How do you factor completely: $4x^3 + 12x^2 + 3x + 9$ ?

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Answer 1 · 2015-07-11T23:14:42

$4x^3+12x^2+3x+9=(4x^2+3)(x+3)$

This has no simpler linear factors with real coefficients, since $4x^2+3 >= 3 > 0$ for all $x in RR$ .

$4x^3+12x^2+3x+9$

$=(4x^3+12x^2)+(3x+9)$

$=4x^2(x+3) + 3(x+3)$

$=(4x^2+3)(x+3)$

The remaining quadratic factor $(4x^2+3)$ cannot be factored into linear factors with real coefficients, since $4x^2+3 >= 3 > 0$ for all $x in RR$ .

How do you factor completely: 4x^3 + 12x^2 + 3x + 94x^3 + 12x^2 + 3x + 9?