I was wondering, is it that you can only add and subtract like terms... but what about multiplication and division..

 

and with powers or exponents like, 6^8, can you call them power coefficients... like the coefficient of terms such as 56t or 4ab etc...

You need to be a little clearer in what you are asking. You can multiply two terms to give a product term

 

[math] 4x . 3y = 12xy [/math]

 

With exponents (ie the 2 in [math] x^2 [/math]) you can only only really work on them if the base (ie the x in [math] x^2 [/math]) are the same and you are multiplying (dividing is the same with negative exponent)

 

ie [math] x^2 . x^3 = x^5[/math] but [math] x^2 . y^3 \neq xy^5 \neq (xy)^5 [/math]

 

whenever you are unsure of stuff like this sub in simple numbers and check it still works (or fails)

 

[math] 2^2 . 2^3 = x^5 [/math]

[math] 4 * 8 = 32 [/math]

 

but

[math] 2^2 . 3^3 \neq 6^5 [/math]

[math]4 * 27 = 108\[/math] but [math]6^5=7776[/math]

Edited June 2, 201114 yr by imatfaal

You need to be a little clearer in what you are asking. You can multiply two terms to give a product term

 

[math] 4x . 3y = 12xy [/math]

 

With exponents (ie the 2 in [math] x^2 [/math]) you can only only really work on them if the base (ie the x in [math] x^2 [/math]) are the same and you are multiplying (dividing is the same with negative exponent)

 

ie [math] x^2 . x^3 = x^5[/math] but [math] x^2 . y^3 \neq xy^5 [/math]

 

whenever you are unsure of stuff like this sub in simple numbers and check it still works (or fails)

 

[math] 2^2 . 2^3 = x^5 [/math]

[math] 4 * 8 = 32 [/math]

 

but

[math] 2^2 * 3^3 \neq 6^5 [/math]

[math]4 * 27 = 108\[/math] but [math]6^5=7776[/math]

 

Or simplify the exponentiation to multiplication

 

[math] x^2 * y^3 = x * x * y * y * y [/math]

[math] xy^5 = xy * xy * xy * xy * xy * xy [/math]

Edited June 2, 201114 yr by baric

Or simplify the exponentiation to multiplication

 

[math] xy^5 = xy * xy * xy * xy * xy * xy [/math]

 

 

ridiculous....

 

[math]xy = x*y[/math]

 

so [math] xy^5 = x*y*y*y*y*y [/math]

 

and, bracket always important.

 

[math] xy * xy * xy * xy * xy * xy = (xy)^5 = x^5y^5[/math]

 

actually, the exponent simplify multiplication... i dun think it's happen the other way

Edited June 2, 201114 yr by Vastor

ridiculous....

 

I thought it was obvious that I was following the notation of the person I quoted who used [math] xy^5 [/math] as a visual shorthand for [math] (xy)^5 [/math]

 

My point about converting the exponentiation to multiplication still holds.

ok, that makes sense, but what about misusing, such as: X^2 - x

 

or such a one as B^5c - 3abc...

 

What i meant before is that, can you take 3abc away from B^5c?. in another question, do you treat the 3 as the quantities of the abc letters or do you associate it just as itself (number 3)? or both..

 

so would you end up with '3ab^4' ?

I thought it was obvious that I was following the notation of the person I quoted who used [math] xy^5 [/math] as a visual shorthand for [math] (xy)^5 [/math]

 

My point about converting the exponentiation to multiplication still holds.

 

No - I was showing why it didn't work, I will edit my initial to make it clearer.

sorry, also what i meant was, can the exponents of terms, ie. X^3 be regarded as power/exponent coefficients ?, therefore when its said that like terms only differ in their coefficients, they mean the powers as well...

Edited June 2, 201114 yr by 1123581321

ok, that makes sense, but what about misusing, such as: X^2 - x

 

or such a one as B^5c - 3abc...

 

What i meant before is that, can you take 3abc away from B^5c?. in another question, do you treat the 3 as the quantities of the abc letters or do you associate it just as itself (number 3)? or both..

 

so would you end up with '3ab^4' ?

 

You can only add and subtract like terms. You can also factor out a term from an addition/subtraction to make it easy to work with. Again try it with real numbers

 

EDIT

 

OK I see your follow-up. No you cannot think of things being like terms that can added if their exponents are different. However you can factorise and take out most of the x term and leave something that looks like this x^2(x+3) which might help.

Edited June 2, 201114 yr by imatfaal

I thought it was obvious that I was following the notation of the person I quoted who used [math] xy^5 [/math] as a visual shorthand for [math] (xy)^5 [/math]

 

My point about converting the exponentiation to multiplication still holds.

 

 

let's me explain you about the use of bracket and others that u should know...

 

[math] xy^5 = x^1y^5 = (x)^1(y)^5[/math]

 

based on the equation, u can see that the '1' actually not shown becoz everything would result the same

 

 

[math] () = * [/math] in different way, let's see

 

[math] a = xy [/math]

 

[math] a^5 = (xy)^5 = (x)^5*(y)^5 \neq xy^5 = (x)^1(y)^5 = x^1*y^5 [/math]

 

() usually used to substitute an unknown while having the same effect of * which is multiply

Edited June 3, 201114 yr by Vastor

let's me explain you about the use of bracket and others that u should know...

 

[math] xy^5 = x^1y^5 = (x)^1(y)^5[/math]

 

 

Please, I understand all of that. That is below basic algebra.

 

I was simply trying to follow the same visual format of the person I was quoting for clarify rather than going back and nitpicking his post over a minor formatting error.

 

And if you had read the text of the post you just quoted, I wouldn't have to repeat that.

Edited June 4, 201114 yr by baric

so, would you be able to take x away from x^2 or only via division?

 

so, if the base is the same, the power difference doesn't matter does it..

yes and yes