What are the solutions to the equation 12x^3+4x^2+3x+1=0 ?

The solutions to the equation 12x^3+4x^2+3x+1=0 are x=-1/3 ,x= 1/2 i,x=-1/2 i

Find the zeros of 12x^3+4x^2+3x+1=0

The zeros of 12x^3+4x^2+3x+1=0 are x=-1/3 ,x= 1/2 i,x=-1/2 i

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12 x^{3} + 4 x^{2} + 3 x + 1 = 0

x = - \frac{1}{3}, x = \frac{1}{2} i, x = - \frac{1}{2} i

Solution steps

12 x^{3} + 4 x^{2} + 3 x + 1 = 0

Factor

12 x^{3} + 4 x^{2} + 3 x + 1 : (3 x + 1) (4 x^{2} + 1)

(3 x + 1) (4 x^{2} + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x + 1 = 0 or 4 x^{2} + 1 = 0

Solve

3 x + 1 = 0 : x = - \frac{1}{3}

Solve

4 x^{2} + 1 = 0 : x = \frac{1}{2} i, x = - \frac{1}{2} i

The solutions are

x = - \frac{1}{3}, x = \frac{1}{2} i, x = - \frac{1}{2} i

3 = x^{3} + 2 x

2 x^{3} - x^{2} + 2 x - 1

6 x^{3} + 36 x - 12 = 0

x^{4} = x^{5}

(2 x + 1)^{3} = 900

What are the solutions to the equation 12x^3+4x^2+3x+1=0 ?
The solutions to the equation 12x^3+4x^2+3x+1=0 are x=-1/3 ,x= 1/2 i,x=-1/2 i
Find the zeros of 12x^3+4x^2+3x+1=0
The zeros of 12x^3+4x^2+3x+1=0 are x=-1/3 ,x= 1/2 i,x=-1/2 i