How to proceed for the following analysis question For functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$ and $x\in\mathbb{R}$ define $$F(fg)=\int\limits_{1}^{x}f(t)g(t)dt$$ If $a(x),b(x),c(x), d(x)$ are real polynomials show that $F(ac)F(bd)-F(ad)F(bc)$ is always divisible by $(x-1)^4$
 A: Let \begin{equation}
    P(x)= F(ac)\,F(bd)-F(ad)\,F(bc)
\end{equation}
Note that, $P(x)$ is a real polynomial. We can also write equation $P(x)= F(ac)\,F(bd)-F(ad)\,F(bc)$ in determinant form as follows $$P(x)=\begin{vmatrix} 
F(ac) & F(ad) \\
F(bc) & F(bd) 
\end{vmatrix}=\begin{vmatrix}
\displaystyle\int\limits_{1}^{x}a(t)c(t)\,dt & \displaystyle\int\limits_{1}^{x}a(t)d(t)\,dt \\
\displaystyle\int\limits_{1}^{x}b(t)c(t)\,dt & \displaystyle\int\limits_{1}^{x}b(t)d(t)\,dt
\end{vmatrix}$$
Now in order to show that $P(x)$ is divisible by $(x-1)^4$, it is enough to show that $x=1$ is a root of $P(x)$ with multiplicity at least $4$. To show this we need the following idea regarding the differentiation of a determinant.
Consider $$f(x)=\begin{vmatrix}
  a_{1}(x) & b_{1}(x) & \cdots & c_{1}(x) \\
  a_{2}(x) & b_{2}(x) & \cdots & c_{2}(x) \\
  \vdots  & \vdots  & \ddots & \vdots  \\
  a_{n}(x) & b_{n}(x) & \cdots & c_{n}(x) 
 \end{vmatrix}$$ Then $$f'(x)=\begin{vmatrix}
  a_{1}'(x) & b_{1}(x) & \cdots & c_{1}(x) \\
  a_{2}'(x) & b_{2}(x) & \cdots & c_{2}(x) \\
  \vdots  & \vdots  & \ddots & \vdots  \\
  a_{n}'(x) & b_{n}(x) & \cdots & c_{n}(x) 
 \end{vmatrix}+\begin{vmatrix}
  a_{1}(x) & b_{1}'(x) & \cdots & c_{1}(x) \\
  a_{2}(x) & b_{2}'(x) & \cdots & c_{2}(x) \\
  \vdots  & \vdots  & \ddots & \vdots  \\
  a_{n}(x) & b_{n}'(x) & \cdots & c_{n}(x) 
 \end{vmatrix}+\cdots+\begin{vmatrix}
  a_{1}(x) & b_{1}(x) & \cdots & c_{1}'(x) \\
  a_{2}(x) & b_{2}(x) & \cdots & c_{2}'(x) \\
  \vdots  & \vdots  & \ddots & \vdots  \\
  a_{n}(x) & b_{n}(x) & \cdots & c_{n}'(x) 
 \end{vmatrix}$$
Using this we can show that $$P(1)=P'(1)=P''(1)=P'''(1)=0$$ Hence $x=1$ is a root of $P(x)$ with multiplicity at least $4$. Thus $(x-1)^4$ occurs as a factor in $P(x)$. Therefore $F(ac)\,F(bd)-F(ad)\,F(bc)$ is always divisible by $(x-1)^4$.
