What are the solutions to the equation 3x^3+2x^2-19x+6=0 ?

The solutions to the equation 3x^3+2x^2-19x+6=0 are x=2,x= 1/3 ,x=-3

Find the zeros of 3x^3+2x^2-19x+6=0

The zeros of 3x^3+2x^2-19x+6=0 are x=2,x= 1/3 ,x=-3

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3 x^{3} + 2 x^{2} - 19 x + 6 = 0

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

Decimal

x = 2, x = 0.33333 \dots, x = - 3

Solution steps

3 x^{3} + 2 x^{2} - 19 x + 6 = 0

Factor

3 x^{3} + 2 x^{2} - 19 x + 6 : (x - 2) (3 x - 1) (x + 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

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

Solve

x + 3 = 0 : x = - 3

The solutions are

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

x^{3} - 3 x^{2} - 25 x + 75 = 0

r^{5} - 3 r^{4} + 3 r^{3} - 3 r^{2} + 2 r = 0

x^{3} + x^{2} - 5 x - 5 = 0

x^{3} - 2 x^{2} - 11 x + 12

x^{4} + 9 x^{3} + 12 x^{2} - 34 x + 12 = 0

What are the solutions to the equation 3x^3+2x^2-19x+6=0 ?
The solutions to the equation 3x^3+2x^2-19x+6=0 are x=2,x= 1/3 ,x=-3
Find the zeros of 3x^3+2x^2-19x+6=0
The zeros of 3x^3+2x^2-19x+6=0 are x=2,x= 1/3 ,x=-3