What are the solutions to the equation 9x^3+54x^2=x+6 ?

The solutions to the equation 9x^3+54x^2=x+6 are x=-6,x=-1/3 ,x= 1/3

Find the zeros of 9x^3+54x^2=x+6

The zeros of 9x^3+54x^2=x+6 are x=-6,x=-1/3 ,x= 1/3

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9 x^{3} + 54 x^{2} = x + 6

x = - 6, x = - \frac{1}{3}, x = \frac{1}{3}

Decimal Notation

x = - 6, x = - 0.33333 \dots, x = 0.33333 \dots

Solution steps

9 x^{3} + 54 x^{2} = x + 6

Move

6

to the left side

9 x^{3} + 54 x^{2} - 6 = x

Move

x

to the left side

9 x^{3} + 54 x^{2} - 6 - x = 0

Write in the standard form

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

9 x^{3} + 54 x^{2} - x - 6 = 0

Factor

9 x^{3} + 54 x^{2} - x - 6 : (x + 6) (3 x + 1) (3 x - 1)

(x + 6) (3 x + 1) (3 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 6 = 0 or 3 x + 1 = 0 or 3 x - 1 = 0

Solve

x + 6 = 0 : x = - 6

Solve

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

Solve

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

The solutions are

x = - 6, x = - \frac{1}{3}, x = \frac{1}{3}

- 4 x^{3} + 6 x - 1 = 0

38 = h (2 h^{2} - 3 h)

(y - 1) (y^{2} + y + 2) = 0

5 = 3 - \frac{1}{10} b^{3} - 101 b

- 6 x^{3} + 36 x^{2} + 42 x = 0

What are the solutions to the equation 9x^3+54x^2=x+6 ?
The solutions to the equation 9x^3+54x^2=x+6 are x=-6,x=-1/3 ,x= 1/3
Find the zeros of 9x^3+54x^2=x+6
The zeros of 9x^3+54x^2=x+6 are x=-6,x=-1/3 ,x= 1/3