What are the solutions to the equation 0=2x^3+8x^2+14x+24 ?

The solutions to the equation 0=2x^3+8x^2+14x+24 are x=-3,x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2

Find the zeros of 0=2x^3+8x^2+14x+24

The zeros of 0=2x^3+8x^2+14x+24 are x=-3,x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2

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0 = 2 x^{3} + 8 x^{2} + 14 x + 24

x = - 3, x = - \frac{1}{2} + i \frac{15}{2}, x = - \frac{1}{2} - i \frac{15}{2}

Solution steps

0 = 2 x^{3} + 8 x^{2} + 14 x + 24

Switch sides

2 x^{3} + 8 x^{2} + 14 x + 24 = 0

Factor

2 x^{3} + 8 x^{2} + 14 x + 24 : 2 (x + 3) (x^{2} + x + 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 3 = 0 : x = - 3

Solve

x^{2} + x + 4 = 0 : x = - \frac{1}{2} + i \frac{15}{2}, x = - \frac{1}{2} - i \frac{15}{2}

The solutions are

x = - 3, x = - \frac{1}{2} + i \frac{15}{2}, x = - \frac{1}{2} - i \frac{15}{2}

- 2 x^{5} + 7 x^{2} - 3 = 0

5 = 2 x^{3} - 3.2

3 x^{4} + 15 x^{2} - 72 = 0

(x - 2) (x + 1) (2 x - 7) = 0

y^{4} - 13 y^{2} = - 36

What are the solutions to the equation 0=2x^3+8x^2+14x+24 ?
The solutions to the equation 0=2x^3+8x^2+14x+24 are x=-3,x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2
Find the zeros of 0=2x^3+8x^2+14x+24
The zeros of 0=2x^3+8x^2+14x+24 are x=-3,x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2