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This is kind of non-specific, but I'm assuming you're looking for the values of x that will make the following true:

x3-2x2-8x = 0

To find these values of x, you must factor the polynomial expression to the left of the equal sign.

x3-2x2-8x = x(x2-2x-8) = x(x-4)(x+2) = 0

By the zero identity, setting each individual factor equal to zero will yield a valid solution to the original equation:

x = 0

x-4 = 0, x = 4

x+2 = 0, x = -2

So the following values of x will solve the equation:

x = 0

x = 4

x = -2

Q: What is x3 -2x2 -8x 0?

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x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0

x3 + 2x2 - 35x = x(x + 7)(x - 5)

2x2 = 8x x = 8x/2x x = 4 Check: 2(4)2 = 8(4) 2(16) = 32 32 = 32

2x2 - 8x + 3 = 0 needs to be solved using the quadratic formula.x = (4 + sqrt(10))/2 or x = (4 - sqrt(10))/2

5x2-8x-4

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x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0

The factor theorem states that for any polynomial function f(x), if f(a) = 0, then (x-a) is a factor of f(x). Let f(x) = x3-2x2-8x-5. If (x+1) is a factor, then f(-1) = 0. (x+1 = x - (-1)) Input x = -1 into f: (-1)3-2(-1)2-8(-1)-5 f(-1) = -1 -2 + 8 - 5 f(-1) = 0. Since f(-1) = 0, (x+1) is a factor of x3-2x2-8x-5. Q.E.D.

x3 + x2 - 8x - 6 = 0 x3 + 3x2 - 2x2 - 6x - 2x - 6 = 0 x2(x + 3) - 2x(x + 3) - 2(x + 3) = 0 (x + 3)(x2 - 2x - 2) = 0 The second bracket cannot be factorised, so applying the quadratic formula to it gives the answers as So x = -3 or x = 1 +/-sqrt(3) that is, x = -3, x = -0.732051 or x = 2.732051

x3 + 2x2 - 35x = x(x + 7)(x - 5)

x3 + 2x2 + 5x + 4 = (x + 1)(x2 + x + 4)

If that's + 2x2, the answer is x(x + 3)(x - 1) x = 0, -3, 1 If that's - 2x2, the answer is x(x - 3)(x + 1) x = 0, 3, -1

x3 - 2x2 + x - 2 =(x - 2)(x2 + 1)

x3 - 2x2 - 25x + 50 = 0

2x2 = 8x x = 8x/2x x = 4 Check: 2(4)2 = 8(4) 2(16) = 32 32 = 32

2x2 - 8x + 3 = 0 needs to be solved using the quadratic formula.x = (4 + sqrt(10))/2 or x = (4 - sqrt(10))/2

2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7

x3 - 2x2 - 4x + 8 = (x2 - 4)(x - 2) = (x + 2)(x - 2)(x - 2)