Solution of cubic equations in terms of quadratic equations

Question

Is there a general way in which a cubic equation of $3$rd degree can be represented by a quadratic equation of $2$nd degree such that $2$ solutions of a cubic equation is equal to $2$ solutions of corresponding quadratic equation.

For example $2$ solutions of equation $Ax^3+Bx^2+Cx-(A+B+C)=0$ is equal to $2$ solutions of quadratic equation $Ax^2+(A+B)x+(A+B+C)=0$.

But here the last term of cubic equation is $-(A+B+C)$.

If the last term is arbitrarily some constant number not equal to $-(A+B+C)$ then what quadratic equation will give the $2$ solution of cubic equation $Ax^3+Bx^2+Cx+D=0$ such that $D$ is not equal to $-(A+B+C)$

Answer 1

$$a x^3+b x^2+c x-(a+b+c)=0\tag{1}$$ has the integer solution $x=1$ thus can be factored as $$(x-1)\left[a x^2+(a+b) x+(a+b+c)\right]=0$$ So two soolutions of $(1)$ are solutions of $$a x^2+(a+b) x+(a+b+c)=0$$

Answer 2

Once you find one root of a cubic, you can take out a linear factor by the factor theorem, leaving a quadratic to solve for the other roots. It's not pretty, but it works. (The weird part is some other quadratics are of interest along the way.)

In analogy with completing the square, divide out the leading coefficient and shifting the unknown to lose the second highest-order term. We want a root of $x^3+px+q=0$. Given any root $x$, some $u,\,v\in\Bbb C$ satisfy $u+v=x,\,uv=-p/3$, namely the roots of $t^2-xt-p/3$. Then$$u^3+v^3=x(x^2-3uv)=x^3+px=-q,\,u^3v^3=(-p/3)^3,$$so $u^3,\,v^3$ are roots of $t^2+qt-p^3/27$. The root we'll take out as aforementioned is $\sum_\pm\sqrt[3]{\frac{-q\pm\sqrt{q^2+4p^3/27}}{2}}$. (Note you have to define these cube roots in just the right way, so their product is $-p/3$; we don't want to be off by a factor of $\exp\tfrac{\pm2\pi i}{3}$.)

Answer 3

Given the equation $\quad x^3 - 4x^2 - 7x + 10 = 0 \quad $ we can guess that it has one or more rational roots and, by the rational root theorem, we can "guess" these by using the "factors" or the first and last term. The factors of the last term are $\pm1,\pm2,\pm5,\pm10$ and if we try the first one, we find

$$\quad ( x^3 - 4x^2 - 7x + 10)/(x-1)=x^2 - 3 x - 10\quad$$

From here, it is easy to use the quadratic equation to solve it. On the other hand it is just as easy to try all the factors of $10$ and then find that

$$( x^3 - 4x^2 - 7x + 10)\quad = \quad (x - 1) (x + 2)(x - 5) = 0$$

In your other example $\quad 100y^3-120y^2+21y-1=0 \quad$ we have $(y-1)$ as a factor and it also yields a solvable quadratic. In both these equations, one factor must be found before the quadratic can be solved. One way to find the [single] factor to be divided (see synthetic division) is to use the cubic equation. If we use your last equation as an example we have coeficients $\quad a=100\quad b=-120\quad c=21\quad d=-1\quad $ and then

$$x=\sqrt[\Large{3}]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)+\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}\\ +\sqrt[\Large{3}]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)-\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}-\frac{b}{3a}$$