Solving brackets does not include forced distribution
Yes it does! 😂
Juxtaposition means multiplication,
No, it doesn't. A Product is the result of Multiplication. If a=2 and b=3, axb=ab, 2x3=6, axb=2x3, ab=6. 3(x-y) is 1 term, 3x-3y is 2 terms...
as such, 2(3+5)² is the same as 2*(3+5)²
No it isn't. 2(3+5)² is 1 term, 2x(3+5)² is 2 terms
so once the brackets result in 8
They don't - you still have an undistributed coefficient, 2(8)
they’re solved
Not until you've Distributed and Simplified they aren't
Distribution needs to happen if you want to remove the brackets
if you want to remove the brackets, YES, that's what the Brackets step is for, duh! 😂 The textbook above says to Distribute first, then Simplify.
while there are still multiple terms inside
As in 2(8)=(2x8) and 2(3+5)=(6+10) is multiple Terms inside 😂
it’s still a part of the multiplication
Nope! The Brackets step, duh 😂 You cannot progress until all Brackets have been removed
which has higher priority.
It doesn't have a higher priority than Brackets! 🤣
Your whole argument hangs on the misinterpretation of textbooks
says person who can't cite any textbooks that agree with them, so their whole argument hangs on all Maths textbooks are wrong but can't say why, 😂 wrongly calls Products "Multiplication", and claimed that I invented a rule that is in an 1898 textbook! 🤣 And has also failed to come up with any alterative "interpretations" of "must" and "Brackets" that don't mean, you know, must and brackets 😂
This is what it feels like to argue against Bible fanatics
says the Bible fanatic, who in this case can't even show me what it says in The Bible (Maths textbooks) that agrees with them 😂
provide me a solver that says 2(3+5)² is 256 and you’ve won, it’s so easy no?
provide me a Maths textbook that says 8/2(1+3)=16 and you’ve won, it’s so easy no? 🤣
And in the meantime, here's one saying it's 1, because x(x-1) is a single Term...
So you're saying there no such rule as 2(ab)²=2a²b². Got it. you're admitting you're wrong then 😂
Says person who just claimed there's no such rule as the one they've been making the basis of their wrong claims 😂
That's right. That's why you cannot separate the coefficient from it 😂
Several of your textbooks are outdated then
So is (2+3)-4 equal to -20 or 1? I'll wait
Wrong Factorisation. ab+ac=a(b+c)
And yet, is still wrong
You were the one who just said there's no such special rule as 2(ab)²=2a²b² 😂
7x(m+n)
No, 7(m+n)x is invalid syntax due to ambiguity when x is negative.
That's right. I never said otherwise.
They happen first because they already are a single Term...
Yes it is! If you haven't solved Brackets then you cannot progress onto Exponents.
Exactly! That's why you cannot separate the a from (b+c) - it's all ONE NUMBER 😂
if it has been written as 2(ab)², not if it has been written a(b+c)². Also, again, there is no exponent in a(b+c), so that rule doesn't apply anyway 😂
6(ab)² <== note: not 6(a+b)², nor 6(a+b) for that matter. You're still desperately trying to make a False Equivalence argument 😂
No it doesn't. a(b-c) is one term
And there's no exponent in a(b+c) 🙄
See page 37
Also see page 282 and answers on page 577. x(x-1) is one term, as taught on page 37
3a² <== note: not 3(a+b), which has no exponent 🙄 It's hilarious how much effort you're putting into such an obvious False Equivalence argument 😂
BWAHAHAHAHAHAHAH! I see you haven't read ANY of the sources I've posted so far then 🤣
Yep, present tense and past tense, since 8(7) is a Product, a number, as per Pages 36 and 37, which you're still conveniently ignoring, despite me having posted it multiple times
Nope. 8(7) is a Product, a single Term, as per Pages 36 and 37. You won't find them writing 8(7)=8x7 anywhere in the whole book, always 8(7)=56, because it's a number
Yes it is, as per the textbook you're quoting from! 🤣
Nope! 8(7) is a number, as per the textbook you are - selectively - quoting from 😂 8(7) is one term, 8x7 is two terms, as per Page 37
Where there are multiple pronumeral factors and you need the brackets to specify which factors the square is applying to, which, again, none of which applies to a(b+c) anyway Mr. False Equivalence 😂
They're not variables if you know what they are - they are constants, literally a number as per pages 36 and 37
That's right. a(1+c)=(a+ac), and you're point is??
and no Pronumerals, and 2(8) is a number as per pages 36 and 37. 🙄And if you had read those pages, you would find it also tells you why you cannot write 2(8) as 28 (in case it's not obvious). if you want 2x8², then you can just write 2x8² 🙄
says someone who is trying to say that a rule about exponents applies to expressions without exponents 🤣