Question 331322
factor the polynomial completely {{{x^3+8x^2+19x+12}}}
<pre><font size = 4 color = "indigo"><b>
All feasible rational zeros are {{{"" +- ""}}} the factors of 12 


 {{{"" +- 1}}}, {{{"" +- 2}}}, {{{"" +- 3}}}, {{{"" +- 4}}}, {{{"" +- 6}}}, {{{"" +- 12}}},   

Also there are no sign changes so there are no positive zeros.

So we begin by trying -1


 -1| 1  8 19  12
   |<u>   -1 -7 -12</u>
     1  7 12   0

So x = -1 is a zero, so therefore (x + 1) is a factor.
So now we have factored the original polynomial as

{{{x^3+8x^2+19x+12}}}

{{{(x+1)(x^2+7x+12)}}}

Now we factor the trinomial on the right and get:

{{{(x+1)(x+4)(x+3)}}}

That's it.

Edwin</pre>