How do you factor #8x^3 - 4y^2-50y+25#?

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Answer 1 · 2015-05-15T19:19:28

I think the question is probably incorrect in mixing #x# and #y#.

It looks like the intended problem was how to factor #8y^3-4y^2-50y+25#, which you can do by grouping:

#8y^3-4y^2-50y+25#

#=(8y^3-4y^2)-(50y-25)#

#=4y^2(2y-1)-25(2y-1)#

#=(4y^2-25)(2y-1)#

The #4y^2-25# factor can be factored further.

Notice that #4y^2-25 = (2y)^2-5^2#, which is of the form #a^2-b^2#.

#a^2-b^2 = (a+b)(a-b)#

So #4y^2-25 = (2y)^2-5^2 = (2y+5)(2y-5)#

So #8y^3-4y^2-50y+25 = (2y+5)(2y-5)(2y-1)#