What is factor 16y^8-1 ?

The solution to factor 16y^8-1 is (2y^2+2y+1)(2y^2-2y+1)(2y^2+1)(2y^2-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

factor 16y^8-1

factor

16 y^{8} - 1

(2 y^{2} + 2 y + 1) (2 y^{2} - 2 y + 1) (2 y^{2} + 1) (2 y^{2} - 1)

Solution steps

16 y^{8} - 1

Apply exponent rules

= (4 y^{4})^{2} - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (4 y^{4} + 1) (4 y^{4} - 1)

Factor

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

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

Factor

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

= (2 y^{2} + 2 y + 1) (2 y^{2} - 2 y + 1) (2 y^{2} + 1) (2 y^{2} - 1)

312 - 212 - 54

2 x^{2} (- x^{2} - 7)

- 6 x^{5} - \frac{1}{8} - x^{3}

\frac{x ^{5} + x ^{3} - 2}{x - 1}

i^{175}

What is factor 16y^8-1 ?
The solution to factor 16y^8-1 is (2y^2+2y+1)(2y^2-2y+1)(2y^2+1)(2y^2-1)