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

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

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

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

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

k = - 1, k = 2, k = 3

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

k + 1 = 0 : k = - 1

Solve

k - 2 = 0 : k = 2

Solve

k - 3 = 0 : k = 3

The solutions are

k = - 1, k = 2, k = 3

x^{3} - 10 x^{2} + 31 x - 30 = 0

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

9 x^{6} = 324 x^{4}

9 x^{3} + 8 x^{6} = - 1

z^{4} - 1 = 0

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