Polynomial function

In mathematics, a polynomial function of degree $n$ is a function of the form:

$f(x)=a_{n}{x^{n}}+a_{n-1}{x^{n-1}}+a_{n-2}{x^{n-2}}+...+a_{0}\,$

where $n$ is an integer, and $a_{n},a_{n-1},a_{n-2},...,a_{0},\,$

are real numbers, and $a_{n}=O\,$

The Factor Theorem

Let P(x) be a polynomial. If P(c)= 0, where c is a real number, the x-c is a factor of P(x). Conversely, if x-c is a factor of P(x), then P(c)= 0

The Remainder Theorem states That when the polynomial P(x) is divided by x-c, the remainder is P(c).

For example, when $P(x)=x^{3}-x^{2}-4x+4\,$ is divided by $x-2\,$ , the remainder is 0; that is, $P(2)=0\,$ .

The same result is obtained using synthetic division File:Synthetic 2.jpg Notice That P(c) in this case is 0. Since P(c) is 0,the equation becomes: File:Factor 2.jpg

Since the theorem has a converse, the proof consists of two parts.

a. If (x-c) is a factor of P(x), then P(c) = 0 .

b. If P(c) = 0, then (x-c) is a factor of P(x).

Proof for a:

Suppose (x-c) is a factor of P(x),then $P(x)=(x-c)\bullet Q(x)\,$ . Since the equation is an identity and is true for any value of x, the it must be true for x = c

Then,

$P(c)=(c-c)\bullet Q(c)\,$

$P(x)=0\bullet Q(c)=0\,$

Proof for b:

Suppose P(c)= 0. By the Remainder Theorem, when P(x) is divided by (x-c), the remainder R = P(c)= 0.

Then,

$P(c)=(x-c)\bullet Q(x)+0\,$

$P(c)=(x-c)\bullet Q(x)\,$

Therefore,(x-c) is a factor of P(x)

References

Soledad Jose-Diloa,Ed.d.,Fernando B. Orines & Julieta G. Bernabe,"Advanced Algebra (Trigonometry and Statistics)" Textbook for Fourth Year ISBN 971-07-2227-1

External links

The Factor Theorem

References

See also

External links