X cubed divided by x-2 is . . .?
since you dont have the value for x I will assign a random number if x=2 then you will have 2^3/2^-2 2*2*2/(1/2^2) 6/(1/2*2) 6/(1/4) 6/.25 24
What is x if x cubed equals 1?
If x^3 = 1, then taking the cube root of both sides gives you x = 1. If you're only looking for real solutions, then stop here. If you're looking for complex solutions, then that's another story. To find the other two solutions, start with x^3 - 1 = 0 and factor as the difference of two cubes: (x - 1) (x^2 + x + 1) = 0 Use the quadratric formula to solve x^2 + x + 1 = 0, which gives you your two complex solutions.
How do you factor X Cubed + 8?
a³ + b³ = (a + b)(a² -ab + b²) Let a = x and b = 2 => x³ + 2³ = (x + 2)(x² - 2x + 2²) => x³ + 8 = (x + 2)(x² - 2x + 4)
Factor x cubed plus 8 over x plus 2?
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How do you factor x cubed minus 8?
1) You might notice that if x = 2, x^3 = 8, so x^3 - 8 = 0 That should tell you that (x - 2) is a factor As this was a cubic the other factor is a quadratic, (which might or might not itself give linear factors). The first and last terms of the quadratic are obvious but we do not yet know the coefficient of the x term. (x - 2)(x^2 + kx + 4) = x^3 - 8 Equating coefficients of x -2k + 4 = 0 k = 2, so now we have x^3 - 8 = (x - 2)(x^2 + 2x + 4) 2. 36x^2 - 16 = 4(9x^2 - 4) Then using a "difference of two squares" 36x^2 - 16 = 4 (3x - 2)(3x + 2) Regards - Ian
What is (x+2)cubed? idk how to foil it three times?
First multiply (x+2)(x+2) = (x^2 + 4x + 4)(x+2) = x^3 + 2x^2 + 4x^2 + 8x + 4x + 8 x^3 + 6x^2 + 12x + 8
How do you factor x cubed minus 3x minus 2?
If you tried to solve x³ - 3x - 2 = 0, then x = -1 is a solution. (use rational root test to get possible choices to check) If x = -1, then x + 1 = 0, so (x + 1) must be a factor. (by factor theorem) so divide x³ - 3x - 2 by (x+1) to get x² - x - 2 so, x³ - 3x - 2 = (x+1)(x² - x - 2) now factor x² - x - 2 to get x³ - 3x - 2 = (x+1)(x - 2)(x + 1)
How do you solve x(cube)-8=0? and what are the answers to this equation?
x = 2 2x2x2 - 8 = 0 to solve: x**3 - 8 = 0 x**3 = 8 cubed root of 8 = 2 -Dio
If x = 1 and x = 2 are zeroes of polynomial x cubed + ax squared +bx+c and a+b=1, then find the value of b?
The answer is a= -4 , b= -5 . You simply have to solve two equations that's it. See the picture