How do you factor completely: $2u^3 w^4 -2u^3$ ?

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Answer 1 · 2015-07-24T17:31:50

$2u^3w^4-2u^3 = 2u^3(w-1)(w+1)(w^2+1)$

First separate out the common factor $2u^3$ to get:

$2u^3w^4-2u^3 = 2u^3(w^4-1)$

Then use the difference of squares identity (twice):

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

$2u^3(w^4-1)$

$=2u^3((w^2)^2-1^2)$

$=2u^3(w^2-1)(w^2+1)$

$=2u^3(w^2-1^2)(w^2+1)$

$=2u^3(w-1)(w+1)(w^2+1)$

$w^2+1$ has no linear factors with real coefficients since $w^2+1 >= 1 > 0$ for all $w in RR$

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