What are the solutions to the equation x^3=32(x+2) ?

The solutions to the equation x^3=32(x+2) are x=-4,x=2(1+sqrt(5)),x=2(1-sqrt(5))

Find the zeros of x^3=32(x+2)

The zeros of x^3=32(x+2) are x=-4,x=2(1+sqrt(5)),x=2(1-sqrt(5))

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x^{3} = 32 (x + 2)

x = - 4, x = 2 (1 + 5), x = 2 (1 - 5)

Decimal Notation

x = - 4, x = 6.47213 \dots, x = - 2.47213 \dots

Solution steps

x^{3} = 32 (x + 2)

Move

32 (x + 2)

to the left side

x^{3} - 32 (x + 2) = 0

Factor

x^{3} - 32 (x + 2) : (x + 4) (x^{2} - 4 x - 16)

(x + 4) (x^{2} - 4 x - 16) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 4 = 0 or x^{2} - 4 x - 16 = 0

Solve

x + 4 = 0 : x = - 4

Solve

x^{2} - 4 x - 16 = 0 : x = 2 (1 + 5), x = 2 (1 - 5)

The solutions are

x = - 4, x = 2 (1 + 5), x = 2 (1 - 5)

x^{3} - 3 x + 4 = 6.1

2 x^{3} - 3 = 0

y^{4} = 36

x^{3} + x^{2} = 49 x + 49

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

What are the solutions to the equation x^3=32(x+2) ?
The solutions to the equation x^3=32(x+2) are x=-4,x=2(1+sqrt(5)),x=2(1-sqrt(5))
Find the zeros of x^3=32(x+2)
The zeros of x^3=32(x+2) are x=-4,x=2(1+sqrt(5)),x=2(1-sqrt(5))