factor x^6+2x^3-3 — step-by-step solution

1) Let's factorize in

ℝ

x^{6} + 2 x^{3} - 3

⇥1.1) Let's define:

X = x^{3}

⇥You can rephrase the expression into:

X^{2} + 2 X - 3

⇥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 = 2

and

c = -3

⇥

Δ = 2^{2} - 4 \cdot 1 \cdot (-3)

⇥ⓘ 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 (-3)

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

- 4 \cdot (-3)

12

⇥

Δ = 4 + 12

⇥ⓘ Start from

4

and move forward

12

on the number line:

4 + 12

16

⇥

Δ = 16

⇥As

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

X_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 2 - \sqrt{16}}{2 \cdot 1}

⇥

X_{1} = \frac{- 2 - \sqrt{16}}{2}

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

\sqrt{16}

4

;

4^{2}

16

⇥

X_{1} = \frac{- 2 - 4}{2}

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

- 2 - 4

-6

⇥

X_{1} = - \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_{1} = -3

⇥The quadratic formula gives

X_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 2 + \sqrt{16}}{2 \cdot 1}

⇥

X_{2} = \frac{- 2 + \sqrt{16}}{2}

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

\sqrt{16}

4

;

4^{2}

16

⇥

X_{2} = \frac{- 2 + 4}{2}

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

- 2 + 4

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

⇥Factors identified:

(X + 3) \cdot (X - 1)

⇥1.2) Let's factorize in

ℝ

(x^{3} + 3) \cdot (x^{3} - 1)

⇥⇥1.2.1) Let's factorize in

ℝ

(x^{3} + 3)

⇥⇥⇥1.2.1.1) The roots of

x^{3} + 3

are the solutions of

x^{3} = -3

⇥⇥⇥The (complex) solutions are:

\sqrt[3]{3} \cdot \exp (i \cdot \frac{2 k π + π}{3})

for k from 0 to 2

⇥⇥⇥

k	Exponential solution	Transformed and simplified as
0	$\sqrt[3]{3} \cdot \exp (i \cdot \frac{2 \cdot 0 \cdot π + π}{3}) = \sqrt[3]{3} \cdot \exp (\frac{i π}{3})$	$\sqrt[3]{3} \cdot (\cos (\frac{π}{3}) + i \cdot \sin (\frac{π}{3})) = \frac{\sqrt[3]{3} + \sqrt[6]{243} \cdot i}{2}$
1	$\sqrt[3]{3} \cdot \exp (i \cdot \frac{2 \cdot 1 \cdot π + π}{3}) = \sqrt[3]{3} \cdot \exp (i π)$	$\sqrt[3]{3} \cdot (\cos (π) + i \cdot \sin (π)) = - \sqrt[3]{3}$
2	$\sqrt[3]{3} \cdot \exp (i \cdot \frac{2 \cdot 2 \cdot π + π}{3}) = \sqrt[3]{3} \cdot \exp (\frac{i \cdot 5 π}{3})$	$\sqrt[3]{3} \cdot (\cos (\frac{5 π}{3}) + i \cdot \sin (\frac{5 π}{3})) = \frac{\sqrt[3]{3} - \sqrt[6]{243} \cdot i}{2}$

⇥⇥⇥In ℝ, factoring gives:

x + \sqrt[3]{3}

⇥⇥⇥1.2.1.2) Compute the quotient and remainder of

(x^{3} + 3)

(x + \sqrt[3]{3})

⇥⇥⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{3} + 3$	$\frac{x^{3}}{x}$	$x^{2}$	$x^{2} \cdot (x + \sqrt[3]{3}) = x^{3} + \sqrt[3]{3} \cdot x^{2}$	$(x^{3} + 3) - (x^{3} + \sqrt[3]{3} \cdot x^{2}) = - \sqrt[3]{3} \cdot x^{2} + 3$
$- \sqrt[3]{3} \cdot x^{2} + 3$	$- \frac{\sqrt[3]{3} \cdot x^{2}}{x}$	$- x \cdot \sqrt[3]{3}$	$- x \cdot \sqrt[3]{3} \cdot (x + \sqrt[3]{3}) = - \sqrt[3]{3} \cdot x^{2} - \sqrt[3]{9} \cdot x$	$(- \sqrt[3]{3} \cdot x^{2} + 3) - (- \sqrt[3]{3} \cdot x^{2} - \sqrt[3]{9} \cdot x) = \sqrt[3]{9} \cdot x + 3$
$\sqrt[3]{9} \cdot x + 3$	$\frac{\sqrt[3]{9} \cdot x}{x}$	$\sqrt[3]{9}$	$\sqrt[3]{9} \cdot (x + \sqrt[3]{3}) = \sqrt[3]{9} \cdot x + 3$	$(\sqrt[3]{9} \cdot x + 3) - (\sqrt[3]{9} \cdot x + 3) = 0$

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

x^{2} - x \cdot \sqrt[3]{3} + \sqrt[3]{9}

⇥⇥ⓘ The other roots,

\frac{\sqrt[3]{3} - \sqrt[6]{243} \cdot i}{2}

\frac{\sqrt[3]{3} + \sqrt[6]{243} \cdot i}{2}

, are not real:

x^{2} - x \cdot \sqrt[3]{3} + \sqrt[3]{9}

cannot be factored in ℝ

▷The factor

(x^{3} + 3)

factors as:

(x + \sqrt[3]{3}) \cdot (x^{2} - x \cdot \sqrt[3]{3} + \sqrt[3]{9})

⇥⇥1.2.2) Let's factor in

ℝ

(x^{3} - 1)

⇥⇥⇥1.2.2.1) The roots of

x^{3} - 1

are the solutions of

x^{3} = 1

⇥⇥⇥Solutions are:

\exp (i \cdot \frac{2 k π}{3})

(with k = 0..2)

⇥⇥⇥

k	Exponential solution	Rewritten and simplified as
0	$\exp (i \cdot \frac{2 \cdot 0 \cdot π}{3}) = 1$	$\cos (0) + i \cdot \sin (0) = 1$
1	$\exp (i \cdot \frac{2 \cdot 1 \cdot π}{3}) = \exp (\frac{i \cdot 2 π}{3})$	$\cos (\frac{2 π}{3}) + i \cdot \sin (\frac{2 π}{3}) = \frac{- 1 + i \cdot \sqrt{3}}{2}$
2	$\exp (i \cdot \frac{2 \cdot 2 \cdot π}{3}) = \exp (\frac{i \cdot 4 π}{3})$	$\cos (\frac{4 π}{3}) + i \cdot \sin (\frac{4 π}{3}) = - \frac{1 + i \cdot \sqrt{3}}{2}$

⇥⇥⇥In ℝ, factoring gives:

x - 1

⇥⇥⇥1.2.2.2) Perform polynomial long division of

(x^{3} - 1)

(x - 1)

⇥⇥⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{3} - 1$	$\frac{x^{3}}{x}$	$x^{2}$	$x^{2} \cdot (x - 1) = x^{3} - x^{2}$	$(x^{3} - 1) - (x^{3} - x^{2}) = x^{2} - 1$
$x^{2} - 1$	$\frac{x^{2}}{x}$	$x$	$x \cdot (x - 1) = x^{2} - x$	$(x^{2} - 1) - (x^{2} - x) = x - 1$
$x - 1$	$\frac{x}{x}$	$1$	$x - 1 = x - 1$	$(x - 1) - (x - 1) = 0$

⇥⇥⇥So the result of the euclidean division is:

x^{2} + x + 1

⇥⇥ⓘ The other roots,

- \frac{1 + i \cdot \sqrt{3}}{2}

\frac{- 1 + i \cdot \sqrt{3}}{2}

, are not real:

x^{2} + x + 1

cannot be factored in ℝ

▷The factor

(x^{3} - 1)

factors as:

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

▶Factored expression:

(x - 1) \cdot (x^{2} + x + 1) \cdot (x + \sqrt[3]{3}) \cdot (x^{2} - x \cdot \sqrt[3]{3} + \sqrt[3]{9})