Question 1081686
 f(x)=3x^4 + 7x^3 + ax^2 + bx -14 
f(1)=0=3+7+a+b-14
a+b=4
f(-1)=-12=3-7+a-b-14
a-b=6
2a=10
a=5
b=1
{{{graph(300,300,-10,10,-10,10,3x^4+7x^3+5x^2-x-14)}}}
(x+2) is a factor
k=2
(x^2+x-2) divides into 3x^4 + 7x^3 + ax^2 + bx -14 and that quotient is
3x^2+4x+7
(x-1)(x+2)(3x^2+4x+7)
The roots of the quadratic term are complex
the graph of it is
{{{graph(300,300,-10,10,-10,10,3x^2+4x+7)}}}
The original polynomial has two real integer roots and two complex roots.