Strategy on factoring quartic polynomials: cross method.

Question

I found a method to factorize quartic polynomials which I don't understand how it works.

It is presented like this:

Cross method This methodology lets to factorize ordered and completed 4th polynomials of the form:

$$F(x) = ax^4+bx^3+cx^2+dx+e$$

Rules:

1. Factor the extreme terms with cross method to get a squared term (generally different from the squared term of the original polynomial).
2. Get $\Delta$ from the difference of the squared term of the polynomial and the term of the first step, and replace the result in the original polynomial.
3. Then, verify the binary combinations as double cross factoring.

I used for different exercises and it works, but I don't understand the basis for this method.

Can anyone explain the reasons?

P.D.

Example

$$x^4+2x^3+3x^2+2x-3$$

1. Factoring the extremes terms; $x^4$ and $-3$:

$$(x^2+3)(x^2-1)$$ The result of the cross method is $3x^2-x^2=2x^2$

2. Calculating $\Delta$:

$$\Delta=\text{(original squared term)}-\text{(step one squared term)}= (3x^2)-(2x^2)=x^2$$

And replaced it in the original polynomial:

$$x^4+2x^3+x^2+2x-3$$

3. Using the cross method for second and fourth term:

For second term: $(x^2+x)(x^2+x)$ is $2x^3$

For fourth term: $(x+3)(x-1)$ is $2x$

So, arranging the terms, the two factors are:

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

Answer 1

I might be misunderstanding, but this just looks like concealed expanding $$(x^2+ax+b)(x^2+cx+d)\tag{1}$$ and comparing coefficients.

1. Factoring extremes

Well, what we want is to choose $b$ and $d$ such that $(x^2+b)(x^2+d)$ matches leading and constant factor. What it amounts to is finding $b$ and $d$ such that $bd = -3$. Let's say we were lucky and chose the winning pair $b = -1,\ d=3.$

2. Calculating $\Delta$

Why is this important? Well, $(x^2-1)(x^2+3) = x^4+2x^2-3$, so we are missing one $x^2$: $\Delta = 3x^2 - 2x^2$.

3. Covering for the missing $\Delta$

There is only one place where $\Delta$ can come from, that is from $a$ and $c$ - we must have $ac x^2 = \Delta$. Why? Well, if you expand $(1)$, you will find that $x^2$ term must be equal to $b + d + ac$. We found $b+d$ in step 1 and 2, so $acx^2$ is equal to $\Delta$. In this case $ac = 1$. There are two possibilities in $\Bbb Z$, either $a=c=1$, or $a=c=-1$. It is easy to decide which one, since it must fit the other terms as well, namely $x^3$ term is equal to $a + c$ and the $x$ term is equal to $ad + bc = 3a - c$. Thus, $a = c = 1$.

4. Collecting terms

We simply write $(x^2+x-1)(x^2+x+3) = x^4+2x^3+3x^2+2x-3$ and feel the magic all around us.

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Now, let us do the same thing, only transparently:

$$(x^2+ax+b)(x^2+cx+d) = x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$

so we need to solve the following system in $\Bbb Z$:

\begin{array}{c r}\begin{align} a+c&=2\\ b+d+ac&=3\\ ad+bc&=2\\ bd &= -3 \end{align} &\tag{2}\end{array}

There are two possibilities, either $b = 1$ and $d = -3$, or $b = -1$ and $d = 3$. We can see that former leads to no solution, so we will pursuit the latter, the system becomes:

\begin{align} a+c&=2\\ ac&=1\\ 3a-c&=2 \end{align}

The first and the third equation form linear system, which has unique solution $a = c = 1$, luckily, consistent with the second equation. Thus, we now know for certain that $$(x^2+x-1)(x^2+x+3) = x^4+2x^3+3x^2+2x-3.$$

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Addendum:

There is a legitimate question why we would consider $(1)$ in the first case. Now, we are trying to factor a quartic over $\Bbb Z$. Let's say $p(x) = x^4 +2x^3+3x^2+2x-3$ and assume that $p = fg$. We have that $\deg p = \deg f + \deg g$, so there are two cases: $4=1+3$ or $4=2+2$. The first case would imply that $p$ has integer root and we can use rational root theorem to rule that out. Thus, only case $4=2+2$ remains, which leads to considering $(1)$.

Finally, if the system $(2)$ had no solutions in $\Bbb Z$, we would conclude that $p$ is irreducible over $\Bbb Z$.

Answer 2

Are you aware of how to factor quadratic equations? This is just a method for that, adapted to work for quartic equations. The reason that this works is that many (but not all) quartic equations factor into the product of two quadratic equations.

To see that this doesn't always work, apply it to $$p(x)=x^4-3x^3+x^2-2x-3=(x-3)(x^3+x+1)$$ The second term doesn't factor any further.

Here's a less confusing way of giving the same instructions:

1. Assume the polynomial, $f(x)$, factors into $(x^2+ax+b)(x^2+cx+d)$
2. $bd$ must give the constant term of the original polynomial, so find two values that multiply to give the constant term and write $g(x)=(x^2+b)(x^2+d)=x^4+(b+d)x^2+bd$. In the computational example, $g(x)=(x^2+3)(x^2-1)$
3. Now let's look at $f(x)-g(x)$, which is $2x^3+x^2+2x$ in your example. This is a degree 3 polynomial with no constant term and represents the parts of the factors we need to still find. In our factorization, we know that the degree $1$ term is given by $(ad+bc)x$, and we have already found $b$ and $d$. Similarly, we know that the degree $3$ term is given by $(a+c)x^3$. This is a system of equations with two variables and two unknowns and so can be solved. In your example, these equations are $-a+3c=2$ and $a+c=2$ respectively.
4. Solving gives the last two coefficients we need, and then putting them together with the original piece gives us the factorization. In your example, the solution $a=c=1$ works. This gives us $(x^2+x+3)(x^2+x-1)$