Yep I have looked at all you've posted, I say you're wrong because what you've posted says things that are true, but you're reading them wrong. For example your last image clearly says a number next to a bracket means the content of the bracket must be multiplied with said number. Nowhere there does anybody speak of distribution taking precedence over other operations. In fact, nowhere in all sources I can find does it say so. Wonder why all screenshots you post use convoluted wording and wonder why you pop up everywhere arguing the same thing and keep getting downvoted? At some point you need to understand that if one old-ass calculator and selective reading of cherry picked passages is all the proof you have, when all modern calculators and algebra solvers go against you, maybe it's time to reconsider.

Juxtaposition taking precedence over other multiplications I can understand and it's an arguable point. Distribution being a mandatory step and taking precedence over even exponents is just silly and unfortunately wrong.

Also another thing: you're a math teacher as you've said, and consistently ask if I think "random programmers" know more about algebra than you. What I say to that is I've met plenty of teachers who are wrong about things in their own fields, for one. And also, people defining the rules of all those algebra solvers aren't the programmers, as you'd know if you looked a bit into product development. It's domain experts, who also define tests and receive feedback on the software's performance and errors. I'm sure (lol) you've sent feedback to them, and they probably looked at it and decided you're wrong. As well all have.

An algebraic expression written as a product or quotient of numerals or variables or both is called a term

So b * c, which is a product of the variables b and c, is a term, according to this textbook.

You seem to be getting confused because none of the examples on this particular page feature the multiplication symbol ×. But that is because on the previous page, the author writes:

When a product involves a variable it is customary to omit the symbol × of multiplication.

That means that the expression bc is just another way of writing b×c; it is treated the same other than requiring fewer strokes of the pen or presses on a keyboard, because this is just a custom. That should clear up your confusion in interpreting this textbook (though really, the language is clear: you don't dispute that b×c - or b * c - are products, do you.)

Elsewhere in this thread you are clearly confused about what brackets mean. They are explained on page 20 of your textbook, where it says that you evaluate the expression inside the (innermost) brackets before doing anything else. Notice that, in its elucidation of several examples, involving addition and multiplication, the "distributive law" is not mentioned, because the distributive law has nothing to do with brackets and is not an operation.

Thus the expression 3 × (2 + 4) can be evaluated by first performing the summation inside the brackets to get 3 × 6 and then the product to get 18. The textbook then says that it is customary to omit the multiplication symbol and instead write 3(2+4), again indicating that these expressions are merely different ways of writing the same thing.

The exact same process of course must be followed whether numbers are represented by numerals or by letters designating a variable. You cannot do algebra if you don't follow the same procedure in both cases. So consider the expression 2(a+b)². You have suggested that you must evaluate this as (2a+2b)² because you must "do brackets first", but this is not what "doing brackets" means. You haven't produced any authority to suggest that it is, and your own textbook makes it clear that "doing brackets" means do what is inside the brackets first. Not what is outside the brackets. Distributing 2 over a+b is not "doing brackets"; it is multiplication and comes afterwards.

If 2(a+b)² were equal to (2a+2b)² let us try with a=b=2. Let us first evaluate (2a+2b)²: it is equal to (2×2+2×2)² = (4+4)² = 8² = 64. Now let us evaluate 2(a+b)²: it is 2(2+2)² and now, following your textbook's instruction to do what is inside the brackets first, this is equal to 2(4)². The next highest-priority operation is the exponent, giving us 2×16 (we now must write the × because it is an expression purely in numerals, with no brackets or variables) which is 32.

The fact that these two answers are different is because your understandings of what it means to "do brackets" and the distributive law are wrong.

Since I'm working off the textbook you gave, and I referred liberally to things in that textbook, I'm sure if you still disagree you will be able to back up your interpretations with reference to it.

By the way, I noticed this statement on page 23, regarding the order of operations:

However, mathematicians have agreed on a rule to fall back on if someone omits punctuation marks.

it does rather seem like this rule is one established not by the fundamental laws of mathematics but by agreement as they say, does it not? Care to comment?

b*c is the product of b and c.

Show me one textbook where a(b+c)^2^ gets an a^2^ term. Here's four in a row that say you're full of shit.

I’ll take that as an admission that you’re wrong. Thanks for playing

So no answer for centuries of textbooks saying you're full of shit.

a=8, b=1, it’s the same thing

No it isn't! 😂 8 is a single numerical factor. ab is a Product of 2 algebraic factors.

False equivalence is you arguing about brackets and exponents

Nope. I was talking about 1/a(b+c) the whole time, as the reason the Distributive Law exists, until you lot decided to drag exponents into it in a False Equivalence argument. I even posted a textbook that showed more than a century ago they were still writing the first set of Brackets. i.e. 1/(a)(b+c). i.e. It's the FOIL rule, (a+b)(c+d)=(ac+ad+bc+bd) where b=0, and these days we don't write (a)(b+c) anymore, we just write a(b+c), which is already a single Term, same as (a+b)(c+d) is a single term, thus doesn't need the brackets around the a to show it's a single Term.

3(x-y) is a single term...

This entire thing is about your lone-fool campaign

Hilarious that all Maths textbooks, Maths teachers, and most calculators agree with me then, isn't it 😂

insist 2(8)2 doesn’t mean 2*82,

Again, you lot were the ones who dragged exponents into it in a False Equivalence argument to 1/a(b+c)

I found four examples, across two centuries

None of which relate to the actual original argument about 1/a(b+c)=1/(ab+ac) and not (b+c)/a

You can’t pivot to pretending this is a division syntax issue

I'm not pivoting, that was the original argument. 😂 The most popular memes are 8/2(2+2) and 6/2(1+2), and in this case they removed the Division to throw a curve-ball in there (note the people who failed to notice the difference initially). We know a(b+c)=(ab+ac), because it has to work when it follows a Division, 1/a(b+c)=1/(ab+ac). It's the same reason that 1/a²=1/(axa), and not 1/axa=a/a=1. It's the reason for the brackets in (ab+ac) and (axa), hence why it's done in the Brackets step (not the MULTIPLY step). It's you lot trying to pivot to arguments about exponents, because you are desperately trying to separate the a from a(b+c), so that it can be ax(b+c), and thus fall to the Multiply step instead of the Brackets step, but you cannot find any textbooks that say a(b+c)=ax(b+c) - they all say a(b+c)=(ab+ac) - so you're trying this desperate False Equivalence argument to separate the a by dragging exponents into it and invoking the special Brackets rule which only applies in certain circumstances, none of which apply to a(b+c) 😂

2(8)2 is (2*8)2.

That's right

are you just full of shit?

says the person making False Equivalence arguments. 🙄 Let me know when you find a textbook that says a(b+c)=ax(b+c), otherwise I'll take that as an admission of being wrong that you keep avoiding the actual original point that a(b+c) is a single Term, as per Maths textbooks