What are the solutions to the equation 2x^3-3x^2+8x-12=0 ?

The solutions to the equation 2x^3-3x^2+8x-12=0 are x= 3/2 ,x=2i,x=-2i

Find the zeros of 2x^3-3x^2+8x-12=0

The zeros of 2x^3-3x^2+8x-12=0 are x= 3/2 ,x=2i,x=-2i

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2 x^{3} - 3 x^{2} + 8 x - 12 = 0

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

Solution steps

2 x^{3} - 3 x^{2} + 8 x - 12 = 0

Factor

2 x^{3} - 3 x^{2} + 8 x - 12 : (2 x - 3) (x^{2} + 4)

(2 x - 3) (x^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 x - 3 = 0 : x = \frac{3}{2}

Solve

x^{2} + 4 = 0 : x = 2 i, x = - 2 i

The solutions are

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

- 2 (3 x - 1) (x^{2} - 9) = 0

m^{4} - m^{3} - 7 m^{2} + m + 6 = 0

12 k^{3} - 48 k^{2} + 12 k + 72 = 0

x - 3 x^{5} = \frac{1}{4}

x^{2} + x - 43 = 3 x^{4} - 1

What are the solutions to the equation 2x^3-3x^2+8x-12=0 ?
The solutions to the equation 2x^3-3x^2+8x-12=0 are x= 3/2 ,x=2i,x=-2i
Find the zeros of 2x^3-3x^2+8x-12=0
The zeros of 2x^3-3x^2+8x-12=0 are x= 3/2 ,x=2i,x=-2i