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

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

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

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

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

x = - 4

Solution steps

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

Switch sides

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

Move

48 x

to the left side

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

Move

12 x^{2}

to the left side

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

Write in the standard form

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

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 4 = 0

Solve

x + 4 = 0 : x = - 4

The solution is

x = - 4

0.75 = 10 x^{9} - 9 x^{10}

x^{4} - 2 x^{2} + 6 = 0

4 x^{3} - 9 x^{2} + 4 = 0

a^{6} = 30

- x^{3} + 2 x^{2} + 7 x + 4 = 0

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