Answer:
Yes
Step-by-step explanation:
Use synthetic division
2 | 1 -8 14 -4
____2 -12_4__
1 -6 2 | 0
Thus, [tex](x^3-8x^2+14x-4)\div(x-2)=x^2-6x+2\:\text{R}0[/tex], so [tex](x-2)[/tex] is indeed a factor of f(x) since there's no remainder.
