Comment on I dunno
mindbleach@sh.itjust.works 10 hours ago3(x-y) is a single term…
So is 3xy, according to that textbook. That doesn’t mean 3xy^2^ is 9*y^2^*x^2^. The power only applies to the last element… like how (8)2^2^ only squares the 2.
Four separate textbooks explicitly demonstrate that that’s how a(b)^c^ works. 6(ab)^3^ is 6(ab)(ab)(ab), not (6ab)(6ab)(6ab). 3(x+1)^2^ for x=-2 is 3, not 9. 2(x-b)^2^ has a 2b^2^ term, not 4b^2^. 15(a-b)^3^x^2^ is not (3375a-3375b)x^2^. If any textbook anywhere shows a(b)^c^ producing (ab)^c^, or x(a-b)^c^ producing (xa-xb)^c^, then reveal it, or shut the fuck up.
2(ab)^2^ is 2(ab)(ab) the same way 6(ab)^3^ is 6(ab)(ab)(ab). For a=8, b=1, that’s 2*(8*1)*(8*1).
SmartmanApps@programming.dev 9 hours ago
That’s right
That’s right. It means 3abb=(3xaxbxb)
Factor yes, hence the special rule about Brackets and Exponents that only applies in that context
It doesn’t do anything, being an invalid syntax to follow brackets immediately with a number. You can do ab, a(b), but not (a)b
Yep, as opposed to 6(a+b), which is (6a+6b)
No it isn’t. See previous point. Do we have an a(b+c), yes we do. Do we have an a(bc)²? No we don’t.
No, it has a a(b-c) term, squared
says someone still trying to make the special case of Exponents and Brackets apply to a Factorised Term when it doesn’t. 😂 I’ll take that as your admission of being wrong about a(b+c)=ax(b+c) then. Thanks for playing
Only if you had defined it as such to begin with, otherwise the Brackets Exponents rule doesn’t apply if you started out with 2(8)², which is different to 2(8²) and 2(ab)²
mindbleach@sh.itjust.works 47 minutes ago
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 PDF you plainly got from Archive.org. “In an expression such as 3a^2^, 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.
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.