How do you factor completely: #2x^3 + 14x^2 + 4x + 28#?

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Answer 1 · 2015-07-24T01:06:24

Separate out the common scalar factor, then factor by grouping to find:

#2x^3+14x^2+4x+28 = 2(x^2+2)(x+7)#

#2x^3+14x^2+4x+28#

#=2(x^3+7x^2+2x+14)#

#=2((x^3+7x^2)+(2x+14))#

#=2(x^2(x+7)+2(x+7))#

#=2(x^2+2)(x+7)#

#x^2+2# has no linear factors with real coefficients since #x^2+2 >= 2 > 0# for all #x in RR#.

If you really want, you can factor #x^2+2 = (x+isqrt(2))(x-isqrt(2))#

and hence:

#2x^3+14x^2+4x+28 = 2(x+isqrt(2))(x-isqrt(2))(x+7)#