What is factor 16u^8-8u^4+1 ?

The solution to factor 16u^8-8u^4+1 is (2u^2+1)^2(sqrt(2)u+1)^2(sqrt(2)u-1)^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

factor 16u^8-8u^4+1

factor

16 u^{8} - 8 u^{4} + 1

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

Solution steps

16 u^{8} - 8 u^{4} + 1

Let

v = u^{2}

= 16 v^{4} - 8 v^{2} + 1

Factor

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

= (2 v + 1) (2 v - 1) (2 v + 1) (2 v - 1)

Substitute back

v = u^{2}

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

Refine

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m},

assuming

a \geq 0, b \geq 0

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

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

Factor

(2 u^{2} + 1) (2 u^{2} - 1) : (2 u^{2} + 1) (2 u + 1) (2 u - 1)

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

x^{2} + 3 - 2

x - y + x z - yz + x z^{2} - y z^{2}

8 + 23

\frac{x ^{2} - 9}{25}

12 b c - 4 b d - 15 x c + 5 x d

What is factor 16u^8-8u^4+1 ?
The solution to factor 16u^8-8u^4+1 is (2u^2+1)^2(sqrt(2)u+1)^2(sqrt(2)u-1)^2