What are the solutions to the equation y^4+4y^3+3y^2+12y+36=0 ?

The solutions to the equation y^4+4y^3+3y^2+12y+36=0 are y=-3,y\approx-2.67588…

Find the zeros of y^4+4y^3+3y^2+12y+36=0

The zeros of y^4+4y^3+3y^2+12y+36=0 are y=-3,y\approx-2.67588…

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y^{4} + 4 y^{3} + 3 y^{2} + 12 y + 36 = 0

y = - 3, y \approx - 2.67588 \dots

Solution steps

y^{4} + 4 y^{3} + 3 y^{2} + 12 y + 36 = 0

Factor

y^{4} + 4 y^{3} + 3 y^{2} + 12 y + 36 : (y + 3) (y^{3} + y^{2} + 12)

(y + 3) (y^{3} + y^{2} + 12) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y + 3 = 0 or y^{3} + y^{2} + 12 = 0

Solve

y + 3 = 0 : y = - 3

Solve

y^{3} + y^{2} + 12 = 0 : y \approx - 2.67588 \dots

The solutions are

y = - 3, y \approx - 2.67588 \dots

x^{2} = x^{4}

2 x^{3} + x^{2} - 8 x - 4 = 0

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

- 4 x^{3} - 12 x^{2} + 7 = 0

2 x^{2} - 3 x + 2 = x^{3} + 4 x + 4 - 4

What are the solutions to the equation y^4+4y^3+3y^2+12y+36=0 ?
The solutions to the equation y^4+4y^3+3y^2+12y+36=0 are y=-3,y\approx-2.67588…
Find the zeros of y^4+4y^3+3y^2+12y+36=0
The zeros of y^4+4y^3+3y^2+12y+36=0 are y=-3,y\approx-2.67588…