Re: [Maxima-discuss] Feature request: step by step

[Stavros M. <mac...@gm...>]

These days, LLMs such as ChatGPT, Perplexity, and Gemini do a very good job
of explaining math operations step by step.

For example:

Q: explain step by step how to factor x^4+1

A: Factoring the expression $x^4 + 1$ is a classic problem because it
cannot be factored using simple real roots (since $x^4 + 1$ is never zero
for any real number $x$). To factor it into polynomials with real
coefficients, we use a technique called *completing the square* in a
slightly unconventional way.

Here is the step-by-step process:
1. Identify the Goal

We want to turn $x^4 + 1$ into a *difference of squares*, which follows the
pattern $a^2 - b^2 = (a - b)(a + b)$. Currently, we have a sum of
squares: $(x^2)^2
+ (1)^2$.
2. Add and Subtract a Middle Term

To make $x^4 + 1$ part of a perfect square trinomial, we need a middle term.

Recall that $(x^2 + 1)^2 = x^4 + 2x^2 + 1$.

We can rewrite our expression by adding and then immediately subtracting
$2x^2$ so that we don't change the value of the original expression:

$$x^4 + 2x^2 + 1 - 2x^2$$
3. Group into a Perfect Square

Now, group the first three terms, which form the perfect square we
identified in step 2:

$$(x^4 + 2x^2 + 1) - 2x^2$$
$$(x^2 + 1)^2 - 2x^2$$
4. Set Up the Difference of Squares

To use the $a^2 - b^2$ formula, we need the second term to be a perfect
square as well. We can write $2x^2$ as $(\sqrt{2}x)^2$:

$$(x^2 + 1)^2 - (\sqrt{2}x)^2$$
5. Apply the Factoring Formula

Now we apply $a^2 - b^2 = (a + b)(a - b)$, where $a = (x^2 + 1)$ and $b =
\sqrt{2}x$:

$$(x^2 + 1 + \sqrt{2}x)(x^2 + 1 - \sqrt{2}x)$$
Final Result

Standard practice is to write the terms in descending order of power:

$$(x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$$
------------------------------

*Note on Complex Roots:* If you were looking for factors using complex
numbers (which might appeal to your math background from MIT!), you would
use the fourth roots of $-1$, resulting in four linear factors involving $i$.
However, the version above is the standard "real" factorization.

Would you like me to show you the version using complex numbers (roots of
unity), or perhaps apply this logic to a different polynomial?

And a more difficult case:

Q: explain step by step how to factor x^7-x^6+x^4-x^3+2*x^2-2*x+1

You will see that it does not get this right. Maxima gets the correct
factorization, (x^3-x+1)*(x^4-x^3+x^2-x+1)

To factor the polynomial $x^7 - x^6 + x^4 - x^3 + 2x^2 - 2x + 1$, we can
use a combination of *factoring by grouping* and the technique for *completing
the square* that we used for $x^4 + 1$.

Given your preference for quality and substance over "flash," let’s break
this down into clear, logical segments.
1. Group the Terms

The first step is to look for common patterns in the terms. Notice how the
coefficients seem to pair up:

$$(x^7 - x^6) + (x^4 - x^3) + (2x^2 - 2x) + 1$$
2. Factor out Common Monomials

>From each of the first three groups, we can pull out the greatest common
factor:

$$x^6(x - 1) + x^3(x - 1) + 2x(x - 1) + 1$$
3. Factor by Grouping (Partially)

Notice that $(x-1)$ is common to almost everything except that final $+1$.
Let's group the terms containing $(x-1)$:

$$(x-1)(x^6 + x^3 + 2x) + 1$$

Wait—this doesn't seem to lead to a clean product. Let's try a different
grouping strategy by looking for a *perfect square* or a *cyclotomic*
pattern.
4. Alternative Approach: Strategic Rearrangement

Let's look at the expression again and group it differently to find a
hidden $x^4+1$ or similar structure:

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

Actually, there is a more elegant "hidden" structure. Let's look for the
expression $x^4+1$ which we factored previously. We can rewrite the
polynomial as:

$$x^3(x^4 - x^3 + x - 1) + 2x^2 - 2x + 1$$

If we check for small integer roots (Rational Root Theorem), $x=1$ and
$x=-1$ do not work (the sum of coefficients is $1 \neq 0$). This suggests
the factors are higher-degree polynomials.
5. Identifying the Quadratic Factor

The pattern of coefficients $(1, -1, 0, 1, -1, 2, -2, 1)$ suggests this
polynomial is related to the factors of $x^4+1$. If we test the factors of
$x^4+1$ that we found earlier: $(x^2 \pm \sqrt{2}x + 1)$, they won't work
because of the $\sqrt{2}$.

However, notice the "tail" of the polynomial: $x^4 - x^3 + x^2$ (if we
split the $2x^2$).

Let's test the cyclotomic-style factor $(x^2 - x + 1)$. This is a common
factor in polynomials with alternating signs.

If we perform polynomial long division by $(x^2 - x + 1)$:

   1.

   $(x^7 - x^6 + x^4 - x^3 + 2x^2 - 2x + 1) \div (x^2 - x + 1)$
   2.

   $x^5(x^2 - x + 1) = x^7 - x^6 + x^5$. Subtracting this leaves: $-x^5 +
   x^4 - x^3 + 2x^2 - 2x + 1$.
   3.

   $-x^3(x^2 - x + 1) = -x^5 + x^4 - x^3$. Subtracting this leaves: $2x^2 -
   2x + 1$.
   4.

   Unfortunately, $(x^2 - x + 1)$ does not divide $2x^2 - 2x + 1$ evenly
   (it would need to be $2x^2 - 2x + 2$).

6. The "Aha!" Moment: Square of a Trinomial

Let's look at the polynomial one more time. It can be compressed into a
very specific form:

$$x^4(x^3 - x^2) + x^2(x^2 - x) + 2x^2 - 2x + 1$$

Notice the recurring $x^2-x+1$ pattern if we manipulate the terms:

$$x^4(x^2-x+1) ... \text{ No.}$$

Let's try:

$$(x^4 - x^3 + x^2) + (x^3 - x^2 + x) + ... \text{ No.}$$

The most efficient way to see this factorization is to recognize it as:

$$(x^4+1)(x^3-x^2+1) + ... \text{ actually simplified to:}$$
$$(x^4+x+1)(x^3-x^2+1) \dots \text{ is not quite it.}$$
The Correct Factorization

The polynomial factors into:

$$(x^3 - x + 1)(x^4 - x^3 + x^2 - x + 1) \dots \text{ (Checking... no)}$$

Let's use the grouping $x^7-x^6$ and $x^4-x^3$ and $x^2-x$ and $x^2-x+1$:

$$x^6(x-1) + x^3(x-1) + 2x(x-1) + 1$$

Actually, this specific polynomial is often used in competitions because it
factors into:

$$(x^2 - x + 1)(x^5 + x^2 - 1) + x^2 \dots \text{ no.}$$

*Correction:* This polynomial is *irreducible* over the integers. In
competitive math or advanced algebra, polynomials of this complexity that
do not have rational roots (tested in Step 4) and do not easily fall into
the $x^4+1$ or $x^n \pm 1$ patterns are typically prime.
Final Result

Since there are no rational roots (by the Rational Root Theorem, only $\pm
1$ are possible, and $f(1)=1, f(-1)=7$) and no simple quadratic factors
like $(x^2+1)$ or $(x^2 \pm x + 1)$ work, the polynomial:

$$x^7 - x^6 + x^4 - x^3 + 2x^2 - 2x + 1$$

is irreducible over the integers.

If this was a problem from a specific textbook or context (like a Boston
College history of math paper!), let me know—sometimes there's a typo in
the prompt that changes a $+1$ to a $-1$, which makes it factor beautifully!



On Wed, Jan 14, 2026 at 11:57 AM Raymond Toy <toy...@gm...> wrote:

> On 1/14/26 11:19 AM, Richard Fateman wrote:
>
> If you are expecting Maxima to improve your math skills as in some of the
> on-line tutoring programs,
> by some nearly automatic process  (like tracing existing programs), that
> is highly unlikely to work.
>
> Elaborating on Stavros' note -- typically Maxima DOES NOT USE THE STEPS
> USED BY BEGINNING
> STUDENTS OF ALGEBRA, CALCULUS, ETC.
>
> Exactly.
>
> But really, what is the goal of seeing the step-by-step solution? If
> you're a calc 1 student, would it be useful to know that uses keyhole
> integration over the complex plane to evaluate some definite integrals? I
> suppose for motivated students, that would encourage them to learn new
> math, but I suspect that would basically turn off most students. (Perhaps
> even worse now when you hear stories about how college students can't do
> basic arithmetic.)
>
> And would this be better done by looking up Youtube videos that explain
> how to solve interesting problems? I think blackpenredpen is really good
> for calc 1/2 level math. For more interesting questions, I like Michael
> Penn. Maths 505 has really interesting integration and diffeq problems.
> Maxima can't do most of them, I think.
>
> There are zillions of math channels on Youtube.
>
>
> 2. There is a mystique of "calculus is hard, uses heuristics, requires
> intelligence and problem-solving" promoted by
> teachers, textbooks, etc.
>
> Well, I think that's kind of true when it comes to integration. If you
> watch Maths 505 doing integrals, many of them uses insights that I would
> not think of doing. Maybe I could, if I spent enough time to thinking about
> them, but probably not.
> &#8203;
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