What are the solutions to the equation -2(3x-1)(x^2-9)=0 ?

The solutions to the equation -2(3x-1)(x^2-9)=0 are x=-3,x=3,x= 1/3

Find the zeros of -2(3x-1)(x^2-9)=0

The zeros of -2(3x-1)(x^2-9)=0 are x=-3,x=3,x= 1/3

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- 2 (3 x - 1) (x^{2} - 9) = 0

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

Decimal Notation

x = - 3, x = 3, x = 0.33333 \dots

Solution steps

- 2 (3 x - 1) (x^{2} - 9) = 0

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 3 = 0 : x = - 3

Solve

x - 3 = 0 : x = 3

Solve

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

The solutions are

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

m^{4} - m^{3} - 7 m^{2} + m + 6 = 0

12 k^{3} - 48 k^{2} + 12 k + 72 = 0

x - 3 x^{5} = \frac{1}{4}

x^{2} + x - 43 = 3 x^{4} - 1

2 x (3 x^{2} - 1) + 5 x = 16 x

What are the solutions to the equation -2(3x-1)(x^2-9)=0 ?
The solutions to the equation -2(3x-1)(x^2-9)=0 are x=-3,x=3,x= 1/3
Find the zeros of -2(3x-1)(x^2-9)=0
The zeros of -2(3x-1)(x^2-9)=0 are x=-3,x=3,x= 1/3