## martian86 2 years ago I was trying to find the roots of the polynomial f(x) = 4x^3+3x^2-6x-1. By Rolle's theorem, If there exists 2 different values of x, say x(1) and x(2) that returns a same value of f(x), then at some point between x(1) and x(2), d/dx[f(x)] = 0. ---> (1) Hence, I differentiate f(x) to get, d/dx[f(x)] = 12x^2+6x-6 By (1) 12x^2+6x-6 = 0 Solving for x, i get 2 roots, x = -1 and 1/2. Now solve for f(x) = 4x^3+3x^2-6x-1 with x = 1 (by changing the sign) and surprisingly, f(1) = 0 !!, seems to be one of the root of f(x). Can someone help me to understand this ?

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I will explain you how to find roots of the f(x). First root we will guess. It will be 1 or -1. It is deviders of the last member of the equation. f(-1)=4*(-1)^3+3*(-1)^2-6*(-1)-1=-4+3+6-1=4 f(1)=4*(1)^3+3*(1)^2-6*(1)-1=4+3-6-1=0 => x_(1)=1 - the root devide column the equation: (4x^3+3x^2-6x-1)/(x-1)=4x^2+7x+1 then we find the roots of 4x^2+7x+1, D=7^2-4*4*1=33 x_(2)=(-1+sqrt(33))/8 x_(3)=(-1-sqrt(33))/8

2. martian86

Thx for the post, but my question was not about the roots of the equation. If u closely observe this, f'(x) = 12x^2+6x-6 has roots one of the roots as -1. and strangely, one of the roots of the equation, f(x) = 4x^3+3x^2-6x-1 is 1 [f(1) = 0] By changing the sign of one of the root of d/dx[f(x)], i actually get one of the real root of f(x). This is cool, but I cant figure out all the cases, where such a relationship can exist !!! Or to put it the other way, why does f(x) has root of 1 and d/dx[f(x)] has root of -1. I some how cant accept that this is just a coincidence..