$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive 4 real roots. 
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive four real roots.


$$(x-a)(x-b)(x-c)(x-d) \\ +(x-a)(x-b)(x-c)(x-e) \\ +(x-a)(x-b)(x-d)(x-e) \\ +(x-a)(x-c)(x-d)(x-e) \\ + (x-b)(x-c)(x-d)(x-e) =0.$$

My attempt:
\begin{align}
& f(x)=\sum_{cyc} (x-a)(x-b)(x-c)(x-d). \\
& f(a)=(a-b)(a-c)(a-d)(a-e). \\
& f(b)=(b-a)(b-c)(b-d)(b-e). \\
& f(c)=(c-a)(c-b)(c-d)(c-e). \\
& f(d)=(d-a)(d-b)(d-c)(d-e). \\
& f(e)=(e-a)(e-b)(e-c)(e-d). \\
\ \\
& f(0)=\sum_{cyc} abcd.\\
\end{align}
I put some arbitrary integers in $a, b, c, d, e$ and drew a graph of $\displaystyle f(x)=\sum_{cyc} (x-a)(x-b)(x-c)(x-d).$
It sure makes 4 distinct roots... How can we prove this?

Values in the graph above:
$a=-1.34, b=-4.67, c=-2.91, d=0.33, e=-6.09$
 A: Let $F(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$.  Then your polynomial is $f(x)=F'(x)$. Because $a,b,c,d,e$ are all distinct, by Rolle's theorem, between any two consecutive roots of $F(x)$ will be a root of $f(x)$.   Thus, $f(x)$ will have 4 distinct real roots because $F(x)$ has 5 distinct real roots.
A: I found an answer from the comment.
\begin{align}
& \text{WLOG } a<b<c<d<e. \\
\ \\
& f(a)>0, f(b)<0, f(c)>0, f(d)<0, f(e)>0. \\
\ \\
& \therefore \text{ There exists 4 roots, one each  in } [a, b], [b, c], [c, d], [d, e] \qquad \text{by the Intermediate Value Theorem}.
\end{align}
A: This can also be approached without theorems from calculus, though much of it uses ideas from pre-calculus that lead into limits and the Intermediate Value Theorem.  For this argument, we will use a simpler function with just three terms,  $ \ g(x) \ = \ (x - a)·(x - b) + (x - a)·(x - c) + (x - b)·(x - c) \ \ , $ from which we will form the rational function
$$ R(x) \ \ = \ \ \frac{g(x)}{(x - a)·(x - b)·(x - c)} \ \ = \ \ \frac{1}{x \ - \ a} \ + \ \frac{1}{x \ - \ b} \ + \ \frac{1}{x \ - \ c} \ \ . $$
We know that $ \ R(x) \ $ has vertical asymptotes at $ \ x = a \ \ , \ \ x = b \ \ , \ \ x = c \ \ $ (or the domain of $ \ R(x) \ $ is $ \ x \ \neq \ a \ , \ b \ , \ c \ ) \ . \ $  It also has the horizontal asymptote $ \ y \ = \ 0 \ \ $ (or $ \ \lim_{ \ x \ \rightarrow \ \pm \ \infty}  R(x) \ = \ 0 \ ) \ . $
For $ \ x \ < \ a \ \ , \ R(x) \ < \ 0  \ \ , \   $ while for $ \ x \ > \ c \ \ , \ R(x) \ > \ 0  \ \ . $
We are particularly interested in the intervals between the vertical asymptotes.  In $ \ (a \ , \ b) \ \ , \ $ we see that the $ \ \frac{1}{x \ - \ a} \ $ term becomes a larger and larger positive number as $ \ x \ $ gets closer to $ \ a \ $, while the sum of the other terms is "nearly constant", so $ \ R(x) \ $ becomes a larger and larger positive number (or $ \ \lim_{ \ x \ \rightarrow \ a^{+}}  R(x) \ = \ +\infty \ ) \ . \ $ By the same token, the $ \ \frac{1}{x \ - \ b} \ $ term becomes a larger and larger negative number as $ \ x \ $ gets closer to $ \ b \ $, while the sum of the other terms is "nearly constant", so $ \ R(x) \ $ becomes a larger and larger negative number (or $ \ \lim_{ \ x \ \rightarrow \ b^{-}}  R(x) \ = \ -\infty \ ) \ . \ $  Since $ \ R(x) \ $ is defined and continuous (however that is expressed in a particular pre-calculus course) "everywhere in the interval", there is (at least) one value $ \ x \ = \ p \ \ , \ \ a \ < \ p \ < \ b \ $ for which $ \ R(p) \ = \ 0 \ \ . $
For the overall purpose of this discussion, we need to show that there is only one such value of $ \ x \ $ in the interval.  We have $ \ \frac{1}{p \ - \ a}  +  \frac{1}{p \ - \ b}  +  \frac{1}{p \ - \ c} \ = \ 0 \ \ . \ $  For $ \ a \ < \ X \ < \ p \ \ , $
$$ R(X) \ = \ \left[ \ \frac{1}{X \ - \ a} \ - \ \frac{1}{p \ - \ a} \ \right] \ + \ \left[ \ \frac{1}{X \ - \ b} \ - \ \frac{1}{p \ - \ b} \ \right] \ + \ \left[ \ \frac{1}{X \ - \ c} \ - \ \frac{1}{p \ - \ c} \ \right] \ \ $$
$$ = \  \frac{p \ - \ X}{(X \ - \ a)·(p \ - \ a)}  \  + \ \frac{p \ - \ X}{(b \ - \ X)·(b \ - \ p)} \ + \ \frac{p \ - \ X}{(c \ - \ X)·(c \ - \ p)} \ \ > \ \ 0 \ \ , $$
since the numerators and denominators of each term are positive.  By a similar argument, we can show that $ \ R(X) \ < \ 0 \ \ $ for $ \ p \ < \ X \ < \ b \ \ . \ $  So there is no other zero of $ \ R(x) \ $ in $ \ ( a \ , \ b) \ \ . \ $  [I was unable to find a convincing proof that $ \ R(x) \ $ is one-to-one in the interval; and while solving $ \ g(x) \ = \ 0 \ $ does show, with some effort, that one of the zeroes of this quadratic polynomial lies in $ \ (a \ , \ b) \ $ and the other does not, that method is not easily generalizable.]
Hence, as the denominator of $ \ R(x) \ $ is not zero at $ x \ = \ p \ $ or $ \ x \ = \ q \ \ , \ \ g(x) \ $ has exactly two zeroes, one each in $ \ (a \ , \ b) \ $ and $ \ (b \ , \ c ) \ \ . \ $  We can generalize this line of reasoning to show that for a function $ \ p(x) \  $ being the sum of every distinct product of $ \ (n - 1) \ $ factors from the set $ \ \large\{ \ (x - a_1) \ , \ (x - a_2) \ , \ldots , \ (x - a_n) \ \large\} \ \ , \ $ so $ \ \binom{n}{n-1} \ = \ n \ $ terms in all,  the $ \ a_i \ $ being distinct, there are $ \ (n - 1) \ $ intervals, $ \ (a_1 \ , \ a_2) \ \ , \ \  (a_2 \ , \ a_3) \ \ , \ldots , \ \  (a_{n-1} \ , \ a_n) \ \ ,  \ $ each of which contains just one zero of $ \ p(x) \ \ . $
