Degree 4 polynomial has 2 points of inflection A&B. The line connecting A&B intersects the function at C&D &→?

Prove that the ratio AB:AC:BD is constant, and what is it?

Degree 4 polynomial has 2 points of inflection A%26amp;B. The line connecting A%26amp;B intersects the function at C%26amp;D %26amp;→?
Let be the polynomial p = x^4 - 6x^2 + 5.





Its second derivative is





p'' = 12 (x^2 - 1), and its roots are +1 and -1.





Then the inflexion point are





A = (-1, 0)


B = (1, 0)





The line connecting A %26amp; B is y=0, then the equation for C %26amp; D is





x^4 - 6x^2 + 5 = 0





Then





C = (-sqrt(5), 0)


D = (sqrt(5), 0)





The ratios are





(AD) / (AB) = (sqrt(5)+1) / 2





(AB) / (AC) = 2 / (sqrt(5)-1) = (sqrt(5) + 1) / 2





I tried numerically for different 4th degree polynomials, and the ratios are always the golden ratio.





For a general 4th degree polynomial with two inflection points there is rotation and translation followed for an expansion of the coordinates that move the inflexion points to (-1,0) and (1,0). The only difference of this "transformed" polynomial and the one of the example above is that the principal coefficient of p is 1, but in the general case might be any value not null. But this multiplicative factor cannot change the solutions of the homogeneous quartic equation, then the distances between A, B, C and D are the same. The rotation and the translation does not change the distances, and the expansion changes them, but not the ratios, then the result is completely general.
Reply:f(x) = ax^4 + bx^3 + cx^2 + dx + f


f'(x) = 4ax^3 + 3bx^2 + 2cx + d


f''(x) = 12ax^2 + 6bx + 2c





Inflection points are when f''(x) = 0


12ax^2 + 6bx + 2c = 0


x = (-6b +/- SQRT(36b^2 - 96ac)) / (24a)


Inflection points


x_1 = -b/(4a) + SQRT(b^2 - 8ac/3) / (4a)


x_1 = (SQRT(b^2 - 8ac/3) - b) / (4a)


x_2 = -b/(4a) - SQRT(b^2 - 8ac/3) / (4a)


x_2 = (-SQRT(b^2 - 8ac/3) - b) / (4a)





sorry, I have to go, I'll try to come back as soon as I can, this is a long problem, you need to substitute x_1 and x_2 into f(x), get the line equation for AB, where A is the point(x_1,f(x_1)) and B is (x_2,f(x_2)), then you need to equalize that line to f(x) so you can get the intersection points C %26amp; D.