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

The solutions to the equation k^3-4k^2-2k+8=0 are k=4,k=-sqrt(2),k=sqrt(2)

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

The zeros of k^3-4k^2-2k+8=0 are k=4,k=-sqrt(2),k=sqrt(2)

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

k = 4, k = - 2, k = 2

Decimal Notation

k = 4, k = - 1.41421 \dots, k = 1.41421 \dots

Solution steps

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

Factor

k^{3} - 4 k^{2} - 2 k + 8 : (k - 4) (k + 2) (k - 2)

(k - 4) (k + 2) (k - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

k - 4 = 0 or k + 2 = 0 or k - 2 = 0

Solve

k - 4 = 0 : k = 4

Solve

k + 2 = 0 : k = - 2

Solve

k - 2 = 0 : k = 2

The solutions are

k = 4, k = - 2, k = 2

x^{3} + 9 x^{2} + 27 x + 19 = 0

x^{3} = 3 x^{2}

2 x^{3} - 24 x^{2} + 64 x = 0

x^{3} + 4 x^{2} + 8 x + 10 = 0

x^{4} - 20 x^{2} + 99 = 0

What are the solutions to the equation k^3-4k^2-2k+8=0 ?
The solutions to the equation k^3-4k^2-2k+8=0 are k=4,k=-sqrt(2),k=sqrt(2)
Find the zeros of k^3-4k^2-2k+8=0
The zeros of k^3-4k^2-2k+8=0 are k=4,k=-sqrt(2),k=sqrt(2)