<s>[INST] Curve of degree four touching the line Let the curve $y=x^4+Ax^3+Bx^2+Cx+D$ touches the line $y=px+q$ at x=2 and x=3 where A, B, C, D, p, q $\in$R. If the area bounded by the curve and the line is $\frac{1}{\alpha}$ then the number of divisors of $\alpha$ are.
My approach was that I used $\frac{dy}{dx}=m=p$
Then I substituted $p=4x^3+3Ax^2+2Bx+C$ at x=2 and x=3
As it touches the line at x=2 and 3 I used the following substitution
$px+q=x^4+Ax^3+Bx^2+Cx+D$ by putting x=2 and x=3 but getting more and more complicated.
Please suggest some short cut
 [/INST] 
\begin{align}
f_1(x)&=x^4+Ax^3+Bx^2+Cx+D
,\\
f_2(x)&=px+q
.
\end{align}  
We have a system of four equations
\begin{align}
f_1(2)&=f_2(2)
,\\
f_1(3)&=f_2(3)
,\\
f'_1(2)&=f'_2(2)
,\\
f'_1(3)&=f'_2(3)
,
\end{align}  
which allows to express $A,B,C,D$
in terms of given numbers $p,q$
as
\begin{align}
A &= -10,\quad
B = 37,\quad 
C = p-60,\quad
D = 36+q. 
\end{align}  
The area in question is
\begin{align}
\int_2^3(f_1(x)-f_2(x))\,dx
&=\frac{65}4\,A+\frac{19}3\,B
+\frac52\,C+D+\frac{211}5-\frac52\,p-q
\\
&=\frac1{30}
,
\end{align}
hence, $\alpha=30$.
</s>