x3 + 2x2 + 3x + 6 = x2(x + 2) + 3(x + 2) = (x + 2)(x2 + 3)
-2x3-2x2+12x (-2)(x3+x2-6x) (-2x)(x2+x-6) (-2x)(x2+3x-2x-6) (-2x)((x)(x+3)-2(x+3)) (-2x)((x-2)(x+3)
-3/2
-2x3 - 2x2 + 12x = -2x(x2 + x - 6) = -2x(x2 - 2x + 3x - 6) = -2x[ x(x - 2) + 3(x - 2) ] = -2x(x + 3)(x - 2)
assuming x*2-3x+2:factors of -3:-1,3 --> sum: 21,-3 --> sum: -2(x-1)(x+3)
x3 + 2x2 + 3x + 6 = x2(x + 2) + 3(x + 2) = (x + 2)(x2 + 3)
(3x + 5)(x2 - 2)
x2+7x=-2x2+6 3x2+7x-6=0 (3x-2)(x+3)=0 3x-2=0 or x+3=0 x=2/3 or x=-3
-2x3-2x2+12x (-2)(x3+x2-6x) (-2x)(x2+x-6) (-2x)(x2+3x-2x-6) (-2x)((x)(x+3)-2(x+3)) (-2x)((x-2)(x+3)
-3/2
-2x3 - 2x2 + 12x = -2x(x2 + x - 6) = -2x(x2 - 2x + 3x - 6) = -2x[ x(x - 2) + 3(x - 2) ] = -2x(x + 3)(x - 2)
assuming x*2-3x+2:factors of -3:-1,3 --> sum: 21,-3 --> sum: -2(x-1)(x+3)
X2 + 3X - 18 = 0What two factors of - 18 add up to 3 ??(X - 3)(X + 6)============so, by the zero sum rule,X = 3X = - 6
(x -3)(2x2 + 3x - 4)
-2x3 + 2x2 + 12x = -2x(x2 - x - 6) = -2x(x2 + 2x - 3x - 6) = -2x[ x(x + 2) - 3(x + 2) ] = -2x(x - 3)(x + 2)
If you mean 2x2 + 3x = (3x - 1)2 - x2 Then you can solve it this way: Expand: 2x2 + 3x = 9x2 - 6x + 1 - x2 Group like terms: 2x2 + 3x = 8x2 - 6x + 1 6x2 - 9x - 1 = 0 Complete the square: 6x2 - 9x = 1 x2 - 3x/2 = 1/6 x2 - 3x/2 + 9/16 = 1/6 + 9/16 (x - 3/4)2 = 8/48 + 27/48 (x - 3/4)2 = 35/48 Solve for x: x - 3/4 = ± √(35/48) x - 3/4 = ± √(35/3) / 4 x = 3/4 ± √(35/3) / 4 x = [3 ± √(35/3)] / 4 If you mean something else, then you'll need to be clearer with your question.
x3 + x2 - 3x - 3 x(x2 + x - 3) - 3