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

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

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

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

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

x = 3, x = - 2, x = 2

Solution steps

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

Move

4 x

to the left side

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

Move

3 x^{2}

to the left side

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

Write in the standard form

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

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

Factor

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

(x - 3) (x + 2) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 3 = 0 or x + 2 = 0 or x - 2 = 0

Solve

x - 3 = 0 : x = 3

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

x = 3, x = - 2, x = 2

21494 = 35000 (1 + r)^{3}

x^{4} - 6 x^{2} - 55 = 0

4 x^{3} = 56

b^{4} = 625

66 = 8 x^{3}

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