What are the solutions to the equation 6x^3+7x^2-1=0 ?

The solutions to the equation 6x^3+7x^2-1=0 are x=-1,x= 1/3 ,x=-1/2

Find the zeros of 6x^3+7x^2-1=0

The zeros of 6x^3+7x^2-1=0 are x=-1,x= 1/3 ,x=-1/2

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6 x^{3} + 7 x^{2} - 1 = 0

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

Decimal Notation

x = - 1, x = 0.33333 \dots, x = - 0.5

Solution steps

6 x^{3} + 7 x^{2} - 1 = 0

Factor

6 x^{3} + 7 x^{2} - 1 : (x + 1) (3 x - 1) (2 x + 1)

(x + 1) (3 x - 1) (2 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or 3 x - 1 = 0 or 2 x + 1 = 0

Solve

x + 1 = 0 : x = - 1

Solve

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

Solve

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

The solutions are

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

x^{3} + x^{2} + 2 x + 5 = 0

b x^{3} = 15

x^{5} - 2 x^{4} - 5 x^{3} + 6 x^{2} = 0

x^{3} - 3 x^{2} - 16 x - 12 = 0

4 x^{3} + 3 x^{2} + 64 x + 48 = 0

What are the solutions to the equation 6x^3+7x^2-1=0 ?
The solutions to the equation 6x^3+7x^2-1=0 are x=-1,x= 1/3 ,x=-1/2
Find the zeros of 6x^3+7x^2-1=0
The zeros of 6x^3+7x^2-1=0 are x=-1,x= 1/3 ,x=-1/2