factorize the expression x^2+x-6 — step-by-step solution

Let's factorize in

ℝ

x^{2} + x - 6

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 = -6

Δ = 1 - 4 \cdot 1 \cdot (-6)

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

Since

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 nonnegative real number is the nonnegative number whose square is that number:

\sqrt{25}

5

because

5^{2}

25

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

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

- 1 - 5

-6

x_{1} = - \frac{6}{2}

ⓘ Simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{6}{2}

-3

x_{1} = -3

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 nonnegative real is defined as the nonnegative number whose square gives that real number:

\sqrt{25}

5

because

5^{2}

25

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

ⓘ Calculate the numerator and denominator before dividing:

- 1 + 5

4

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

ⓘ Simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{4}{2}

2

x_{2} = 2

▶Factors identified:

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