What are the solutions to the equation (2x+1)^3=-8 ?

The solutions to the equation (2x+1)^3=-8 are x=-3/2 ,x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i

Find the zeros of (2x+1)^3=-8

The zeros of (2x+1)^3=-8 are x=-3/2 ,x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i

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(2 x + 1)^{3} = - 8

x = - \frac{3}{2}, x = \frac{3}{2} i, x = - \frac{3}{2} i

Solution steps

(2 x + 1)^{3} = - 8

Expand

(2 x + 1)^{3} : 8 x^{3} + 12 x^{2} + 6 x + 1

8 x^{3} + 12 x^{2} + 6 x + 1 = - 8

Move

8

to the left side

8 x^{3} + 12 x^{2} + 6 x + 9 = 0

Factor

8 x^{3} + 12 x^{2} + 6 x + 9 : (2 x + 3) (4 x^{2} + 3)

(2 x + 3) (4 x^{2} + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 x + 3 = 0 : x = - \frac{3}{2}

Solve

4 x^{2} + 3 = 0 : x = \frac{3}{2} i, x = - \frac{3}{2} i

The solutions are

x = - \frac{3}{2}, x = \frac{3}{2} i, x = - \frac{3}{2} i

n^{3} = (n)^{2}

32 m^{3} - 5 m^{2} + 26 m - 1 = 0

a^{4} - 14 a^{2} + 24 = 0

15 x^{3} + 27 x^{2} + 3 x - 3 = 0

(\frac{x + 1}{2})^{3} = \frac{- 1}{8}

What are the solutions to the equation (2x+1)^3=-8 ?
The solutions to the equation (2x+1)^3=-8 are x=-3/2 ,x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i
Find the zeros of (2x+1)^3=-8
The zeros of (2x+1)^3=-8 are x=-3/2 ,x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i