How do you factor $x^3+8x^2+19x+12$ ?

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Answer 1 · 2016-04-27T21:49:03

$x^3+8x^2+19x+12 = (x+1)(x+3)(x+4)$

$f(x) = x^3+8x^2+19x+12$

Note that the coefficients of $f(-x) = -x^3+8x^2-19x+12$ add up to $0$ .

So $f(-1) = 0$ and $(x+1)$ is a factor:

$x^3+8x^2+19x+12 = (x+1)(x^2+7x+12)$

Then notice that $3+4=7$ and $3xx4$ =12#

Hence:

$x^2+7x+12 = (x+3)(x+4)$

Putting it all together:

$x^3+8x^2+19x+12 = (x+1)(x+3)(x+4)$

How do you factor x^3+8x^2+19x+12 x^3+8x^2+19x+12?