Answer :
If this is the case, then you know by the polynomial remainder theorem that [tex]x-a[/tex] is a factor of [tex]p(x)[/tex], so there is some polynomial [tex]q(x)[/tex] such that
[tex]p(x)=(x-a)q(x)[/tex]
[tex]p(x)=(x-a)q(x)[/tex]