What are the solutions to the equation 8x^3+12x^2+6x+1=0 ?

The solutions to the equation 8x^3+12x^2+6x+1=0 are x=-1/2

Find the zeros of 8x^3+12x^2+6x+1=0

The zeros of 8x^3+12x^2+6x+1=0 are x=-1/2

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8 x^{3} + 12 x^{2} + 6 x + 1 = 0

x = - \frac{1}{2}

Decimal Notation

x = - 0.5

Solution steps

8 x^{3} + 12 x^{2} + 6 x + 1 = 0

Factor

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

(2 x + 1)^{3} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x + 1 = 0

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

The solution is

x = - \frac{1}{2}

x, a = (2 x^{5} \cdot y^{3})^{3}

x^{3} - 4 x^{2} - 20 x - 35 = 0

(x - 8)^{3} = 0

x^{4} - 5 x^{3} + 8 x^{2} + 1 = 0

4 x^{4} - 13 x^{2} + 3 = 0

What are the solutions to the equation 8x^3+12x^2+6x+1=0 ?
The solutions to the equation 8x^3+12x^2+6x+1=0 are x=-1/2
Find the zeros of 8x^3+12x^2+6x+1=0
The zeros of 8x^3+12x^2+6x+1=0 are x=-1/2