What is factor x^{12}-1 ?

The solution to factor x^{12}-1 is (x^2+1)(x+1)(x-1)(x^2+x+1)(x^2-x+1)(x^4-x^2+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

factor x^{12}-1

factor

x^{12} - 1

(x^{2} + 1) (x + 1) (x - 1) (x^{2} + x + 1) (x^{2} - x + 1) (x^{4} - x^{2} + 1)

Solution steps

x^{12} - 1

Apply exponent rules

= (x^{4})^{3} - 1^{3}

Apply Difference of Cubes Formula:

x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

(x^{4})^{3} - 1^{3} = (x^{4} - 1) (x^{8} + x^{4} + 1)

= (x^{4} - 1) (x^{8} + x^{4} + 1)

Factor

x^{4} - 1 : (x^{2} + 1) (x + 1) (x - 1)

= (x^{2} + 1) (x + 1) (x - 1) (x^{8} + x^{4} + 1)

Factor

x^{8} + x^{4} + 1 : (x^{2} + 1 + x) (x^{2} + 1 - x) (x^{4} + 1 - x^{2})

= (x^{2} + 1) (x + 1) (x - 1) (x^{2} + 1 + x) (x^{2} + 1 - x) (x^{4} + 1 - x^{2})

= (x^{2} + 1) (x + 1) (x - 1) (x^{2} + x + 1) (x^{2} - x + 1) (x^{4} - x^{2} + 1)

lo g_{4} (49)

216 x - 12 x^{2} = 0

2 + 2 u \leq 6

x + \frac{5}{3} = \frac{2}{3}

(2 x^{3} y)^{6}

What is factor x^{12}-1 ?
The solution to factor x^{12}-1 is (x^2+1)(x+1)(x-1)(x^2+x+1)(x^2-x+1)(x^4-x^2+1)