Find roots in cubic equation $x^3+(1-2a)x^2-(3a^2+a+2)x-6a^2+6a=0$

Question

In the equation

$$x^3+(1-2a)x^2-(3a^2+a+2)x-6a^2+6a=0$$

How should I go to find the roots of $~x~$? Is it possible to apply the factor theorem somehow?

Answer 1

Hint We can think of the polynomial as a quadratic polynomial equation in $a$ with parameter $x$:

$$(-3x - 6) a^2 + (-2 x^2 - x + 6) a + (x^3 + x^2 - 2 x) = 0. $$ Factoring coefficients gives $$0 = -3(x + 2) a^2 - (x + 2) (2 x - 3) a + (x + 2) x (x - 1) .$$

Additional hint Factoring the entire expression gives: $$0 = -(x + 2) [3 a^2 + (2 x - 3) a - x (x - 1)] ,$$ so (1) $x = -2$ is a solution, and (2) to find the other solutions we need only factor the quadratic $3 a^2 + (2 x - 3) a - x (x - 1)$.

Answer 2

Observe that, in the cubic equation, the constant term is $a$-dependent and the terms free of $a$ form the equation below,

$$x^3+x^2-2x=x(x-1)(x+2)=0$$

which admits the roots $0$, $1$ and $-2$. Plug them into the full equation

$$x^3+(1-2a)x^2-(3a^2+a+2)x-6a^2+6a=0$$ to identify that $x=-2$ is an actual root and $x+2$ is a factor. Thus, the equation can be factorized as

$$(x+2)(x-3a)(x+a-1)=0$$

Then, the three roots are $-2$, $3a$ and $1-a$.