What are the solutions to the equation 3x^3-8x^2+3x+2=0 ?

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

Find the zeros of 3x^3-8x^2+3x+2=0

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

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3 x^{3} - 8 x^{2} + 3 x + 2 = 0

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

Decimal Notation

x = 1, x = - 0.33333 \dots, x = 2

Solution steps

3 x^{3} - 8 x^{2} + 3 x + 2 = 0

Factor

3 x^{3} - 8 x^{2} + 3 x + 2 : (x - 1) (3 x + 1) (x - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

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

Solve

x - 2 = 0 : x = 2

The solutions are

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

x^{4} - 2 x^{3} + 3 x - 1 = 0

x^{11} = 2

λ^{3} - 6 λ^{2} + 11 λ - 6 = 0

x^{3} - 5 x - 10 = 0

x^{4} - 10 x^{2} + 1 = 0

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