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Popular >

factor 16y^8+8y^4+1

Solution

factor

16 y^{8} + 8 y^{4} + 1

Solution

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

Solution steps

16 y^{8} + 8 y^{4} + 1

Let

u = y^{2}

= 16 u^{4} + 8 u^{2} + 1

Factor

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

= (4 u^{2} + 1)^{2}

Substitute back

u = y^{2}

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

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))^{2}

Apply exponent rule:

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

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

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

Popular Examples

factor n^4+6n^3+13n^2+12n+4 factor

n^{4} + 6 n^{3} + 13 n^{2} + 12 n + 4

factor a^2-2a factor

a^{2} - 2 a

x-(2x-1)/3 =(3x-5)/5

x - \frac{2 x - 1}{3} = \frac{3 x - 5}{5}

3x(x-3)+5(x+7)-x(x+1)-2(x^2+7)+4=0

3 x (x - 3) + 5 (x + 7) - x (x + 1) - 2 (x^{2} + 7) + 4 = 0

simplify 13/2-5 simplify

\frac{13}{2} - 5

Frequently Asked Questions (FAQ)

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