Understanding the Basics of Algebraic Equations: Is 2x(x) = 3x or 2x(x) = 2x^2?

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The discussion clarifies that the expression 2x(x) simplifies to 2x^2, as it represents 2 multiplied by x and then by x again. It emphasizes that in algebra, multiplication is typically implied when variables are placed next to each other, without needing an explicit multiplication sign. Examples are provided to illustrate how algebraic expressions are structured, such as 2xy equating to 2*x*y. The conversation reinforces the importance of understanding these foundational concepts to avoid mistakes in algebra. Overall, the thread serves as a helpful guide for grasping basic algebraic principles.
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I know this is a dumb question but dumb questions are easier to fix than dumb mistakes so here goes.

Is 2x(x)= 3x

or 2x(x) = 2x^2
 
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2x^2
 
Cornraker said:
I know this is a dumb question but dumb questions are easier to fix than dumb mistakes so here goes.

Is 2x(x)= 3x

or 2x(x) = 2x^2

2x(x) means 2*x*x, which can be written as 2x2, as Tevion stated. This could also be written as 2xx, but is usually not written that way.

In algebraic expressions, multiplication is not usually shown explicitly, while addition (and subtraction and division) always are shown explicitly. If two expressions are next to each other, with or without parentheses, the two expressions are understood to be multiplied.

2xy = 2*x*y.
3xy + 2x = 3*x*y + 2*x.
3(xy + 2x) = 3*(xy + 2x) = 3xy + 6x.
2x + x = x(2 + 1) = x(3) = 3x
 
2x+ x= 3x
 
thanks everybody
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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