What are the solutions to the equation 4x^3+23x^2-2x=0 ?

The solutions to the equation 4x^3+23x^2-2x=0 are x=0,x=(-23+sqrt(561))/8 ,x=(-23-sqrt(561))/8

Find the zeros of 4x^3+23x^2-2x=0

The zeros of 4x^3+23x^2-2x=0 are x=0,x=(-23+sqrt(561))/8 ,x=(-23-sqrt(561))/8

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4 x^{3} + 23 x^{2} - 2 x = 0

x = 0, x = \frac{- 23 + 561}{8}, x = \frac{- 23 - 561}{8}

Decimal Notation

x = 0, x = 0.08567 \dots, x = - 5.83567 \dots

Solution steps

4 x^{3} + 23 x^{2} - 2 x = 0

Factor

4 x^{3} + 23 x^{2} - 2 x : x (4 x^{2} + 23 x - 2)

x (4 x^{2} + 23 x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 4 x^{2} + 23 x - 2 = 0

Solve

4 x^{2} + 23 x - 2 = 0 : x = \frac{- 23 + 561}{8}, x = \frac{- 23 - 561}{8}

The solutions are

x = 0, x = \frac{- 23 + 561}{8}, x = \frac{- 23 - 561}{8}

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

4 x^{4} - 33 x^{2} + 50 = 0

a^{4} - a^{2} = 2

2 x^{3} - 31 = 0

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

What are the solutions to the equation 4x^3+23x^2-2x=0 ?
The solutions to the equation 4x^3+23x^2-2x=0 are x=0,x=(-23+sqrt(561))/8 ,x=(-23-sqrt(561))/8
Find the zeros of 4x^3+23x^2-2x=0
The zeros of 4x^3+23x^2-2x=0 are x=0,x=(-23+sqrt(561))/8 ,x=(-23-sqrt(561))/8