x^3>2x+x^2 — step-by-step solution

1) Let's solve in

ℝ

x^{3} > 2 x + x^{2}

You can factor:

x^{3} - x^{2} - 2 x

into:

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

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

x^{2} - x - 2

by widening the sign to ≥ and solving in

ℝ

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

Given the quadratic form

a x^{2} + b x + c

, let's compute 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

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 sign of a =

1

outside its roots, so it is indeed positive outside its roots; the roots, where it is zero, are included.

Solution found:

x \in (-∞, -1] \cup [2, +∞)

⇥So the expression

x^{2} - x - 2

is positive or zero on

(-∞, -1] \cup [2, +∞)

and negative otherwise

2) Let's draw up the sign table:

▶Solution obtained:

x \in (-1, 0) \cup (2, +∞)