I was able to easily prove that the tangent at the average of two roots of a real cubic polynomial passed through the third root of the function. But I have only done this for functions with three distinct real roots using $y = (x-a) (x-b) (x-c)$ as the function.
I calculated the derivative of the function in terms of ($x$= midpoint of $a$ and $b$) and then just subbed in ($x = c$) in the equation of the tangent that I worked out to prove this conjecture.
BUT!!
Now, I am trying to prove the same for cubics with one real and two complex roots.
I am using $y = (x -d) (x-(a+bi) (x- (a-bi))$ where $d$ is the one real root.
So far, I have found the derivative of this to be
\begin{align}
dy/dx &= 2(x-d)(x-a) + (x-a-bi) (x-a+bi)\\
&= 2(x-d)(x-a) + x^2 + a^2 – bi^2\\
&= 2(x-a) (x-d) (x^2 + a^2 + b^2)
\end{align}
The average of the two roots I have used is $((a+bi/2) + (a – bi/2))/2 = a$
I have already proved it for this average of roots!
But!!
I also tried the other possible average of roots $((d + a +bi)/2)$
where the Point of contact I used is $((a+bi+d)/2), -((a+bi-d)/2)^3$
Why is that the equation of the tangent I worked out for this second possible pair of roots using ($m =ystep/xstep$) is not equalling zero when I sub in the third root into the equation of tangent I am getting??
Maybe I have calculated some things wrong?
Can you please help me prove the conjecture?
Plus, another mini question:
Can cubic polynomials be non-real? because when we say all cubic polynomials must have at least one real root, it implies that the cubic is real.
So can I say that a cubic with an imaginary coefficient is non-real?
I won't try to prove this conjecture for non-real cubics though.
Can you please help me?!
I would really appreciate even a bit of help
Sorry and Thanks!! =)
Best Answer
To the last question, traditionally if one does not qualify the coefficients directly or from context, rational or integer coefficients are assumed, so the polynomial is a real function.
The calculations for the case with complex roots should be the same as with the real roots, just take $b,c$ complex instead of real, with $c=\bar b$.
$f(x)=(x-a)(x-b)(x-c)$, $f'(x)=(x-a)(x-b)+(x-a)(x-c)+(x-b)(x-c)$, the tangent at $x_0=\frac12(a+b)$ is $$ t(x)=f(x_0)+(x-x_0)f'(x_0)=(x_0-a)(x_0-b)(x_0-c)+(x-x_0)(x_0-a)(x_0-b) =(x-c)(x_0-a)(x_0-b) $$ so it has a root at $x=c$.
However, you would get the third root more quickly by using that $-(a+b+c)$ is the quadratic coefficient of the polynomial.