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

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

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

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

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

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

Solution steps

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

Move

32

to the left side

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

Factor

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

(x - 8) (x^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 8 = 0 or x^{2} + 4 = 0

Solve

x - 8 = 0 : x = 8

Solve

x^{2} + 4 = 0 : x = 2 i, x = - 2 i

The solutions are

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

x^{6} + 3 x^{3} - 5 = 0

2 y^{4} - 9 y^{3} + 14 y^{2} + 6 y - 63 = 0

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

5 z^{3} + 22 z^{2} = 15 z

6 + x = x^{3}

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