What are the solutions to the equation 2x^3+7x^2-12x-42=0 ?

The solutions to the equation 2x^3+7x^2-12x-42=0 are x=-7/2 ,x=-sqrt(6),x=sqrt(6)

Find the zeros of 2x^3+7x^2-12x-42=0

The zeros of 2x^3+7x^2-12x-42=0 are x=-7/2 ,x=-sqrt(6),x=sqrt(6)

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2 x^{3} + 7 x^{2} - 12 x - 42 = 0

x = - \frac{7}{2}, x = - 6, x = 6

Decimal Notation

x = - 3.5, x = - 2.44948 \dots, x = 2.44948 \dots

Solution steps

2 x^{3} + 7 x^{2} - 12 x - 42 = 0

Factor

2 x^{3} + 7 x^{2} - 12 x - 42 : (2 x + 7) (x + 6) (x - 6)

(2 x + 7) (x + 6) (x - 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x + 7 = 0 or x + 6 = 0 or x - 6 = 0

Solve

2 x + 7 = 0 : x = - \frac{7}{2}

Solve

x + 6 = 0 : x = - 6

Solve

x - 6 = 0 : x = 6

The solutions are

x = - \frac{7}{2}, x = - 6, x = 6

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

x, x^{3} + y^{3} = 126

x^{3} = - 10

x^{3} = - 15

x^{4} = 5

What are the solutions to the equation 2x^3+7x^2-12x-42=0 ?
The solutions to the equation 2x^3+7x^2-12x-42=0 are x=-7/2 ,x=-sqrt(6),x=sqrt(6)
Find the zeros of 2x^3+7x^2-12x-42=0
The zeros of 2x^3+7x^2-12x-42=0 are x=-7/2 ,x=-sqrt(6),x=sqrt(6)