How do you factor $(w^2-1)^2-23(w^2-1)+120$ ?

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Answer 1 · 2016-04-28T12:56:07

$(w^2-1)^2-23(w^2-1)+120=(w+3)(w-3)(w+4)(w-4)$

$(w^2-1)^2-23(w^2-1)+120$

= $(w^2-1)^2-15(w^2-1)-8(w^2-1)+120$

= $(w^2-1)((w^2-1)-15)-8((w^2-1)-15)$

= $((w^2-1)-8)((w^2-1)-15)$

= $(w^2-9)(w^2-16)$

but as $a^2-b^2=(a+b)(a-b)$ , above is equal to

$(w+3)(w-3)(w+4)(w-4)$

How do you factor (w^2-1)^2-23(w^2-1)+120(w^2-1)^2-23(w^2-1)+120?