What are the solutions to the equation -2x(3x+1)(x-8)(x+2)=0 ?

The solutions to the equation -2x(3x+1)(x-8)(x+2)=0 are x=0,x=-1/3 ,x=8,x=-2

Find the zeros of -2x(3x+1)(x-8)(x+2)=0

The zeros of -2x(3x+1)(x-8)(x+2)=0 are x=0,x=-1/3 ,x=8,x=-2

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- 2 x (3 x + 1) (x - 8) (x + 2) = 0

x = 0, x = - \frac{1}{3}, x = 8, x = - 2

Decimal Notation

x = 0, x = - 0.33333 \dots, x = 8, x = - 2

Solution steps

- 2 x (3 x + 1) (x - 8) (x + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 3 x + 1 = 0 or x - 8 = 0 or x + 2 = 0

Solve

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

Solve

x - 8 = 0 : x = 8

Solve

x + 2 = 0 : x = - 2

The solutions are

x = 0, x = - \frac{1}{3}, x = 8, x = - 2

9 x^{4} + 27 x^{3} - 4 x^{2} - 12 x = 0

x^{3} - 9 x^{2} + 27 x - 27 = 0

2 x^{3} + 3 x^{2} = 8 x + 12

x^{3} - 8 = x - 2

(3^{3})^{2} = n^{6}

What are the solutions to the equation -2x(3x+1)(x-8)(x+2)=0 ?
The solutions to the equation -2x(3x+1)(x-8)(x+2)=0 are x=0,x=-1/3 ,x=8,x=-2
Find the zeros of -2x(3x+1)(x-8)(x+2)=0
The zeros of -2x(3x+1)(x-8)(x+2)=0 are x=0,x=-1/3 ,x=8,x=-2