What are the solutions to the equation 2x^3-3x^2-23x+12=0 ?

The solutions to the equation 2x^3-3x^2-23x+12=0 are x=-3,x= 1/2 ,x=4

Find the zeros of 2x^3-3x^2-23x+12=0

The zeros of 2x^3-3x^2-23x+12=0 are x=-3,x= 1/2 ,x=4

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2 x^{3} - 3 x^{2} - 23 x + 12 = 0

x = - 3, x = \frac{1}{2}, x = 4

Decimal Notation

x = - 3, x = 0.5, x = 4

Solution steps

2 x^{3} - 3 x^{2} - 23 x + 12 = 0

Factor

2 x^{3} - 3 x^{2} - 23 x + 12 : (x + 3) (2 x - 1) (x - 4)

(x + 3) (2 x - 1) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 3 = 0 or 2 x - 1 = 0 or x - 4 = 0

Solve

x + 3 = 0 : x = - 3

Solve

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

Solve

x - 4 = 0 : x = 4

The solutions are

x = - 3, x = \frac{1}{2}, x = 4

4 = x^{5} + 2 x + 1

t, M = \frac{5}{8} π t^{5}

\frac{x ^{3} - 2}{2} = 11

x, - y = 2 (- x)^{3}

x^{2} (x + 5) - 12 (x + 3) = 0

What are the solutions to the equation 2x^3-3x^2-23x+12=0 ?
The solutions to the equation 2x^3-3x^2-23x+12=0 are x=-3,x= 1/2 ,x=4
Find the zeros of 2x^3-3x^2-23x+12=0
The zeros of 2x^3-3x^2-23x+12=0 are x=-3,x= 1/2 ,x=4