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

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

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

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

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

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

Decimal Notation

x = - 3, x = - 0.5, x = 0.5

Solution steps

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

Move

3

to the left side

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

Move

x

to the left side

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

Write in the standard form

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

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 3 = 0 : x = - 3

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

2 x - 1 = 0 : x = \frac{1}{2}

The solutions are

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

(6 x^{4}) + (6 x^{1}) = x (4 + 1)

x^{3} - 4 x^{2} + 5 x = 0

x^{6} + 18 x^{3} + 81

x^{4} - 2 x^{3} + 3 x^{2} - 2 x + 1 = 0

x^{1} x^{2} = 4 + (- \frac{8}{6})

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