What are the solutions to the equation 12k^3-48k^2+12k+72=0 ?

The solutions to the equation 12k^3-48k^2+12k+72=0 are k=-1,k=2,k=3

Find the zeros of 12k^3-48k^2+12k+72=0

The zeros of 12k^3-48k^2+12k+72=0 are k=-1,k=2,k=3

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12 k^{3} - 48 k^{2} + 12 k + 72 = 0

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

Solution steps

12 k^{3} - 48 k^{2} + 12 k + 72 = 0

Factor

12 k^{3} - 48 k^{2} + 12 k + 72 : 12 (k + 1) (k - 2) (k - 3)

12 (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 x^{5} = \frac{1}{4}

x^{2} + x - 43 = 3 x^{4} - 1

2 x (3 x^{2} - 1) + 5 x = 16 x

x^{4} - 2 x^{3} = 16 + 8 x - 4 x^{3}

4 x^{3} - 4 = 0

What are the solutions to the equation 12k^3-48k^2+12k+72=0 ?
The solutions to the equation 12k^3-48k^2+12k+72=0 are k=-1,k=2,k=3
Find the zeros of 12k^3-48k^2+12k+72=0
The zeros of 12k^3-48k^2+12k+72=0 are k=-1,k=2,k=3