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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!! =)

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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.

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  • $\begingroup$ LutzL, Thankyou. I really appreciate your help with this. Just another $\endgroup$ – angel May 18 '14 at 20:18
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A real cubic polynomial with one real and two complex roots is of the form $$f(x)=(x-\alpha)\bigl((x-m)^2+\beta^2\bigr)\ .$$ One computes $f(m)=(m-\alpha)\beta^2$ and $f'(m)=1\cdot\beta^2+(m-\alpha)\cdot 0=\beta^2$; therefore the tangent in question is the line $\ell$ with the equation $$\ell:\qquad y=(m-\alpha)\beta^2+\beta^2(x-m)\ .$$ This line passes through the point $(\alpha,0)$.

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