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

Question

1137 views around the world

You can reuse this answer
Creative Commons License

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)#