What are the solutions to the equation x^3-8x^2-4x+32=0 ?

The solutions to the equation x^3-8x^2-4x+32=0 are x=8,x=-2,x=2

Find the zeros of x^3-8x^2-4x+32=0

The zeros of x^3-8x^2-4x+32=0 are x=8,x=-2,x=2

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x^{3} - 8 x^{2} - 4 x + 32 = 0

x = 8, x = - 2, x = 2

Solution steps

x^{3} - 8 x^{2} - 4 x + 32 = 0

Factor

x^{3} - 8 x^{2} - 4 x + 32 : (x - 8) (x + 2) (x - 2)

(x - 8) (x + 2) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 8 = 0 or x + 2 = 0 or x - 2 = 0

Solve

x - 8 = 0 : x = 8

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

x = 8, x = - 2, x = 2

85 = 13 \cdot (1 + x)^{3}

x^{4} = 3 x^{2}

4 (x + 6)^{4} - 13 (x + 6)^{2} = - 9

15 X^{4} - 445 X^{2} + 8000 = 0

4 x - 4 x^{6} = 0

What are the solutions to the equation x^3-8x^2-4x+32=0 ?
The solutions to the equation x^3-8x^2-4x+32=0 are x=8,x=-2,x=2
Find the zeros of x^3-8x^2-4x+32=0
The zeros of x^3-8x^2-4x+32=0 are x=8,x=-2,x=2