What are the solutions to the equation k^4-3k^3-3k^2+5k=0 ?

The solutions to the equation k^4-3k^3-3k^2+5k=0 are k=0,k=1,k=1+sqrt(6),k=1-sqrt(6)

Find the zeros of k^4-3k^3-3k^2+5k=0

The zeros of k^4-3k^3-3k^2+5k=0 are k=0,k=1,k=1+sqrt(6),k=1-sqrt(6)

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k^{4} - 3 k^{3} - 3 k^{2} + 5 k = 0

k = 0, k = 1, k = 1 + 6, k = 1 - 6

Decimal Notation

k = 0, k = 1, k = 3.44948 \dots, k = - 1.44948 \dots

Solution steps

k^{4} - 3 k^{3} - 3 k^{2} + 5 k = 0

Factor

k^{4} - 3 k^{3} - 3 k^{2} + 5 k : k (k - 1) (k^{2} - 2 k - 5)

k (k - 1) (k^{2} - 2 k - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

k = 0 or k - 1 = 0 or k^{2} - 2 k - 5 = 0

Solve

k - 1 = 0 : k = 1

Solve

k^{2} - 2 k - 5 = 0 : k = 1 + 6, k = 1 - 6

The solutions are

k = 0, k = 1, k = 1 + 6, k = 1 - 6

4 x (x^{2} - 2) = 0

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

(x + 1)^{4} = 0

49 x^{4} - 175 x^{2} + 144 = 0

12 x^{3} - 20 x^{2} - 5 x + 1 = 0

What are the solutions to the equation k^4-3k^3-3k^2+5k=0 ?
The solutions to the equation k^4-3k^3-3k^2+5k=0 are k=0,k=1,k=1+sqrt(6),k=1-sqrt(6)
Find the zeros of k^4-3k^3-3k^2+5k=0
The zeros of k^4-3k^3-3k^2+5k=0 are k=0,k=1,k=1+sqrt(6),k=1-sqrt(6)