Given; f(x)=3x4+17x3+9x2−7x−10;
g(x)=x+5
g(x)=x+5=x−(−5)
If f(x) is completely divided by g(x)=(x−a) then by factor theorem (x−a) is factor of f(x) and f(a)=0.
∵f(−5)=3×(−5)4+17×(−5)3+9×(−5)2
−7×(−5)−10
=1875−2125+225+35−10
=2135−2135
=0
∵f(−5)=0
Yes, g(x)=(x+5) is a factor of f(x)=3x4+17x3+9x2−7x−10.