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

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

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

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

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

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

Decimal Notation

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

Solution steps

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

Move

48 x

to the left side

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

Move

x^{2}

to the left side

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

(5 x - 3)^{2} (2 x + 3)^{3} = 100

x^{3} - x^{2} - 20 x = 0

x^{6} + 117 x^{3} - 1000 = 0

(2 x - 1)^{3} = 27

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

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