x3 - x2 + 2x = x*(x2 - x + 2) which cannot be factored further.
x3-x2
x2+x-6 = (x-2)(x+3) when factored
x3 + 8 = x3 + 23 = (x + 2)(x2 + 2x + 22) = (x + 2)(x2 + 2x + 4)
x2 + 5x + 6 = (x + 2)(x + 3)
x3 - x2 + 2x = x*(x2 - x + 2) which cannot be factored further.
x3-x2
x3 - x2 - 2x +2 you can divide this into two parts for now x3 - x2 and -2x +2 factor out everything you can from both x2(x-1) and -2(x+1) since the insides of the parentheses are the same, we can put them together, then multiply it by the sum of the two parts outside the parentheses. (x-1)(x2-2)
x2+x-6 = (x-2)(x+3) when factored
x3 + 8 = x3 + 23 = (x + 2)[x2 - (x)(2) + 22] = (x + 2) (x2 - 2x + 4)
x3 + 8 = x3 + 23 = (x + 2)(x2 + 2x + 22) = (x + 2)(x2 + 2x + 4)
-x3 -10x2 + 24x 1) -x can be factored out to give, -x(x2 + 10x - 24) 2) The bracketed expression can be factored as (x + 12)(x - 2), which now gives, -x(x + 12)(x - 2) = -x3 -10x2 + 24x
x2+7x-18 = (x+9)(x-2) when factored
x2 - 2 + 4 = x2 + 2This term can not be factored any further
x2 + 5x + 6 = (x + 2)(x + 3)
Factor x3 + x2 + 2x + 2, by grouping. Group the first two terms and the last two terms. Then factor. First, factor x3 + x2 by pulling out an x2 term: x2(x + 1) Second, factor 2x + 2 by pulling out a 2: 2(x + 1) So, you now have: x2(x + 1) + 2(x + 1) If you have factored correctly, the terms inside the parentheses should be the same. Now regroup. ANS: (x + 1)(x2 + 2)
(xn+2-1)/(x2-1)