What are the solutions to the equation z^3-4z^2+z+6=0 ?

The solutions to the equation z^3-4z^2+z+6=0 are z=-1,z=2,z=3

Find the zeros of z^3-4z^2+z+6=0

The zeros of z^3-4z^2+z+6=0 are z=-1,z=2,z=3

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z^{3} - 4 z^{2} + z + 6 = 0

z = - 1, z = 2, z = 3

Solution steps

z^{3} - 4 z^{2} + z + 6 = 0

Factor

z^{3} - 4 z^{2} + z + 6 : (z + 1) (z - 2) (z - 3)

(z + 1) (z - 2) (z - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

z + 1 = 0 : z = - 1

Solve

z - 2 = 0 : z = 2

Solve

z - 3 = 0 : z = 3

The solutions are

z = - 1, z = 2, z = 3

48 x^{3} + 40 x^{2} - x - 3 = 0

y^{9} = - 8

2 x^{3} + 14 x^{2} + 24 x = 0

x, f (x^{3}) = x^{2}

x = x^{4} - 1

What are the solutions to the equation z^3-4z^2+z+6=0 ?
The solutions to the equation z^3-4z^2+z+6=0 are z=-1,z=2,z=3
Find the zeros of z^3-4z^2+z+6=0
The zeros of z^3-4z^2+z+6=0 are z=-1,z=2,z=3