What are the solutions to the equation 12(k(k^2-4k+1)+6)=0 ?

The solutions to the equation 12(k(k^2-4k+1)+6)=0 are k=-1,k=2,k=3

Find the zeros of 12(k(k^2-4k+1)+6)=0

The zeros of 12(k(k^2-4k+1)+6)=0 are k=-1,k=2,k=3

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12 (k (k^{2} - 4 k + 1) + 6) = 0

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

Solution steps

12 (k (k^{2} - 4 k + 1) + 6) = 0

Expand

12 (k (k^{2} - 4 k + 1) + 6) : 12 k^{3} - 48 k^{2} + 12 k + 72

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

Find one solution for

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

using Newton-Raphson:

k \approx - 1.00000 \dots

Apply long division:

\frac{12 k ^{3} - 48 k ^{2} + 12 k + 72}{k + 1} = 12 k^{2} - 60 k + 72

12 k^{2} - 60 k + 72 \approx 0

Find one solution for

12 k^{2} - 60 k + 72 = 0

using Newton-Raphson:

k \approx 2.00000 \dots

Apply long division:

\frac{12 k ^{2} - 60 k + 72}{k - 2} = 12 k - 36

12 k - 36 \approx 0

k \approx 3

Approximate to algebraic form

k = 3

Approximate to algebraic form

k \approx - 1.00000 \dots : k = - 1

k \approx 2.00000 \dots : k = 2

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

The solutions are

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

x^{3} - 5 x^{2} + 14 x = 0

x^{3} + 3 x^{2} - 3 x - 1 = 0

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

(x - 5) (x^{2} - 4) = 0

x^{3} - 9 x^{2} - 81 x = - 729

What are the solutions to the equation 12(k(k^2-4k+1)+6)=0 ?
The solutions to the equation 12(k(k^2-4k+1)+6)=0 are k=-1,k=2,k=3
Find the zeros of 12(k(k^2-4k+1)+6)=0
The zeros of 12(k(k^2-4k+1)+6)=0 are k=-1,k=2,k=3