# Proof of roots of polynomials

1. Dec 14, 2008

### Mentallic

It is basic knowledge that if a polynomial P(x) of nth degree has a root or zero at P(a), then (x-a) is a factor of the polynomial. However, can this be proved? or is this more of a definition of roots of polynomials?

2. Dec 14, 2008

### liuixing

there are polynomials Q(x),R(x) ,such that degR(x)<deg(x-a) and P(x)=(x-a)Q(x)+R(x),if
P(a)=0,we have R(a)=0,for degR(x)<deg(x-a),R(x) is constant

3. Dec 14, 2008

If p(a) = 0, use the division algorithm to write p(x) = q(x)(x - a) + r(x), where either deg r(x) < deg (x - a) = 1, or r(x) = 0; in either case r(x) = b is a constant polynomial, so p(x) = q(x)(x - a) + b, so 0 = p(a) = q(a)(a - a) + b = b, which means p(x) = q(x)(x - a). Thus, x - a divides p(x).

4. Dec 14, 2008

### lurflurf

This can be proved. Let us say P(x) has a root at a if P(a)=0. Let us say (x-a) is a factor of the polynomial P(x) is there exist a polynomial Q(x) of degree one less such that P(x)=(x-a)Q(x).

Theorem Factor Theorem

Let P(x) be a polynomial of the nth degree and let P(a)=0, then (x-a) is a factor of P(x).
Proof
This proof will follow from the remainder theorem.

Theorem Remainder Theorem
Let P(x) be a polynomial of the nth degree , then P(x)=(x-a)Q(x)+R where Q(x) is a polynomial of degee one less and R is a constant.
Proof
let (x-a)^k be writtten in exppanded form for k=1,...,n
(x-a)^0=1
(x-a)^1=x-a
(x-a)^2=x^2-2ax+a^2
(x-a)^3=x^3-3ax^2+3a^2x-a^3
and so on
Since Q(x) polynommial we may write it in the form
Q(x)=a+bx+cx^2+dx^3+...
we can select the coefficients as follows
Express P(x) in standard form choose the coefficient of x^n in P(x) for the x^(n-1) term of Q(x).
Express [P(x)-(coeffcicent chosen in previous step)x^n] in standard form choose the coefficient of x^(n-1) for the x^(n-2) term of Q(x).
Express [P(x)-(coeffcicent chosen two steps back)x^n-(coeffcicent chosen one step back)x^(n-2)] in standard form choose the coefficient of x^(n-2) for the x^(n-3) term of Q(x).
Continue in this patern to find all coefficients of Q(x).
chose
R=P(0)+aQ(0)
QED

Proof Theorem Factor Theorem (continued)
Write P(x) as in the remainder theorem

P(x)=(x-a)Q(x)+R

The factor theorem will be proved if we show R=0.
set x=a
P(x)=(x-a)Q(x)+R
P(a)=(a-a)Q(x)+R
P(a)=(0)Q(x)+R
P(a)=R

Since a is a root of P(x),
P(a)=0 hence R=0.

QED

5. Dec 14, 2008

### Mentallic

Ahh I see We simply need to express the polynomial in the form $$P(x)=A(x)Q(x)+R(x)$$ and it all falls into place