What are the solutions to the equation y^3+2y^2+4y+2=0 ?

The solutions to the equation y^3+2y^2+4y+2=0 are y\approx-0.63889…

Find the zeros of y^3+2y^2+4y+2=0

The zeros of y^3+2y^2+4y+2=0 are y\approx-0.63889…

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y^{3} + 2 y^{2} + 4 y + 2 = 0

y \approx - 0.63889 \dots

Solution steps

y^{3} + 2 y^{2} + 4 y + 2 = 0

Find one solution for

y^{3} + 2 y^{2} + 4 y + 2 = 0

using Newton-Raphson:

y \approx - 0.63889 \dots

Apply long division:

\frac{y ^{3} + 2 y ^{2} + 4 y + 2}{y + 0.63889 \dots} = y^{2} + 1.36110 \dots y + 3.13039 \dots

y^{2} + 1.36110 \dots y + 3.13039 \dots \approx 0

Find one solution for

y^{2} + 1.36110 \dots y + 3.13039 \dots = 0

using Newton-Raphson:

No Solution for

y \in R

The solution is

y \approx - 0.63889 \dots

18 x^{5} - 21 x^{4} - 9 x^{3} = 0

2 x^{4} + 8 x^{2} + 4 = 0

99 = c^{3}

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

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

What are the solutions to the equation y^3+2y^2+4y+2=0 ?
The solutions to the equation y^3+2y^2+4y+2=0 are y\approx-0.63889…
Find the zeros of y^3+2y^2+4y+2=0
The zeros of y^3+2y^2+4y+2=0 are y\approx-0.63889…