What are the solutions to the equation 12n^2+48n=-n^3-64 ?

The solutions to the equation 12n^2+48n=-n^3-64 are n=-4

Find the zeros of 12n^2+48n=-n^3-64

The zeros of 12n^2+48n=-n^3-64 are n=-4

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12 n^{2} + 48 n = - n^{3} - 64

n = - 4

Solution steps

12 n^{2} + 48 n = - n^{3} - 64

Switch sides

- n^{3} - 64 = 12 n^{2} + 48 n

Move

48 n

to the left side

- n^{3} - 64 - 48 n = 12 n^{2}

Move

12 n^{2}

to the left side

- n^{3} - 64 - 48 n - 12 n^{2} = 0

Write in the standard form

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

- n^{3} - 12 n^{2} - 48 n - 64 = 0

Factor

- n^{3} - 12 n^{2} - 48 n - 64 : - (n + 4)^{3}

- (n + 4)^{3} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

n + 4 = 0

Solve

n + 4 = 0 : n = - 4

The solution is

n = - 4

x^{4} + 12 x^{3} + 54 x^{2} + 108 x + 81

131946.9 x^{3} - 2968.8 x^{1} - 164.8 = 0

4 x^{4} + 35 x^{2} - 9 = 0

4 x^{3} - 12 x^{2} - 36 x = 0

m^{3} - 5 m^{2} + 3 m + 9 = 0

What are the solutions to the equation 12n^2+48n=-n^3-64 ?
The solutions to the equation 12n^2+48n=-n^3-64 are n=-4
Find the zeros of 12n^2+48n=-n^3-64
The zeros of 12n^2+48n=-n^3-64 are n=-4