There is no special case. You made it up by confusing yourself about “dismissing a bracket.” To everyone else in the world, brackets are just another term. Several of the textbooks I’ve linked will freely juxtapose brackets and variables before or after, because it makes no difference.
Here’s yet another example, PDF page 27: (6+5)x+(-2+10)y. And that’s as factorization. This Maths textbook you plainly didn’t read was published this decade. Still waiting on any book ever that demonstrates your special bullshit.
7bx with b=(m+n) becomes 7(m+n)x and it’s the same damn thing. Splitting it like 7xm+7xn is no different from splitting (m+n)/7 into m/7+n/7. Brackets only happen first because they have to be reduced to a single term. A bracket with one number is not “unsolved” - it’s one number. Squaring a bracket with one number is squaring that number.
The base of an exponent is whatever’s in the symbols of inclusion. Hence: 6(ab)3 = 6(ab)(ab)(ab).
No, it has a a(b-c) term, squared
It has a (b-c) term, squared. The base of an exponent is whatever’s in the symbols of inclusion. See page 121 of 696, in the Modern Algebra: Structure And Method PDF you plainly got from Archive.org. “In an expression such as 3a2, the 2 is the exponent of the base a. In an expression such as (3a)2, the 2 is the exponent of the base 3a, because you enclosed the expression in a symbol of inclusion.” You will never find a published example that makes an exception for distribution first.
On the page before your screenshot - 116 of 696 - this specific Maths textbook refers to both 8x7 and 8(7) as “symbols of multiplication.” It’s just multiplication. It’s only ever multiplication. It’s not special, you crank. 8(7) is a product identical to 8x7. Squaring either factor only squares that factor.
So you’re saying there no such rule as 2(ab)²=2a²b². Got it. you’re admitting you’re wrong then 😂
You made it up by confusing yourself about “dismissing a bracket.”
Says person who just claimed there’s no such rule as the one they’ve been making the basis of their wrong claims 😂
To everyone else in the world, brackets are just another term
That’s right. That’s why you cannot separate the coefficient from it 😂
Several of the textbooks I’ve linked will freely juxtapose brackets and variables before or after
Several of your textbooks are outdated then
because it makes no difference
So is (2+3)-4 equal to -20 or 1? I’ll wait
And that’s as factorization.
Wrong Factorisation. ab+ac=a(b+c)
This Maths textbook you plainly didn’t read was published this decade
And yet, is still wrong
Still waiting on any book ever that demonstrates your special bullshit
You were the one who just said there’s no such special rule as 2(ab)²=2a²b² 😂
7bx with b=(m+n) becomes
7x(m+n)
and it’s the same damn thing
No, 7(m+n)x is invalid syntax due to ambiguity when x is negative.
Splitting it like 7xm+7xn is no different from splitting (m+n)/7 into m/7+n/7
That’s right. I never said otherwise.
Brackets only happen first because they have to be reduced to a single term
They happen first because they already are a single Term…
A bracket with one number is not “unsolved”
Yes it is! If you haven’t solved Brackets then you cannot progress onto Exponents.
it’s one number
Exactly! That’s why you cannot separate the a from (b+c) - it’s all ONE NUMBER 😂
Squaring a bracket with one number is squaring that 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 😂
Hence: 6(ab)3
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 😂
It has a (b-c) term
No it doesn’t. a(b-c) is one term
The base of an exponent is whatever’s in the symbols of inclusion.
And there’s no exponent in a(b+c) 🙄
See page 121 of 696
See page 37
Also see page 282 and answers on page 577. x(x-1) is one term, as taught on page 37
In an expression such as 3a2,
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 😂
You will never find a published example that makes an exception for distribution first
BWAHAHAHAHAHAHAH! I see you haven’t read ANY of the sources I’ve posted so far then 🤣
refers to both 8x7 and 8(7) as “symbols of multiplication.”
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
It’s just multiplication
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
It’s not special, you crank
Yes it is, as per the textbook you’re quoting from! 🤣
8(7) is a product identical to 8x7
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
Squaring either factor only squares that factor
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 😂
Variables don’t work differently when you know what they are.
They’re not variables if you know what they are - they are constants, literally a number as per pages 36 and 37
b=1 is not somehow an exception that isn’t allowed, remember?
That’s right. a(1+c)=(a+ac), and you’re point is??
There’s an exponent in 2(8)2
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 🙄
it concisely demonstrates to anyone who passed high school that you can’t do algebra
says someone who is trying to say that a rule about exponents applies to expressions without exponents 🤣
There is no special case. You made it up by confusing yourself about “dismissing a bracket.” To everyone else in the world, brackets are just another term. Several of the textbooks I’ve linked will freely juxtapose brackets and variables before or after, because it makes no difference.
Here’s yet another example, PDF page 27: (6+5)x+(-2+10)y. And that’s as factorization. This Maths textbook you plainly didn’t read was published this decade. Still waiting on any book ever that demonstrates your special bullshit.
7bx with b=(m+n) becomes 7(m+n)x and it’s the same damn thing. Splitting it like 7xm+7xn is no different from splitting (m+n)/7 into m/7+n/7. Brackets only happen first because they have to be reduced to a single term. A bracket with one number is not “unsolved” - it’s one number. Squaring a bracket with one number is squaring that number.
The base of an exponent is whatever’s in the symbols of inclusion. Hence: 6(ab)3 = 6(ab)(ab)(ab).
It has a (b-c) term, squared. The base of an exponent is whatever’s in the symbols of inclusion. See page 121 of 696, in the Modern Algebra: Structure And Method PDF you plainly got from Archive.org. “In an expression such as 3a2, the 2 is the exponent of the base a. In an expression such as (3a)2, the 2 is the exponent of the base 3a, because you enclosed the expression in a symbol of inclusion.” You will never find a published example that makes an exception for distribution first.
On the page before your screenshot - 116 of 696 - this specific Maths textbook refers to both 8x7 and 8(7) as “symbols of multiplication.” It’s just multiplication. It’s only ever multiplication. It’s not special, you crank. 8(7) is a product identical to 8x7. Squaring either factor only squares that factor.
Variables don’t work differently when you know what they are. b=1 is not somehow an exception that isn’t allowed, remember?
There’s an exponent in 2(8)2 and it concisely demonstrates to anyone who passed high school that you can’t do algebra.
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 🤣