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

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Answer 1 · 2015-07-24T17:31:50

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

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

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

Then use the difference of squares identity (twice):

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

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

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

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

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

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

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

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