factorize the expression x^3+8 — step-by-step solution

1) Let's factor in

ℝ

x^{3} + 8

⇥1.1) The roots of

x^{3} + 8

are the solutions of

x^{3} = -8

⇥The (complex) solutions are:

\sqrt[3]{8} \cdot \exp (i \cdot \frac{2 k π + π}{3})

for k from 0 to 2

⇥

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

⇥For the real solutions, factorization gives:

x + 2

⇥1.2) Perform the Euclidean division of

(x^{3} + 8)

(x + 2)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{3} + 8$	$\frac{x^{3}}{x}$	$x^{2}$	$x^{2} \cdot (x + 2) = x^{3} + 2 x^{2}$	$(x^{3} + 8) - (x^{3} + 2 x^{2}) = - 2 x^{2} + 8$
$- 2 x^{2} + 8$	$- \frac{2 x^{2}}{x}$	$- 2 x$	$- 2 x \cdot (x + 2) = - 2 x^{2} - 4 x$	$(- 2 x^{2} + 8) - (- 2 x^{2} - 4 x) = 4 x + 8$
$4 x + 8$	$\frac{4 x}{x}$	$4$	$4 \cdot (x + 2) = 4 x + 8$	$(4 x + 8) - (4 x + 8) = 0$

⇥So the result of the Euclidean division is:

x^{2} - 2 x + 4

ⓘ The other roots,

1 - i \cdot \sqrt{3}

1 + i \cdot \sqrt{3}

, are not real:

x^{2} - 2 x + 4

cannot be factored in ℝ

▶Factors found:

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