Factor Theorem

Since the theorem is a biconditional statement, the proof will consists of two parts.

Part I

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 equation at

x = k .

P (x) = (x - k) Q (x)

$x = k$

P (k) = (k - k) Q (k)

Subtract term

P (k) = 0 \cdot Q (k)

Zero Property of Multiplication

P (k) = 0

Consequently,

k

is a zero of

P (x) .

This completes the proof of the first part.

Consider the division of

P (x)

and

x - k .

\frac{P ( x )}{x - k}

The division can be rewritten in terms of the quotient and the remainder by using polynomial long division.

\frac{P ( x )}{x - k} = Q (x) + \frac{r}{x - k}

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 .

\frac{P ( x )}{x - k} = 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.