# Common root of two cubic equations

Respected all.

I am willing to know if my following computation is correct or not. I request you to please give me your suggestion how to improve in case any error is found.

The question is : If $$a_1 x^3+b_1x^2+c_1x+d_1=0$$ and $$a_2x^3+b_2x^2+c_2x+d_2=0$$ be two cubic polynomials with real coefficients and they have a common root say $$x=m\in \mathbb{R}$$, how to determine it ?

My try:

Since $$x=m$$ is a common root so we must have \begin{align*} &a_1m^3+b_1m^2+c_1m+d_1=0\\ &a_2m^3+b_2m^2+c_2m+d_2=0 \end{align*} Hence we must have: $$\frac{m^3}{b_1c_2-b_2c_1}=\frac{m^2}{c_1d_2-c_2d_1}=\frac{m}{d_1a_2-d_2a_1}=\frac{1}{a_1b_2-a_2b_1}$$

And from here, we get the value of $$m$$.

Please tell me if it is correct. In case if not, please correct it through editing.

p.S. I know about Sylvester matrix which enables us to verify if two given polynomials of degree $$m,n$$ will share a common root by the Sylvester matrix which determinant is called the Resultant of those two polynomials. But I do not know how to use Sylvester matrix to determine that common root. Hence I tried the above method.

• Note that $m$ is also a root of the gcd of the two polynomials. That gcd can be found with euclidean algorithm and has degree less than 3 (if the two cubic are not equivalent) – Fabio Lucchini Oct 13 '18 at 22:31
By putting \begin{align*} &f(m)=a_1m^3+b_1m^2+c_1m+d_1=0\\ &g(m)=a_2m^3+b_2m^2+c_2m+d_2=0 \end{align*} we get $$a_2f(m)-a_1g(m)=(a_2b_1-a_1b_2)m^2+(a_2c_1-a_1c_2)m+(a_2d_1-a_1d_2)=0$$ hence if $$a_2b_1-a_1b_2\neq 0$$ then $$m=\frac{-(a_2c_1-a_1c_2)\pm\sqrt{(a_2c_1-a_1c_2)^2-4(a_2b_1-a_1b_2)(a_2d_1-a_1d_2)}}{2(a_2b_1-a_1b_2)}$$ otherwise $$m=-\frac{a_2d_1-a_1d_2}{a_2c_1-a_1c_2}$$
Note that if $$a_2b_1-a_1b_2=a_2c_1-a_1c_2=0$$, then also $$a_2d_1-a_1d_2=0$$, hence $$a_2f=a_1g$$, that's $$f$$ and $$g$$ have the same roots.
• what if $a_2b_1-a_1b_2=0$ ? – Anjan3 Oct 13 '18 at 22:40