What are the solutions to the equation 2x^3-7x^2+12x+8=0 ?

The solutions to the equation 2x^3-7x^2+12x+8=0 are x=-1/2 ,x=2+2i,x=2-2i

Find the zeros of 2x^3-7x^2+12x+8=0

The zeros of 2x^3-7x^2+12x+8=0 are x=-1/2 ,x=2+2i,x=2-2i

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2 x^{3} - 7 x^{2} + 12 x + 8 = 0

x = - \frac{1}{2}, x = 2 + 2 i, x = 2 - 2 i

Solution steps

2 x^{3} - 7 x^{2} + 12 x + 8 = 0

Factor

2 x^{3} - 7 x^{2} + 12 x + 8 : (2 x + 1) (x^{2} - 4 x + 8)

(2 x + 1) (x^{2} - 4 x + 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x + 1 = 0 or x^{2} - 4 x + 8 = 0

Solve

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

Solve

x^{2} - 4 x + 8 = 0 : x = 2 + 2 i, x = 2 - 2 i

The solutions are

x = - \frac{1}{2}, x = 2 + 2 i, x = 2 - 2 i

k^{3} = 64

m^{3} + 2 m^{2} + 5 m + 6 = 0

(- x + 2) (x^{2} - 1) = 0

2 x^{3} - 11 x^{2} + 17 x - 6 = 0

x (2 x - 3) - (2 x + 6) (3 - 2 x)^{2} = 0

What are the solutions to the equation 2x^3-7x^2+12x+8=0 ?
The solutions to the equation 2x^3-7x^2+12x+8=0 are x=-1/2 ,x=2+2i,x=2-2i
Find the zeros of 2x^3-7x^2+12x+8=0
The zeros of 2x^3-7x^2+12x+8=0 are x=-1/2 ,x=2+2i,x=2-2i