What are the solutions to the equation 3x^3+x^2-8x+4=0 ?

The solutions to the equation 3x^3+x^2-8x+4=0 are x=1,x= 2/3 ,x=-2

Find the zeros of 3x^3+x^2-8x+4=0

The zeros of 3x^3+x^2-8x+4=0 are x=1,x= 2/3 ,x=-2

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3 x^{3} + x^{2} - 8 x + 4 = 0

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

Decimal Notation

x = 1, x = 0.66666 \dots, x = - 2

Solution steps

3 x^{3} + x^{2} - 8 x + 4 = 0

Factor

3 x^{3} + x^{2} - 8 x + 4 : (x - 1) (3 x - 2) (x + 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

3 x - 2 = 0 : x = \frac{2}{3}

Solve

x + 2 = 0 : x = - 2

The solutions are

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

(X - 2) (X + 4) (X - 5) = 0

5 x^{4} = 80 x^{2}

- 3 x^{3} + 75 x = 0

x^{3} = \frac{1}{64}

9 x^{4} - 8 x^{2} - 1 = 0

What are the solutions to the equation 3x^3+x^2-8x+4=0 ?
The solutions to the equation 3x^3+x^2-8x+4=0 are x=1,x= 2/3 ,x=-2
Find the zeros of 3x^3+x^2-8x+4=0
The zeros of 3x^3+x^2-8x+4=0 are x=1,x= 2/3 ,x=-2