What is factor 32y^5-1 ?

The solution to factor 32y^5-1 is (2y-1)(16y^4+8y^3+4y^2+2y+1)

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factor 32y^5-1

factor

32 y^{5} - 1

(2 y - 1) (16 y^{4} + 8 y^{3} + 4 y^{2} + 2 y + 1)

Solution steps

32 y^{5} - 1

Apply exponent rules

= (2 y)^{5} - 1^{5}

Apply factoring rule:

x^{n} - y^{n} = (x - y) (x^{n - 1} + x^{n - 2} y + \dots + x y^{n - 2} y^{n - 1})

(2 y)^{5} - 1^{5} = (2 y - 1) (2^{4} y^{4} + 2^{3} y^{3} + 2^{2} y^{2} + 2 y + 1)

= (2 y - 1) (2^{4} y^{4} + 2^{3} y^{3} + 2^{2} y^{2} + 2 y + 1)

Simplify

(2 y - 1) (2^{4} y^{4} + 2^{3} y^{3} + 2^{2} y^{2} + 2 y + 1) : (2 y - 1) (16 y^{4} + 8 y^{3} + 4 y^{2} + 2 y + 1)

= (2 y - 1) (16 y^{4} + 8 y^{3} + 4 y^{2} + 2 y + 1)

(\frac{x y ^{- 1}}{2 x ^{2} y ^{- 2}})^{- 3}

v (t)

\frac{2}{5} \cdot \frac{15}{4}

\frac{2}{1} + \frac{1}{3}

- 3 x y \cdot 4 x^{2}

What is factor 32y^5-1 ?
The solution to factor 32y^5-1 is (2y-1)(16y^4+8y^3+4y^2+2y+1)