What are the solutions to the equation 60x^3-22x^2-4x=0 ?

The solutions to the equation 60x^3-22x^2-4x=0 are x=0,x=-2/15 ,x= 1/2

Find the zeros of 60x^3-22x^2-4x=0

The zeros of 60x^3-22x^2-4x=0 are x=0,x=-2/15 ,x= 1/2

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60 x^{3} - 22 x^{2} - 4 x = 0

x = 0, x = - \frac{2}{15}, x = \frac{1}{2}

Decimal Notation

x = 0, x = - 0.13333 \dots, x = 0.5

Solution steps

60 x^{3} - 22 x^{2} - 4 x = 0

Factor

60 x^{3} - 22 x^{2} - 4 x : 2 x (15 x + 2) (2 x - 1)

2 x (15 x + 2) (2 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 15 x + 2 = 0 or 2 x - 1 = 0

Solve

15 x + 2 = 0 : x = - \frac{2}{15}

Solve

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

The solutions are

x = 0, x = - \frac{2}{15}, x = \frac{1}{2}

x^{3} - 7 x^{2} + 8 x + 10 = 0

x^{4} - 8 x^{9} - 9 = 0

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

x^{3} + x = - 1

10 x^{3} + 3 x - 4 = 0

What are the solutions to the equation 60x^3-22x^2-4x=0 ?
The solutions to the equation 60x^3-22x^2-4x=0 are x=0,x=-2/15 ,x= 1/2
Find the zeros of 60x^3-22x^2-4x=0
The zeros of 60x^3-22x^2-4x=0 are x=0,x=-2/15 ,x= 1/2