What are the solutions to the equation 3x^3+x^2-48x=16 ?

The solutions to the equation 3x^3+x^2-48x=16 are x=-1/3 ,x=-4,x=4

Find the zeros of 3x^3+x^2-48x=16

The zeros of 3x^3+x^2-48x=16 are x=-1/3 ,x=-4,x=4

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3 x^{3} + x^{2} - 48 x = 16

x = - \frac{1}{3}, x = - 4, x = 4

Decimal Notation

x = - 0.33333 \dots, x = - 4, x = 4

Solution steps

3 x^{3} + x^{2} - 48 x = 16

Move

16

to the left side

3 x^{3} + x^{2} - 48 x - 16 = 0

Factor

3 x^{3} + x^{2} - 48 x - 16 : (3 x + 1) (x + 4) (x - 4)

(3 x + 1) (x + 4) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x + 1 = 0 or x + 4 = 0 or x - 4 = 0

Solve

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

Solve

x + 4 = 0 : x = - 4

Solve

x - 4 = 0 : x = 4

The solutions are

x = - \frac{1}{3}, x = - 4, x = 4

- x^{3} = 0

x^{3} - 6 x = 0

12 x^{2} - 9 x - 3 x^{3} = 0

(x^{3} - 1) = 0

z^{3} = 216

What are the solutions to the equation 3x^3+x^2-48x=16 ?
The solutions to the equation 3x^3+x^2-48x=16 are x=-1/3 ,x=-4,x=4
Find the zeros of 3x^3+x^2-48x=16
The zeros of 3x^3+x^2-48x=16 are x=-1/3 ,x=-4,x=4