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You cannot. You can only combine terms with common terms and the exponents are part of the terms. So this would yield (in simplest form) x^2 + x
So if I am looking to solve for X and the equation looks like X^2+2X=8 how do I do that?
you could subtract 8 from both sides to get x^2 + 2x - 8 = 0 and then factor the right side to (x - 2)(x + 4) = 0. If you set each factor equal to 0 you get: x - 2 = 0 x + 4 = 0 Solving each of these yields x = 2 and x = -4
I'm sorry, I understand how you went from X^2+2X=8 to X^2+2X -8 = 0 but I don't understand how you got to (x-2)(x+4) = 0 How does that work?
That's what's considered factoring a polynomial. By writing it in the form (x+a)(a+b) it is easy to solve by setting each factor equal to 0. To find a and b you want two number that multiply together to get -8 and add together to get +2. The two numbers that work are a = -2 and b = 4.
Hi, thanks so much for your reply. I'm still having trouble. I don't really see how we are getting -2 and 4 except through guessing... I mean, what is the next line if you were showing your work to show how you got that so that I might make more sense of it. I'm really not trying to cheat on homework, I want to understand
I know. Factoring can be a difficult concept to get. Let me start at the beginning. All binomials (the type of polynomials this works well for) come in the form ax^2+bx+c. When factoring it is sometimes easier to divide the whole problem by "a" to begin with. The example you used had a=1 so it was ready for the next step. In the next step you look at the factors of "c." The factors of 8 are: 1, 8 2, 4 and those are the only ones. Next you look at the factors and find the pair that will add or subtract to give you the "b" value. In this case we had b=2 so the two factors that can be combined to equal 2 are 2 and 4. Now we can write our factored form as (x 2)(x 4) and we are going to put + and - signs in there. In this case we wanted +2 so 4-2=2 and that's how we got the final answer of (x-2)(x+4).
Thank you very much for your help that's really great of you