3x^2+3x-60=3(x^2+x+20)=3(x+5)(x-4)
It is: (3x+4)(2x-3) when factored
(x+4)(x-7)
Your factors must begin 3x and x to give 3x2. The factors of 12 are 1/12, 2/6 and 3/4. Which pair, together with 3x and x, can add to -5? 3 and 4 seem to fit the bill: (3x + 4)(x - 3)
3x2 - 11x - 4 = 0(3x + 1)(x - 4)x = -1/3 and 4
3x2 - 11x - 4 = (x-4)(3 x + 1)
3x^2+3x-60=3(x^2+x+20)=3(x+5)(x-4)
It is: (3x+4)(2x-3) when factored
(x+4)(x-7)
3x2 -14x -24 can be factored as (3x + 4)(x - 6).
Your factors must begin 3x and x to give 3x2. The factors of 12 are 1/12, 2/6 and 3/4. Which pair, together with 3x and x, can add to -5? 3 and 4 seem to fit the bill: (3x + 4)(x - 3)
Yes, but not in rational terms. The factorisation is: {x - [3-sqrt(57)]/6} * {x - [3+sqrt(57)]/6}
The factors are -1(2x + 1)(3x + 4)
For any x ≠ 0, x2 -10/x2 - 4 +3x/x2 - 4 LCD = x2, multiply each term by their missing element of LCD = (x4 + 10 +3x - 8x2)/x = (x4 - 8x2 + 3x + 10)/x2
3x2 - 11x - 4 = 0(3x + 1)(x - 4)x = -1/3 and 4
Let square = [x] -4+3*(-2)*[-4]+[3x-2x]=-4-6-2+[x]=-12+[x]The answer is negative 12 plus squared x.
The minimum value of the parabola is at the point (-1/3, -4/3)