How do you factor #y= 8t^4+ 32t ^3 + t + 4 # ?

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Answer 1 · 2015-12-24T05:42:36

#y=(4t^2-2t+1)(2t+1)(t+4)#

Factor by grouping. (Split into two groups of two.)

#y=(8t^4+32t^3)+(t+4)#

#y=8t^3(t+4)+1(t+4)#

Factor out a common #(t+4)#.

#y=(8t^3+1)(t+4)#

Recognize that #(8t^3+1)# is a sum of cubes.

This is the following identity:

#a^3+b^3=(a+b)(a^2-ab+b^2)#

Thus, #a=2t,b=1#, so

#8t^3+1=(2t+1)(4t^2-2t+1)#

Replace this in the factored equation:

#y=(4t^2-2t+1)(2t+1)(t+4)#