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:

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

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