What are the solutions to the equation y^4+2y^3+y^2+8y-12=0 ?

The solutions to the equation y^4+2y^3+y^2+8y-12=0 are y=1,y=-3,y=2i,y=-2i

Find the zeros of y^4+2y^3+y^2+8y-12=0

The zeros of y^4+2y^3+y^2+8y-12=0 are y=1,y=-3,y=2i,y=-2i

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y^{4} + 2 y^{3} + y^{2} + 8 y - 12 = 0

y = 1, y = - 3, y = 2 i, y = - 2 i

Solution steps

y^{4} + 2 y^{3} + y^{2} + 8 y - 12 = 0

Factor

y^{4} + 2 y^{3} + y^{2} + 8 y - 12 : (y - 1) (y + 3) (y^{2} + 4)

(y - 1) (y + 3) (y^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y - 1 = 0 or y + 3 = 0 or y^{2} + 4 = 0

Solve

y - 1 = 0 : y = 1

Solve

y + 3 = 0 : y = - 3

Solve

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

The solutions are

y = 1, y = - 3, y = 2 i, y = - 2 i

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

2 x^{3} x^{3} = 1 8^{2}

x^{3} - x^{2} - 36 x + 36 = 0

\frac{2}{3} x^{3} = 1

x^{4} + x^{3} - 11 x^{2} - 9 x + 18

What are the solutions to the equation y^4+2y^3+y^2+8y-12=0 ?
The solutions to the equation y^4+2y^3+y^2+8y-12=0 are y=1,y=-3,y=2i,y=-2i
Find the zeros of y^4+2y^3+y^2+8y-12=0
The zeros of y^4+2y^3+y^2+8y-12=0 are y=1,y=-3,y=2i,y=-2i