factor x^3+5x^2+4x — step-by-step solution

Let's factorize in

ℝ

x^{3} + 5 x^{2} + 4 x

This expression can be factorized into:

x \cdot (x^{2} + 5 x + 4)

Let's factorize in

ℝ

(x^{2} + 5 x + 4)

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 = 5

and

c = 4

Δ = 5^{2} - 4 \cdot 1 \cdot 4

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

5^{2}

5 \cdot 5

25

Δ = 25 - 4 \cdot 4

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 4

16

Δ = 25 - 16

ⓘ Start from

25

and move back

16

on the number line:

25 - 16

9

Δ = 9

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 5 - \sqrt{9}}{2 \cdot 1}

x_{1} = \frac{- 5 - \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{- 5 - 3}{2}

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

- 5 - 3

-8

x_{1} = - \frac{8}{2}

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

2

, their greatest common divisor:

- \frac{8}{2}

-4

x_{1} = -4

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 5 + \sqrt{9}}{2 \cdot 1}

x_{2} = \frac{- 5 + \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{- 5 + 3}{2}

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

- 5 + 3

-2

x_{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

x_{2} = -1

▷The factor

(x^{2} + 5 x + 4)

factors as:

(x + 1) \cdot (x + 4)

ⓘ Factor out the common factor

x

▶Factored expression:

x \cdot (x + 1) \cdot (x + 4)