What are the solutions to the equation -3x^3-4x^2+5x+2=0 ?

The solutions to the equation -3x^3-4x^2+5x+2=0 are x=1,x=-1/3 ,x=-2

Find the zeros of -3x^3-4x^2+5x+2=0

The zeros of -3x^3-4x^2+5x+2=0 are x=1,x=-1/3 ,x=-2

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

Factor

- 3 x^{3} - 4 x^{2} + 5 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

4 x^{4} - 39 x^{3} + 54 x^{2} + 16 x

(x^{2} + 1) (4 + x) = 0

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

x^{3} + 125

x^{3} = - 8 i

What are the solutions to the equation -3x^3-4x^2+5x+2=0 ?
The solutions to the equation -3x^3-4x^2+5x+2=0 are x=1,x=-1/3 ,x=-2
Find the zeros of -3x^3-4x^2+5x+2=0
The zeros of -3x^3-4x^2+5x+2=0 are x=1,x=-1/3 ,x=-2