Prove that if a fifth grade polinomial function has five (diferent) zeros, than its derivative has 4 zeros (diferent)?

1 Answer
George C.
Dec 7, 2017

See explanation...

Explanation:

Without loss of generality, suppose that the #5# zeros of our quintic polynomial are:

#r_1 < r_2 < r_3 < r_4 < r_5#

and that the leading coefficient is #1#.

Note that any quintic with #5# distinct real zeros would be a non-zero multiple of such a monic quintic with derivative the same non-zero multiple of a quartic polynomial - with the same zeros.

Then:

#f(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)#

Note that:

#f'(r_1) = lim_(h->0) (f(r_1+h)-f(r_1))/h#

#color(white)(f'(r_1)) = (r_1-r_2)(r_1-r_3)(r_1-r_4)(r_1-r_5) > 0#

#f'(r_2) = (r_2-r_1)(r_2-r_3)(r_2-r_4)(r_2-r_5) < 0#

#f'(r_3) = (r_3-r_1)(r_3-r_2)(r_3-r_4)(r_3-r_5) > 0#

#f'(r_4) = (r_4-r_1)(r_4-r_2)(r_4-r_3)(r_4-r_5) < 0#

#f'(r_5) = (r_5-r_1)(r_5-r_2)(r_5-r_3)(r_5-r_4) > 0#

Now #f'(x)# is a continuous function, so by Bolzano's theorem, we find that it has a zero in each of:

#(r_1, r_2)#, #(r_2, r_3)#, #(r_3, r_4)# and #(r_4, r_5)#

That is, it has #4# distinct real zeros.

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