You can't distribute into a bracket, that's raised to the power of anything other than 1, like this. To do this you need to raise distributed number to the bracket's power's inverse. in this case 1/2.

2(4)^2=(2^(1/2)*4)^2=(sqrt(2)*4)^2=2*4^2=2*16=32
or y'know 2*16=32

Maybe if we look at it with roots you'd get it. wolfram syntax

2(4)^2=2Surd[4,1/2]
2Surd[4,1/2]= Surd[4*2^(1/2),1/2]= (4*sqrt(2))^2= 4^2*2= 16*2= 32

I hope you don't get scared from this math, you're a teacher afterall. I have no Idea how you could have gotten a degree or not kicked from school on day 1. Unless.. you are trolling me, fuck you for that. If you respond with more bullshiting, I'll block you.

[–] SmartmanApps@programming.dev 1 points 7 months ago

2(4)^2=(2x4)^2=8^2=64

Yes, that's right. Brackets before Exponents, as per the order of operations rules.

You can’t distribute into a bracket

You know that's literally what The Distributive Law says you must do, right? Unless you have a source somewhere saying there's an exception?

Apparently you didn't bother reading any of the links I gave you, so here's one of the many textbooks which says you must distribute...

In case that's unclear, that means that 2x² and 2(x)² aren't the same thing (since 2(x)=(2x) by definition).

wolfram syntax

You know Wolfram disobeys The Distributive Law, right? I know I'm not the only one who knows this. Is that why you're insisting your way is right? Cos they're known to be wrong about this.

If you respond with more bullshiting,

You call quoting Maths textbooks "bullshiting"?