Question

Prove that if p is a polynomial with degree n and k is a real number so that p(k) = 0,

then p(x)=(x-k)q(x) where q is a polynomial of degree n-1. Must use the fact that

a^n-b^n=(a-b)(a^(n-1) + a^(n-2)*b + ... + ab^(n-2) + b^(n-1)).

Answer 1

Theorem (Polynomial Division): Suppose d(x) and p(x) are nonzero polynomials where the degree of p is greater than or equal to the degree of d. There exist two unique polynomials, q(x) and r(x), such that p(x) = d(x) q(x) + r(x), where either r(x) = 0 or the degree of r is strictly less than the degree of d.

Theorem (The Remainder Theorem): Suppose p is a polynomial of degree at least 1 and c is a real number. When p(x) is divided by x − c the remainder is p(c).
The proof of Theorem 3.5 is a direct consequence of above theorem. When a polynomial is divided
by x − c, the remainder is either 0 or has degree less than the degree of x − c. Since x − c is degree
1, the degree of the remainder must be 0, which means the remainder is a constant. Hence, in either case, p(x) = (x − c) q(x) + r, where r, the remainder, is a real number, possibly 0. It follows
that p(c) = (c − c) q(c) + r = 0 · q(c) + r = r, so we get r = p(c) as required.

MAIN THEOREM:

If P is a polynomial of degree n, such that P(k) = 0, for any real number k. Then to prove that P(x) = (x - k)q(x), where q(x) is a polynomial of degree n - 1.

Proof: Since k is a zero of P, then P(k) = 0. In this case, The Remainder Theorem tells us the remainder when P(x) is divided by (x − k), namely p(k), is 0, which means (x − c) is a factor of p, i.e. P(x) = (x - k)q(x), where q(x) is a polynomial of degree k - 1.

We can prove this theorem in other way also....