User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 September 20th, 2012, 05:18 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 2X - 2x^2 = ? Hi 1 = 2X - 2x^2 = ? 2 = 2X + 2x^2 = ? 3 = 2X x 2x^2 = ? I face problem in power ? 1 = 2x^-1 2 = 4x^3 3 = 4x2 September 21st, 2012, 12:23 AM #2 Member   Joined: May 2012 Posts: 78 Thanks: 0 Re: 2X - 2x^2 = ? May you explain your problem? September 21st, 2012, 04:05 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2207 For terms that are added or subtracted, identifying what they have in common allows you to factorize the overall expression. 2x - 2x� = 2x(1) - 2x(x) = 2x(1 - x) 2x + 2x� = 2x(1) + 2x(x) = 2x(1 + x) To combine powers of x that are multiplied, you add the exponents, so the product of 2x and 2x� is , which equals 4x�. September 23rd, 2012, 04:04 PM   #4
Math Team

Joined: Sep 2007

Posts: 2,409
Thanks: 6

Re: 2X - 2x^2 = ?

Quote:
 Originally Posted by r-soy Hi 1 = 2X - 2x^2 = ? 2 = 2X + 2x^2 = ? 3 = 2X x 2x^2 = ?
These are 3 problems, right? The "1", "2", and "3" in front of them are not part of the equation? I any case I see no statement of what it is you want to do. I would guess 'factor'. If so then 2x- 2x^2= 2(x- x^2)= 2x(1- x), 2x+ 2x^2= 2(x+ x^2)= 2x(1+ x), and (2x)(2x^2) (it's not a good idea to use "x" to indicate a multiplication when you are already using it to mean a variable) is already factored so I guess you mean to multiply: (2x)(2x^2)= 4x^3.

Quote:
 I face problem in power ? 1 = 2x^-1
2x^(-1)= 2(1/x)= 2/x

Quote:
 2 = 4x^3 3 = 4x2
4x^3 is already in its simplest form so I am not sure what you want here. And if, by "4x2" you mean 4x^2, the same- it is already in simplest form.

It would have helped if you had told us exactly what the problems asked you to do instead of just handing us expressions with no instructions! Tags 2x2 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      