Since the theorem is a , the proof will consists of two parts.
- Part I: If x−k is a factor of P(x), then P(k)=0.
- Part II: If P(k)=0, then x−k is a factor of P(x).
Part I
If
x−k is a factor of
P(x), then
P(x) can be written as the product of
x−k and a certain polynomial
Q(x).
P(x)=(x−k)Q(x)
Now, evaluate the at
x=k.
P(x)=(x−k)Q(x)
P(k)=(k−k)Q(k)
P(k)=0⋅Q(k)
P(k)=0
Consequently,
k is a zero of
P(x). This completes the proof of the first part.
Part II
Consider the division of
P(x) and
x−k.
x−kP(x)
The division can be rewritten in terms of the and the by using .
x−kP(x)=Q(x)+x−kr
By the Remainder Theorem, the remainder of the previous division is equal to
P(k). Since
P(k)=0, the remainder of the division is
0. Therefore, the rightmost term of the previous equation is
0.
x−kP(x)=Q(x)
Finally, multiply both sides of the equation by
x−k.
P(x)=(x−k)Q(x)
Consequently,
x−k is a factor of
P(x). This completes the proof of the second part.