Question 1057809
Yes.  There will be no more than three possible linear binomial factors, not including any multiplicity.


In this example, one factor will be real and the other two factors will be complex.   ( I ran the function as written in Google search engine to look at the graph.)




Using Rational Roots Theorem and working on paper,

Roots to check may be  -1, -19, 1, 19.


Synthetic Division check for Factor Theorem goes this way:
<pre>
-1    |     1    -7    11    19
      |          -1     8    -19
      |______________________________
            1    -8   19     0
</pre>
Meaning {{{f(x)=(x+1)(x^2-8x+19)}}}, so  the ONE rational real root is  NEGATIVE 1.


Finding the zero or roots for the quadratic factor for  {{{x^2-8x+19=0}}} using general solution formula of quadratic equation will give you, after simplification:
{{{system(4+i*sqrt(3),and,4-i*sqrt(3))}}}.



Try to finish as the product of binomials  (if that is what the factors may be called).
{{{f(x)=(x+1)(x-(4+i*sqrt(3)))(x-(4-i*sqrt(3)))}}}
which you could equivalently express as
{{{f(x)=(x+1)(x-4-i*sqrt(3))(x-4+i*sqrt(3))}}}.