How do you factor $5y^6-5y^2$ ?

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Answer 1 · 2015-06-28T15:57:01

First separate the common factor $5y^2$ then factor the remaining well known quartic...

$5y^6-5y^2$

$=5y^2(y^4-1)$

$=5y^2(y-1)(y+1)(y^2+1)$

The difference of squares identity is:

$a^2-b^2 = (a-b)(a+b)$

We use this twice below...

First separate the common factor $5y^2$ then factor the remaining quartic:

$5y^6-5y^2$

$=5y^2(y^4-1)$

The quartic factor is a difference of squares...

$=5y^2((y^2)^2-1^2)$

$=5y^2(y^2-1)(y^2+1)$

The first quadratic factor is a difference of squares...

$=5y^2(y^2-1^2)(y^2+1)$

$=5y^2(y-1)(y+1)(y^2+1)$

The remaining quadratic factor has no linear factors with real coefficients since $y^2+1 >= 1 > 0$ for all $y in RR$

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