factor x^4+2x^3-5x^2-12x-4 — step-by-step solution

1) Let's factor in

ℝ

x^{4}

+ 2 x^{3}

- 5 x^{2}

- 12 x

- 4

⇥1.1) Let's start with the obvious candidate:

x = -2

⇥Let's compute the value of:

{(-2)}^{4}

+ 2 \cdot {(-2)}^{3}

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

- 12 \cdot (-2)

- 4

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

{(-2)}^{4}

(-2) \cdot (-2) \cdot (-2) \cdot (-2)

16

{(-2)}^{3}

(-2) \cdot (-2) \cdot (-2)

-8

{(-2)}^{2}

(-2) \cdot (-2)

4

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

- 12 \cdot (-2)

12 \cdot 2

(10 + 2) \cdot 2

10 \cdot 2 + 2 \cdot 2

20 + 4

24

⇥

16

+ 2 \cdot (-8)

- 5 \cdot 4

+ 24

- 4

⇥ⓘ Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-8)

-16

■Multiplications are done before additions and subtractions:

5 \cdot 4

20

⇥

16

- 16

- 20

+ 24

- 4

⇥ⓘ Start from

16

and move back

16

on the number line, down to zero:

16 - 16

0

⇥

0

- 20

+ 24

- 4

⇥ⓘ Adding 0 changes nothing:

0 - 20

-20

■Start from

-20

and move forward

24

on the number line, going past zero:

- 20 + 24

4

⇥

4 - 4

⇥ⓘ Start from

4

and move back

4

on the number line, down to zero:

4 - 4

0

⇥

0

⇥Root found:

x = -2

⇥1.2) Perform polynomial long division of

(x^{4}

+ 2 x^{3}

- 5 x^{2}

- 12 x

- 4)

(x + 2)

⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{4}$ $+ 2 x^{3}$ $- 5 x^{2}$ $- 12 x$ $- 4$	$\frac{x^{4}}{x}$	$x^{3}$	$x^{3} \cdot (x + 2) = x^{4} + 2 x^{3}$	$(x^{4}$ $+ 2 x^{3}$ $- 5 x^{2}$ $- 12 x$ $- 4) - (x^{4} + 2 x^{3}) = - 5 x^{2} - 12 x - 4$
$- 5 x^{2} - 12 x - 4$	$- \frac{5 x^{2}}{x}$	$- 5 x$	$- 5 x \cdot (x + 2) = - 5 x^{2} - 10 x$	$(- 5 x^{2} - 12 x - 4) - (- 5 x^{2} - 10 x) = - 2 x - 4$
$- 2 x - 4$	$- \frac{2 x}{x}$	$-2$	$-2 \cdot (x + 2) = - 2 x - 4$	$(- 2 x - 4) - (- 2 x - 4) = 0$

⇥Therefore, the result of the euclidean division is:

x^{3} - 5 x - 2

⇥1.3) Let's factor in

ℝ

(x^{3} - 5 x - 2)

⇥⇥1.3.1) Let's start with the obvious candidate:

x = -2

⇥⇥Let's compute the value of:

{(-2)}^{3} - 5 \cdot (-2) - 2

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

{(-2)}^{3}

(-2) \cdot (-2) \cdot (-2)

-8

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

- 5 \cdot (-2)

10

⇥⇥

- 8 + 10 - 2

⇥⇥ⓘ Start from

-8

and move forward

10

on the number line, going past zero:

- 8 + 10

2

⇥⇥

2 - 2

⇥⇥ⓘ Start from

2

and move back

2

on the number line, down to zero:

2 - 2

0

⇥⇥

0

⇥⇥Root found:

x = -2

⇥⇥1.3.2) Perform polynomial long division of

(x^{3} - 5 x - 2)

(x + 2)

⇥⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{3} - 5 x - 2$	$\frac{x^{3}}{x}$	$x^{2}$	$x^{2} \cdot (x + 2) = x^{3} + 2 x^{2}$	$(x^{3} - 5 x - 2) - (x^{3} + 2 x^{2}) = - 2 x^{2} - 5 x - 2$
$- 2 x^{2} - 5 x - 2$	$- \frac{2 x^{2}}{x}$	$- 2 x$	$- 2 x \cdot (x + 2) = - 2 x^{2} - 4 x$	$(- 2 x^{2} - 5 x - 2) - (- 2 x^{2} - 4 x) = - x - 2$
$- x - 2$	$- \frac{x}{x}$	$-1$	$-1 \cdot (x + 2) = - x - 2$	$(- x - 2) - (- x - 2) = 0$

⇥⇥Therefore, the result of the euclidean division is:

x^{2} - 2 x - 1

⇥⇥1.3.3) Let's factorize in

ℝ

(x^{2} - 2 x - 1)

⇥⇥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 = -2

and

c = -1

⇥⇥

Δ = {(-2)}^{2} - 4 \cdot 1 \cdot (-1)

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

{(-2)}^{2}

(-2) \cdot (-2)

4

⇥⇥

Δ = 4 - 4 \cdot (-1)

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

- 4 \cdot (-1)

4

⇥⇥

Δ = 4 + 4

⇥⇥ⓘ Start from

4

and move forward

4

on the number line:

4 + 4

8

⇥⇥

Δ = 8

⇥⇥Since

Δ > 0

, the equation has two roots:

⇥⇥The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-2) - \sqrt{8}}{2 \cdot 1}

⇥⇥

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

⇥⇥ⓘ We look for the largest square dividing

8

4

2^{2}

, and take it out of the root:

\sqrt{8}

\sqrt{4 \cdot 2}

\sqrt{4} \cdot \sqrt{2}

2 \cdot \sqrt{2}

⇥⇥

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

⇥⇥ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

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

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

1 - \sqrt{2}

⇥⇥

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

⇥⇥The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-2) + \sqrt{8}}{2 \cdot 1}

⇥⇥

x_{2} = \frac{2 + \sqrt{8}}{2}

⇥⇥ⓘ We look for the largest square dividing

8

4

2^{2}

, and take it out of the root:

\sqrt{8}

\sqrt{4 \cdot 2}

\sqrt{4} \cdot \sqrt{2}

2 \cdot \sqrt{2}

⇥⇥

x_{2} = \frac{2 + 2 \cdot \sqrt{2}}{2}

⇥⇥ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{2 + 2 \cdot \sqrt{2}}{2}

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

1 + \sqrt{2}

⇥⇥

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

▷The factor

(x^{2} - 2 x - 1)

factors as:

(x - (1 + \sqrt{2})) \cdot (x - (1 - \sqrt{2}))

⇥1.4) Let's replace each factor by its factorization:

(x + 2) \cdot (x - (1 + \sqrt{2})) \cdot (x - (1 - \sqrt{2}))

⇥ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (1 + \sqrt{2})

- 1 - \sqrt{2}

- (1 - \sqrt{2})

- 1 + \sqrt{2}

⇥

(x + 2) \cdot (x - 1 - \sqrt{2}) \cdot (x - 1 + \sqrt{2})

▷The factor

(x^{3} - 5 x - 2)

factors as:

(x + 2) \cdot (x + \sqrt{2} - 1) \cdot (x - \sqrt{2} - 1)

▶Factored expression:

{(x + 2)}^{2} \cdot (x + \sqrt{2} - 1) \cdot (x - \sqrt{2} - 1)