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

The solutions to the equation 3x^3+4=x^2+12x are x= 1/3 ,x=-2,x=2

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

The zeros of 3x^3+4=x^2+12x are x= 1/3 ,x=-2,x=2

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

x = \frac{1}{3}, x = - 2, x = 2

Decimal Notation

x = 0.33333 \dots, x = - 2, x = 2

Solution steps

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

Move

12 x

to the left side

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

Move

x^{2}

to the left side

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

Write in the standard form

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

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

Factor

3 x^{3} - x^{2} - 12 x + 4 : (3 x - 1) (x + 2) (x - 2)

(3 x - 1) (x + 2) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x - 1 = 0 or x + 2 = 0 or x - 2 = 0

Solve

3 x - 1 = 0 : x = \frac{1}{3}

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

x = \frac{1}{3}, x = - 2, x = 2

8 x^{3} - 6 x - 1 = 0

x^{4} = 64

3 a^{5} - a^{2} = 120

2 x^{3} = - 98 x

2 x^{3} - 16 x^{2} + 30 x = 0

What are the solutions to the equation 3x^3+4=x^2+12x ?
The solutions to the equation 3x^3+4=x^2+12x are x= 1/3 ,x=-2,x=2
Find the zeros of 3x^3+4=x^2+12x
The zeros of 3x^3+4=x^2+12x are x= 1/3 ,x=-2,x=2