What are the solutions to the equation z^4+4z^3+2z^2-12z-15=0 ?

The solutions to the equation z^4+4z^3+2z^2-12z-15=0 are z=-sqrt(3),z=sqrt(3)

Find the zeros of z^4+4z^3+2z^2-12z-15=0

The zeros of z^4+4z^3+2z^2-12z-15=0 are z=-sqrt(3),z=sqrt(3)

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z^{4} + 4 z^{3} + 2 z^{2} - 12 z - 15 = 0

z = - 3, z = 3

Solution steps

z^{4} + 4 z^{3} + 2 z^{2} - 12 z - 15 = 0

Find one solution for

z^{4} + 4 z^{3} + 2 z^{2} - 12 z - 15 = 0

using Newton-Raphson:

z \approx - 1.73205 \dots

Apply long division:

\frac{z ^{4} + 4 z ^{3} + 2 z ^{2} - 12 z - 15}{z + 1.73205 \dots} = z^{3} + 2.26794 \dots z^{2} - 1.92820 \dots z - 8.66025 \dots

z^{3} + 2.26794 \dots z^{2} - 1.92820 \dots z - 8.66025 \dots \approx 0

Find one solution for

z^{3} + 2.26794 \dots z^{2} - 1.92820 \dots z - 8.66025 \dots = 0

using Newton-Raphson:

z \approx 1.73205 \dots

Apply long division:

\frac{z ^{3} + 2.26794 \dots z ^{2} - 1.92820 \dots z - 8.66025 \dots}{z - 1.73205 \dots} = z^{2} + 4 z + 5

z^{2} + 4 z + 5 \approx 0

Find one solution for

z^{2} + 4 z + 5 = 0

using Newton-Raphson:

No Solution for

z \in R

Approximate to algebraic form

z \approx - 1.73205 \dots : z = - 3

z \approx 1.73205 \dots : z = 3

z = - 3, z = 3

The solutions are

z = - 3, z = 3

3 a^{3} - 147 a = - 2 a^{2} + 98

695 = 5 x^{3}

z^{3} + 1 - i = 0

2 x^{5} + 5 x = 2215

12 x^{3} - 26 x^{2} - 12 x = 0

What are the solutions to the equation z^4+4z^3+2z^2-12z-15=0 ?
The solutions to the equation z^4+4z^3+2z^2-12z-15=0 are z=-sqrt(3),z=sqrt(3)
Find the zeros of z^4+4z^3+2z^2-12z-15=0
The zeros of z^4+4z^3+2z^2-12z-15=0 are z=-sqrt(3),z=sqrt(3)