Understanding Cubic Factorization: Solving for Roots with a and -2a

[Argonaut]

Homework Statement

Find the roots of the following equation:
$$x^3-3a^2x+2a^3=0$$
where $a$ is a constant.

Relevant Equations

None

This is part of a longer exercise I struggled with. I checked the solutions manual, and there was a bit where they performed the following steps:

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

And then concluded that the roots were $a$ and $-2a$, which is clear. What I can't work out is how they factorised the original equation.

I've got as far as figuring out that $a$ is a root so that $(x-a)$ must be a factor, but I didn't get any further than that.

What am I missing?

 

[topsquark]

Homework Statement: Find the roots of the following equation:
$$x^3-3a^2x+2a^3=0$$
where $a$ is a constant.
Relevant Equations: None

This is part of a longer exercise I struggled with. I checked the solutions manual, and there was a bit where they performed the following steps:

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

And then concluded that the roots were $a$ and $-2a$, which is clear. What I can't work out is how they factorised the original equation.

I've got as far as figuring out that $a$ is a root so that $(x-a)$ must be a factor, but I didn't get any further than that.

What am I missing?

The quick answer to this is a variation on the rational root theorem. First try $x = \pm a, \pm 1$ and see if there are solutions. If there is one, or more, then you can find the rest by doing a long division on the polynomial, and use the quadratic formula or just solve the linear equation. It turns out that you can easily find solutions to this cubic this way.

The long answer is Cardano's method. This is already a depressed cubic, so it shouldn't be too awful hard, but it is going to get a bit involved.

-Dan

 

[pasmith]

Homework Statement: Find the roots of the following equation:
$$x^3-3a^2x+2a^3=0$$
where $a$ is a constant.
Relevant Equations: None

This is part of a longer exercise I struggled with. I checked the solutions manual, and there was a bit where they performed the following steps:

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

And then concluded that the roots were $a$ and $-2a$, which is clear. What I can't work out is how they factorised the original equation.

I've got as far as figuring out that $a$ is a root so that $(x-a)$ must be a factor, but I didn't get any further than that.

What am I missing?

Write $$x^3 - 3a^2 x + 2a^3 = (x - a)(x^2 + Ax + B)$$ and expand the right hand side. Compare coefficients of powers of $x$ to find $A$ and $B$. Or use $$(x - a)^3 = x^3 - 3ax^2 + 3a^2 x - a^3$$ to write $$\begin{split} x^3 - 3a^2 x + 2a^3 &= \left((x - a)^3 + 3ax^2 - 3a^2 x + a^3\right) - 3a^2 x + 2a^3 \\ &= (x - a)^3 + 3ax^2 - 6a^2 x + 3a^3\end{split}$$ and further simplify the final expression.

 

[fresh_42]

If you encounter such problems in artificial situations like homework, exams, or examples in books, then it is almost always the case that one solution can be guessed. We see in our case that $x=a$ is a zero of $x^3-3a^2x+2a^3.$ This means that $(x-a)$ divides the polynomial $x^3-3a^2x+2a^3.$ You can perform a long division like you do with numbers
$$
(x^3-3a^2x+2a^3) \, : \,(x-a) = q(x)
$$
to receive the resulting polynomial $q(x)$ which has to be of degree $2$ and $q(x)=0$ can be solved by e.g. completing the square, or solve the equation
$$
q(x)\cdot (x-a)=(x^2+Ax+B)\cdot (x-a)=x^3-3a^2x+2a^3
$$
as @pasmith suggested. Both works. I like the long division because it is only one step and I need the practice for more complicated cases, but that is a matter of taste (and the degree of the polynomials).

If we have $q(x)=x^2+ax-2a^2$ then we see that $x=a$ is another zero and we can divide again by $(x-a)$ or use the known formula
$$
q(x)=0\Longrightarrow x_{1,2}=-\dfrac{a}{2}\pm \sqrt{\dfrac{a^2}{4}+2a^2}=-\dfrac{a}{2}\pm \dfrac{3}{2}a\Longrightarrow x\in \left\{a\, , \,-2a\right\}
$$
And $x^3-3a^2x+2a^3=(x-a)\cdot (x-a) \cdot (x-(-2a))=(x-a)^2(x+2a).$

 

[Argonaut]

Thanks, @topsquark, @pasmith and @fresh_42, for the insightful replies. I'm much obliged!

I now realise that the long division of polynomials has been a glaring blind spot in my mathematical repertoire.

(An amusing aside:

I blame the inconsistency of mathematical notation across nations!

You see, as a pupil, I was taught long division using the German method. However, I only learnt the long division of polynomials much later and using the 'English method', and the whole thing felt cumbersome because of the new notation. I didn't realise they were the same algorithm, just different notation. I drilled the English method with polynomials for a while, but it quickly fell out of my mathematical toolbox, because I still tend to reach for the German long division whenever needed. Now I know I can use the familiar and comfortable German method to carry out the long division of polynomials. It seems like a silly oversight.)

 

[fresh_42]

Here is an example of the more complicated cases I spoke of:
https://www.physicsforums.com/threa...r-a-polynomial-over-z-z3.889140/#post-5595083
It assumes the existence of an arbitrary number $a$ that is a zero. It additionally shows
$$
p(a)=0 \Longrightarrow (x-a)\,|\,p(x)
$$
for the given polynomial $p(x).$

 

[Argonaut]

Here is an example of the more complicated cases I spoke of:
https://www.physicsforums.com/threa...r-a-polynomial-over-z-z3.889140/#post-5595083
It assumes the existence of an arbitrary number $a$ that is a zero. It additionally shows
$$
p(a)=0 \Longrightarrow (x-a)\,|\,p(x)
$$
for the given polynomial $p(x).$

Thanks! This instantly made me look forward to Abstract Algebra more.