What are the solutions to the equation roots 2x^3-3x^2-11x+6 ?

The solutions to the equation roots 2x^3-3x^2-11x+6 are x=-2,x= 1/2 ,x=3

Find the zeros of roots 2x^3-3x^2-11x+6

The zeros of roots 2x^3-3x^2-11x+6 are x=-2,x= 1/2 ,x=3

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roots

2 x^{3} - 3 x^{2} - 11 x + 6

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

Solution steps

2 x^{3} - 3 x^{2} - 11 x + 6

The roots are the intercepts with the x-axis

(y = 0)

2 x^{3} - 3 x^{2} - 11 x + 6 = 0

Factor

2 x^{3} - 3 x^{2} - 11 x + 6 : (x + 2) (2 x - 1) (x - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 = 0 : x = - 2

Solve

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

Solve

x - 3 = 0 : x = 3

The solutions are

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

x^{4} = 5 x + 6

x^{3} + 8 x^{2} - 36 x = 288

x^{3} - 9 x^{2} + 27 x = 20

4 x^{4} - 14 x^{2} + 12 = 0

x^{3} - 12 x^{2} + 256 = 0

What are the solutions to the equation roots 2x^3-3x^2-11x+6 ?
The solutions to the equation roots 2x^3-3x^2-11x+6 are x=-2,x= 1/2 ,x=3
Find the zeros of roots 2x^3-3x^2-11x+6
The zeros of roots 2x^3-3x^2-11x+6 are x=-2,x= 1/2 ,x=3