What are the solutions to the equation 2x^3-3x^2+32x+17=0 ?

The solutions to the equation 2x^3-3x^2+32x+17=0 are x=-1/2 ,x=1+4i,x=1-4i

Find the zeros of 2x^3-3x^2+32x+17=0

The zeros of 2x^3-3x^2+32x+17=0 are x=-1/2 ,x=1+4i,x=1-4i

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2 x^{3} - 3 x^{2} + 32 x + 17 = 0

x = - \frac{1}{2}, x = 1 + 4 i, x = 1 - 4 i

Solution steps

2 x^{3} - 3 x^{2} + 32 x + 17 = 0

Factor

2 x^{3} - 3 x^{2} + 32 x + 17 : (2 x + 1) (x^{2} - 2 x + 17)

(2 x + 1) (x^{2} - 2 x + 17) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x + 1 = 0 or x^{2} - 2 x + 17 = 0

Solve

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

Solve

x^{2} - 2 x + 17 = 0 : x = 1 + 4 i, x = 1 - 4 i

The solutions are

x = - \frac{1}{2}, x = 1 + 4 i, x = 1 - 4 i

x^{2} (x - 8) = 0

b^{3} = b

- x^{3} + x^{2} - x + 1 = 0

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

8 x^{3} - 50 x = 0

What are the solutions to the equation 2x^3-3x^2+32x+17=0 ?
The solutions to the equation 2x^3-3x^2+32x+17=0 are x=-1/2 ,x=1+4i,x=1-4i
Find the zeros of 2x^3-3x^2+32x+17=0
The zeros of 2x^3-3x^2+32x+17=0 are x=-1/2 ,x=1+4i,x=1-4i