factor 2m^3+2m^2-4m — step-by-step solution

Let's factorize in

ℝ

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

You can factorize this expression into:

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

Let's factor in

ℝ

(2 m^{2} + 2 m - 4)

We factor out

2

, the greatest common divisor of the coefficients:

2 m^{2} + 2 m - 4

2 \cdot (m^{2} + m - 2)

Given the quadratic form

a m^{2} + b m + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 1

and

c = -2

Δ = 1 - 4 \cdot 1 \cdot (-2)

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

m_{1}

\frac{- b - \sqrt{Δ}}{2 a}

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

m_{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

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

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

- 1 - 3

-4

m_{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

m_{1} = -2

The quadratic formula gives

m_{2}

\frac{- b + \sqrt{Δ}}{2 a}

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

m_{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

m_{2} = \frac{- 1 + 3}{2}

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

- 1 + 3

2

m_{2} = \frac{2}{2}

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

2

, their greatest common divisor:

\frac{2}{2}

1

m_{2} = 1

Factors identified:

(m + 2) \cdot (m - 1)

▷The factor

(2 m^{2} + 2 m - 4)

factors as:

2 \cdot (m + 2) \cdot (m - 1)

ⓘ Factor out the common factor

m

▶Factored expression:

2 m \cdot (m + 2) \cdot (m - 1)