = - 11}

But the textbook gives an answer of $7$.

Have I come to the wrong conclusion about how to work out ${p}$?
Hint: Note that $f(x)=(x-3)\cdot g(x)+r(x)$ implies $f(3)=r(3)$.
Let $$P(x)= x^4-3x^3+px-5, \quad g(x)=x-3,$$
  so,
$$
\begin{align}
      P(3)&amp;= 3^4-(3)(3)^3+p(3)-5\\
      P(3)&amp;= 81-81+3p-5\\
      P(3)&amp;= 3p-5.
\end{align}
$$
Note that the remainder is $16$.
Hence,
$$
\begin{align}
     P(3) &amp;= 16\\
     3p-5 &amp;= 16\\
     3p/3 &amp;= 21/3.
\end{align}
$$
So,     $$p = 7.$$