factorize x^2-x-6 — step-by-step solution

Let's factor in

ℝ

x^{2} - x - 6

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 = -6

Δ = {(-1)}^{2} - 4 \cdot 1 \cdot (-6)

ⓘ 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 (-6)

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

- 4 \cdot (-6)

24

Δ = 1 + 24

ⓘ Start from

1

and move forward

24

on the number line:

1 + 24

25

Δ = 25

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-1) - \sqrt{25}}{2 \cdot 1}

x_{1} = \frac{1 - \sqrt{25}}{2}

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

\sqrt{25}

5

;

5^{2}

25

x_{1} = \frac{1 - 5}{2}

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

1 - 5

-4

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

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-1) + \sqrt{25}}{2 \cdot 1}

x_{2} = \frac{1 + \sqrt{25}}{2}

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

\sqrt{25}

5

;

5^{2}

25

x_{2} = \frac{1 + 5}{2}

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

1 + 5

6

x_{2} = \frac{6}{2}

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

2

, their greatest common divisor:

\frac{6}{2}

3

x_{2} = 3

▶Factors identified:

(x + 2) \cdot (x - 3)