What are the solutions to the equation 12+x^3=4x^2+12 ?

The solutions to the equation 12+x^3=4x^2+12 are x=0,x=4

Find the zeros of 12+x^3=4x^2+12

The zeros of 12+x^3=4x^2+12 are x=0,x=4

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

x = 0, x = 4

Solution steps

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

Move

12

to the right side

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

Move

4 x^{2}

to the left side

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

Factor

x^{3} - 4 x^{2} : x^{2} (x - 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 4 = 0

Solve

x - 4 = 0 : x = 4

The solutions are

x = 0, x = 4

x^{6} = 1 0^{24}

7 y^{3} + 25 y^{2} = 12 y

8 x - 6 x^{2} - 2 x^{3} = 0

(x + 3) (x - 1) (x - 2) = 96

a^{3} = 0

What are the solutions to the equation 12+x^3=4x^2+12 ?
The solutions to the equation 12+x^3=4x^2+12 are x=0,x=4
Find the zeros of 12+x^3=4x^2+12
The zeros of 12+x^3=4x^2+12 are x=0,x=4