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:

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

How do you factor completely: 2x^3 + 14x^2 + 4x + 282x^3 + 14x^2 + 4x + 28?