4x-6 = -2x-3 4x+2x = 6-3 6x = 3 x = 3/6 = 1/2
2x2 + 7x + 6 = 2x2 + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (x + 2)(2x + 3)
2x-3=4x+9 -2x-12=0 2x=-12 x=-6
6-4x = 7x-9x-2 6-4x = -2x-2 Add 4x to both sides: 6 = 2x-2 Add 2 to both sides: 8 = 2x Divide both sides by 2: 4 = x
It is 2(2x+3) when factorised
4x-6 = -2x-3 4x+2x = 6-3 6x = 3 x = 3/6 = 1/2
As written, that's 2(x - 6) You may have meant 4x^2 - 2x - 12 which would be 2(2x + 3)(x - 2)
-6 + 4x + (9-2x) = -6 + 4x + 9 - 2x = [ 3 + 2x ]
( 8x2 - 14x - 30 ) divided by ( 2x - 6 ) = 4x + 5
Not at all my dear, this is more like an equation where you solve for x, which is 2. 2x-3+(4x+3)=9-4x+(4x+3) 6x/6=12/6 x=2
4x-6=6x. subtract 4x so -6=2x. divide by 2 so -3=x
2(2x + 3)
# ## 2x > -6 and x - 4 < 3 ## x + 3 > 2x + 1 and -4x < -8 ## -6 < x + 3 < 6 ## -3x > -6 or x - 5 < 2 ## x - 2 > 2x + 1 or 10 > -2x + 2
2x2 + 7x + 6 = 2x2 + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (x + 2)(2x + 3)
2 - 4x - 2x + 4 = 6 - 6x
X = 3 4(3) - 2(3) + 3 = 2(3) + 2(3) - 3 12 - 6 + 3 = 9 6 + 6 - 3 = 9
If you mean integral[(2x^2 +4x -3)(x+2)], then multiply them out to get: Integral[2x^3+8x^2+5x-6]. This is then easy to solve and is = 2/4x^4+8/3x^3+5/2x^2-6x +c