-2x^3+2x^2+4x>0 — step-by-step solution

1) Let's solve in

ℝ

- 2 x^{3} + 2 x^{2} + 4 x > 0

You can factorize the expression:

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

into:

x \cdot (- 2 x^{2} + 2 x + 4)

We study the sign of each factor of the product

⇥The expression

x

is zero at

x = 0

, it is negative before and positive after

Let's study the sign of

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

by widening the sign to ≥ and solving in

ℝ

- 2 x^{2} + 2 x + 4 \geq 0

Since

-2 < 0

, divide both sides by

-2

and reverse the inequality:

x^{2} - x - 2 \leq 0

Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -1

and

c = -2

Δ = {(-1)}^{2} - 4 \cdot 1 \cdot (-2)

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-1)}^{2}

(-1) \cdot (-1)

1

Δ = 1 - 4 \cdot (-2)

ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (-2)

8

Δ = 1 + 8

ⓘ Start from

1

and move forward

8

on the number line:

1 + 8

9

Δ = 9

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-1) - \sqrt{9}}{2 \cdot 1}

x_{1} = \frac{1 - \sqrt{9}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

x_{1} = \frac{1 - 3}{2}

ⓘ The numerator and the denominator are computed before the division:

1 - 3

-2

x_{1} = - \frac{2}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{2}{2}

-1

x_{1} = -1

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-1) + \sqrt{9}}{2 \cdot 1}

x_{2} = \frac{1 + \sqrt{9}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

x_{2} = \frac{1 + 3}{2}

ⓘ The numerator and the denominator are computed before the division:

1 + 3

4

x_{2} = \frac{4}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{4}{2}

2

x_{2} = 2

The quadratic has the opposite sign to a =

1

between its roots, so it is indeed negative between its roots; the roots, where it is zero, are included.

Solution found:

x \in [-1, 2]

⇥So the expression

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

is positive or zero on

[-1, 2]

and negative otherwise

2) Let's draw up the sign table:

▶Solution obtained:

x \in (-∞, -1) \cup (0, 2)