What are the solutions to the equation n^3-n=4n^2-4 ?

The solutions to the equation n^3-n=4n^2-4 are n=4,n=-1,n=1

Find the zeros of n^3-n=4n^2-4

The zeros of n^3-n=4n^2-4 are n=4,n=-1,n=1

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n^{3} - n = 4 n^{2} - 4

n = 4, n = - 1, n = 1

Solution steps

n^{3} - n = 4 n^{2} - 4

Move

4

to the left side

n^{3} - n + 4 = 4 n^{2}

Move

4 n^{2}

to the left side

n^{3} - n + 4 - 4 n^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

n^{3} - 4 n^{2} - n + 4 = 0

Factor

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

(n - 4) (n + 1) (n - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

n - 4 = 0 or n + 1 = 0 or n - 1 = 0

Solve

n - 4 = 0 : n = 4

Solve

n + 1 = 0 : n = - 1

Solve

n - 1 = 0 : n = 1

The solutions are

n = 4, n = - 1, n = 1

(x^{2} + 2)^{2} = 0

4 x^{3} + 12 x^{2} = x + 3

(6 x^{4}) + (6 x^{1}) = x (4 + 1)

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

x^{6} + 18 x^{3} + 81

What are the solutions to the equation n^3-n=4n^2-4 ?
The solutions to the equation n^3-n=4n^2-4 are n=4,n=-1,n=1
Find the zeros of n^3-n=4n^2-4
The zeros of n^3-n=4n^2-4 are n=4,n=-1,n=1