5, if the numbers after the "x" are supposed to be powers
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Fifth degree - the highest power of x that appears.
This is how I did it:8x3+4x2+6x+3[8x3+4x2]+[6x+3][(4x2)(2x+1)]+[(3)(2x+1)]....as you can see, this is in the general form of ab+cb, which factored is b(a+c), so:(2x+1)(4x2+3)
Provided that x- 2 is not zero: 2x3 + x2 - 13x + 6 = 2x3 - 4x2 + 5x2 - 10x - 3x + 6 = 2x2*(x - 2) + 5x*(x - 2) - 3*(x - 2) = (x - 2)*(2x2 + 5x - 3) Therefore, (2x3 + x2 - 13x + 6)/(x - 2) = (x - 2)*(2x2 + 5x - 3)/(x - 2) = 2x2 + 5x - 3 If x = 2 so that x - 2 = 0, then the expression is not defined.
(2x - 3)(2x + 3)(2x-3)(2x+3)
-2x3 + 2x2 + 12x =(-2x) (x2 - x - 6) =(-2x) (x+2) (x-3)