What are the solutions to the equation -x^3+8x^2-4x-48=0 ?

The solutions to the equation -x^3+8x^2-4x-48=0 are x=-2,x=4,x=6

Find the zeros of -x^3+8x^2-4x-48=0

The zeros of -x^3+8x^2-4x-48=0 are x=-2,x=4,x=6

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- x^{3} + 8 x^{2} - 4 x - 48 = 0

x = - 2, x = 4, x = 6

Solution steps

- x^{3} + 8 x^{2} - 4 x - 48 = 0

Factor

- x^{3} + 8 x^{2} - 4 x - 48 : - (x + 2) (x - 4) (x - 6)

- (x + 2) (x - 4) (x - 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x - 4 = 0 or x - 6 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x - 4 = 0 : x = 4

Solve

x - 6 = 0 : x = 6

The solutions are

x = - 2, x = 4, x = 6

12 = + x x^{3}

m^{3} = 5

z^{4} = - 1 + i 3

x^{3} - 4 x^{2} - 53 x + 168 = 0

x^{3} - 3 x = 1

What are the solutions to the equation -x^3+8x^2-4x-48=0 ?
The solutions to the equation -x^3+8x^2-4x-48=0 are x=-2,x=4,x=6
Find the zeros of -x^3+8x^2-4x-48=0
The zeros of -x^3+8x^2-4x-48=0 are x=-2,x=4,x=6