a queer problem on repeated root if $f(x),g(x),h(x),\phi(x)$ are  distinct polynomials and let $$l(x)=\left(\int_{1}^{x}f(x)h(x)dx\right)\left(\int_{1}^{x}g(x)\phi(x)dx\right)-\left(\int_{1}^{x}f(x)\phi(x)dx\right)\left(\int_{1}^{x}g(x)h(x)dx\right)$$ is divisible by $(x-1)^a$ find $\max(a)$.($l(x)$ is not identically zero)
Answer given is $5$.
ofcourse  $l(1)=0$. Also i know that for repeated  root (in this case number of repetitions $5$) we can say
$$l(1),l'(1),l''(1).l'''(1),l''''(1)=0$$.i was able to prove $l'(1)=0$ but the more i differentiate the more ugly it gets.
I am sure there is a simple trick to solve this problem.can anyone provide a hint?thanks!!
 A: I'm not sure the problem is correctly stated. One that would make more sense would be to find the maximal value of $a$ for fixed polynomials (as there is no upper bound if you change the polynomials).
The answer will be highly dependent on the polynomials you chose, but we can still say something in general (namely we can get a lower bound) : instead of differentiating, you can reason with the multiplicity of 1 as a root in each term of your expression.
Let's denote $v_1(P)$ for the multiplicity of $1$ as a root of $P$; or equivalently, $v_1(P)$ is the biggest $a$ such that $(x-1)^a$ divides $P$. We take the convention $v_1(0) = +\infty$. Notice that for any polynomials $P$, $Q$, we have :
$$v_1(P Q) = v_1(P) + v_1(Q)$$
and
$$v_1(P + Q) \ge \min(v_1(P), v_1(Q))$$
with equality if $v_1(P) \ne v_1(Q)$. And finally,
$$v_1\left(\int_1^x P(t) dt\right) = v_1(P) + 1$$
since the derivative of $\int_1^x P(t) dt$ is $P$.
Now getting back to the problem : we can use the above properties to show
$$v_1\left(\int_1^x P_1 P_2 \int_1^x P_3 P_4 -  \int_1^x P_1 P_4 \int_1^x P_2 P_3 \right) \ge \min \left[v_1\left(\int_1^x P_1 P_2 \int_1^x P_3 P_4\right), v_1\left(\int_1^x P_1 P_4 \int_1^x P_2 P_3\right)\right]$$
and
$$v_1\left(\int_1^x P_1 P_2 \int_1^x P_3 P_4\right) = v_1\left(\int_1^x P_1 P_2 \right) + v_1\left(\int_1^x P_3 P_4\right) = (v_1(P_1 P_2) + 1) + (v_1(P_3 P_4) + 1)= v_1(P_1) + v_1(P_2) + v_1(P_3) + v_1(P_4) + 2$$
and similarly
$$ v_1\left(\int_1^x P_1 P_4 \int_1^x P_2 P_3\right) = v_1(P_1) + v_1(P_2) + v_1(P_3) + v_1(P_4) + 2$$
which gives finally
$$v_1\left(\int_1^x P_1 P_2 \int_1^x P_3 P_4 -  \int_1^x P_1 P_4 \int_1^x P_2 P_3 \right) \ge v_1(P_1) + v_1(P_2) + v_1(P_3) + v_1(P_4) + 2$$
To go any further, one would have to know the polynomials themselves ($v_1\left(\int_1^x P_1 P_2 \int_1^x P_3 P_4 -  \int_1^x P_1 P_4 \int_1^x P_2 P_3 \right)$ can get arbitrarily bigger than $v_1(P_1) + v_1(P_2) + v_1(P_3) + v_1(P_4) + 2$, it can even get infinite in some cases, as in the comments).
