For some integer m, the polynomial $x^3-2011x+m$ has the three integer roots a, b, and c. The value of $\left|a\right|+\left|b\right|+\left|c\right|$  is equal to ___________

Starting off like the previous solution, we know that a + b + c = 0 and 
ab + bc + ac = - 2011.
Therefore, c = - b – a.
Substituting,  $ab+b(-b-a)+a(-b-a)=ab-b^2-ab-ab-a^2=-2011$.
Factoring the perfect square, we get :  $ab-{(b+a)}^2=-2011or{(b+a)}^2-ab=2011.$
Therefore, a sum (a + b) squared minus a product (a b) gives 2011.
We can guess and check different a + b’s starting with 45 since  ${44}^2<2011$.
${45}^2=2025$ therefore ab = 2025 – 2011 = 14.
Since no factors of 14 can sum to 45 (1+ 14 being the largest sum), a + b cannot equal 45.
${46}^2=2116$ making ab = 105 = 3*5*13.
5* 7+ 3 < 46 and 3 * 5 * 7 > 46 so 46 cannot work either.
We can continue to do this until we reach 49.
${49}^2=2401$ making ab = 390 = 2 * 3 * 5 * 13.
3 * 13 + 2 * 5 = 49, so one root is 10 and another is 39. The roots sum to zero, so the last root must be – 49.
 $|-49|+10+39=\boxed{098}$