What are the solutions to the equation (z+1)^4-4(z+1)^2+3=0 ?

The solutions to the equation (z+1)^4-4(z+1)^2+3=0 are z=0,z=-2,z=-1+sqrt(3),z=-1-sqrt(3)

Find the zeros of (z+1)^4-4(z+1)^2+3=0

The zeros of (z+1)^4-4(z+1)^2+3=0 are z=0,z=-2,z=-1+sqrt(3),z=-1-sqrt(3)

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

z = 0, z = - 2, z = - 1 + 3, z = - 1 - 3

Decimal Notation

z = 0, z = - 2, z = 0.73205 \dots, z = - 2.73205 \dots

Solution steps

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

Expand

(z + 1)^{4} - 4 (z + 1)^{2} + 3 : z^{4} + 4 z^{3} + 2 z^{2} - 4 z

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

Factor

z^{4} + 4 z^{3} + 2 z^{2} - 4 z : z (z + 2) (z^{2} + 2 z - 2)

z (z + 2) (z^{2} + 2 z - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

z = 0 or z + 2 = 0 or z^{2} + 2 z - 2 = 0

Solve

z + 2 = 0 : z = - 2

Solve

z^{2} + 2 z - 2 = 0 : z = - 1 + 3, z = - 1 - 3

The solutions are

z = 0, z = - 2, z = - 1 + 3, z = - 1 - 3

x^{5} = 625

x^{3} + 6 x^{2} - 27 x = 0

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

(x + 2)^{3} = 1

64 y^{3} - 7 = y - 448 y^{2}

What are the solutions to the equation (z+1)^4-4(z+1)^2+3=0 ?
The solutions to the equation (z+1)^4-4(z+1)^2+3=0 are z=0,z=-2,z=-1+sqrt(3),z=-1-sqrt(3)
Find the zeros of (z+1)^4-4(z+1)^2+3=0
The zeros of (z+1)^4-4(z+1)^2+3=0 are z=0,z=-2,z=-1+sqrt(3),z=-1-sqrt(3)