What are the solutions to the equation -2x^3+x^2+13x+6=0 ?

The solutions to the equation -2x^3+x^2+13x+6=0 are x=-2,x=-1/2 ,x=3

Find the zeros of -2x^3+x^2+13x+6=0

The zeros of -2x^3+x^2+13x+6=0 are x=-2,x=-1/2 ,x=3

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- 2 x^{3} + x^{2} + 13 x + 6 = 0

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

Decimal Notation

x = - 2, x = - 0.5, x = 3

Solution steps

- 2 x^{3} + x^{2} + 13 x + 6 = 0

Factor

- 2 x^{3} + x^{2} + 13 x + 6 : - (x + 2) (2 x + 1) (x - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 = 0 : x = - 2

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

x - 3 = 0 : x = 3

The solutions are

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

x x^{2} = 4

x^{2} (4 x + 8)^{2} = 14 0^{2}

x^{8} = 81 x

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

- λ^{3} + 8 λ^{2} - 19 λ + 12 = 0

What are the solutions to the equation -2x^3+x^2+13x+6=0 ?
The solutions to the equation -2x^3+x^2+13x+6=0 are x=-2,x=-1/2 ,x=3
Find the zeros of -2x^3+x^2+13x+6=0
The zeros of -2x^3+x^2+13x+6=0 are x=-2,x=-1/2 ,x=3