Is it possible to factor #y= 2x^3-50x #? If so, what are the factors?

1 Answer
Mar 9, 2018

Answer:

#y=2x(x+5)(x-5)#

Explanation:

Well, we can already see that both terms have an #x#, and are a multiple of #2# so we can take #2x# out to get #y=2x(x^2-25)#

Difference of two squares tell us that #a^2-b^2=(a+b)(a-b)#.

#x^2-25=(x+5)(x-5)# since #x^2=(x)^2# and #25=5^2#

This gives us #y=2x((x+5)(x-5))=2x(x+5)(x-5)#