by Integr88 » Thu 14 April, 2011 15:49
to factor `32u^3-2u^3x^4`
firs factor `u^3` because it is in both terms and we have 32 in first therm and 2 on the second so we can have 2 as common factor so use `2u^3` as factor and we have:
`2u^3(16-x^4)`
Now `16-x^4` can be difference of square meaning it can be factored as `(4-x^2)(4+x^2)` so we write:
`2u^3(4-x^2)(4+x^2)`
now for `4-x^2` is also difference of square so we can write it as `(2-x)(2+x)` now writing this we have:
`2u^3(2-x)(2+x)(4+x^2)`