How do you factor #2p^3+6p^2+3p+9 #?

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Answer 1 · 2015-10-29T22:29:27

Factor by grouping to find:

#2p^3+6p^2+3p+9 = (2p^2+3)(p+3)#

#2p^3+6p^2+3p+9 = (2p^3+6p^2)+(3p+9)#

#=2p^2(p+3)+3(p+3)#

#=(2p^2+3)(p+3)#

#2p^2+3 >= 3 > 0# for all #p in RR#, so #(2p^2+3)# has no linear factors with Real coefficients.

If you allow Complex coefficients you can factor it as:

#(2p^2+3) = (sqrt(2)p-sqrt(3)i)(sqrt(2)p+sqrt(3)i)#

Hence:

#2p^3+6p^2+3p+9 = (sqrt(2)p-sqrt(3)i)(sqrt(2)p+sqrt(3)i)(p+3)#