Question 1089948
{{{2(x^3+5x^2+3x-9)=0}}}
and then by Rational Roots Theorem some possible roots to try testing are  -3, -1, 1, 3.


<pre>
1   |   1   5   3   -9
    |       1   6   9
    |_____________________
       1    6   9    0

     Meaning the next factor to break further if possible is  {{{x^2+6x+9}}}.
</pre>

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{{{2(x-1)(x^2+6x+9)=0}}}, and the quadratic factor is easily factored further:


{{{2(x-1)(x+3)(x+3)=0}}}