This is a problem from ISI BMath 2014 Subjective Solution based on Mulitple roots or Real root. Try to solve this problem.
Problem: Multiple roots or real root
Let $latex \mathbf { y = x^4 + ax^3 + bx^2 + cx +d , a,b,c,d,e \in \mathbb{R}}$. it is given that the functions cuts the x axis at least 3 distinct points. Then show that it either cuts the x axis at 4 distinct point or 3 distinct point and at any one of these three points we have a maxima or minima.
Discussion:
Since all the coefficients are real, complex roots occur as conjugates. Hence the fourth root (it is a four degree polynomial hence has a fourth root), must be real (if it is complex then we must have at least one more complex root, but all the other three roots are given to be real).
Let l, m, n be the three roots. Then the fourth root is either distinct from l, m, n or it is equivalent to exactly one of them say 'n'.
If it is equal to n then we may rewrite the polynomial as $latex \mathbf { y = (x-n)^2 (x-l)(x-m) }$
We take first and second derivative of the y with respect to x.
$latex \mathbf { y' = (x-n)^2 (x-l) + (x-n)^2 (x-m) + 2(x-n) (x-l)(x-m) }$. At x=n the first derivative vanishes. Hence x=n is a critical point. We want to show that this is also a point of maxima or minima. For that we must show that the second derivative at x=n is positive or negative (not zero).
$latex \mathbf { y'' = 2(x-n)(x-l) + (x-n)^2 + (x-n)^2+ 2(x-n) (x-m) + 2(x-n) (x-m) + 2(x-l)(x-m) + 2(x-n) (x-l) }$
Hence at x=n $latex \mathbf { y'' = 2(n-l)(n-m) }$ . Since n is distinct from m or l, hence the second derivative is either positive or negative and not zero. Hence we have maxima or minima at that point.
Proved.
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[…] Let . it is given that the functions cuts the x axis at least 3 distinct points. Then show that it either cuts the x axis at 4 distinct point or 3 distinct point and at any one of these three points we have a maxima or minima.Solution […]