What are the solutions to the equation roots x^3+5x^2+4x+20 ?

The solutions to the equation roots x^3+5x^2+4x+20 are x=-5,x=2i,x=-2i

Find the zeros of roots x^3+5x^2+4x+20

The zeros of roots x^3+5x^2+4x+20 are x=-5,x=2i,x=-2i

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roots

x^{3} + 5 x^{2} + 4 x + 20

x = - 5, x = 2 i, x = - 2 i

Solution steps

x^{3} + 5 x^{2} + 4 x + 20

The roots are the intercepts with the x-axis

(y = 0)

x^{3} + 5 x^{2} + 4 x + 20 = 0

Factor

x^{3} + 5 x^{2} + 4 x + 20 : (x + 5) (x^{2} + 4)

(x + 5) (x^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 5 = 0 or x^{2} + 4 = 0

Solve

x + 5 = 0 : x = - 5

Solve

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

The solutions are

x = - 5, x = 2 i, x = - 2 i

x, x^{3} = a t^{2}

12 x^{3} - 20 x^{2} + x + 3

3 x^{4} + 5 x^{3} - 12 x^{2} - 20 x = 0

5 x^{3} + 29 x^{2} + 19 x - 5 = 0

x^{3} + x - 130 = 0

What are the solutions to the equation roots x^3+5x^2+4x+20 ?
The solutions to the equation roots x^3+5x^2+4x+20 are x=-5,x=2i,x=-2i
Find the zeros of roots x^3+5x^2+4x+20
The zeros of roots x^3+5x^2+4x+20 are x=-5,x=2i,x=-2i