Polynomial division question.

When the polynomial f(x) is divided by a quadratic polynomial the remainder will be a polynomial of degree less than 2. Show that, when f(x) is divided by (x-a)(x-b) . a does not equal b, the remainder has a constant term of zero, f and only if af(b) = bf(a)

Division algorithm: When a polynomial f(x) is divided by a polynomial g(x), then there exists unique polynomials q(x) and r(x) such that f(x) = q(x) g(x) + r(x) where deg r(x) < deg g(x)

Now let f(x) be given and let g(x) = (x - a)(x - b) where a =/= b. Since deg g(x) = 2 we know that r(x) = Mx + N for some constants M and N. We need to show that N = 0 if and only if a*f(b) = b*f(a). By Division algorithm, we know there exists unique polynomials q(x) and r(x) such that f(x) = q(x) (x - a)(x - b) + r(x) where deg r(x) < deg g(x).

First let's assume N = 0. Need to show that a*f(b) = b*f(a).
f(x) = q(x) (x - a)(x - b) + r(x)
f(x) = q(x) (x - a)(x - b) + Mx

Let x = a.

f(a) = q(a) (a - a)(a - b) + Ma = q(a) * 0 * (a - b) + Ma = Ma
Mulitply both sides by b. Then b*f(a) = Mab.

Let x = b.

f(b) = q(b) (b - a)(b - b) + Mb = q(b) * (b - a) * 0 + Mb = Mb
Multiply both sides by a. Then a*f(b) = Mab.

Thus, a*f(b) = Mab = b*f(a) as needed to show.

Now let a*f(b) = b*f(a) and try to show that N = 0.

f(x) = q(x) (x - a) (x - b) + r(x)
f(x) = q(x) (x - a) (x - b) + Mx + N

Let x = a.

f(a) = q(a)(a - a) (a - b) + Ma + N
f(a) = q(a)*0*(a-b) + Ma + N
f(a) = Ma + N
Multiply each term by b. Then b*f(a) = Mab + Nb.

Let x = b.

f(b) = q(b)(b - a)(b - b) + Mb + N
f(b) = q(b)(b-a)(0) + Mb + N
f(b) = Mb + N
Multiply each term by a. Then a*f(b) = Mab + Na

Since a*f(b) = b*f(a) we have that

Mab + Nb = Mab + Na

Subtract Mab + Nb from both sides.

Mab + Na - Mab - Nb = 0
Na - Nb = 0
N(a - b) = 0

Either N = 0 or a - b = 0

If a - b = 0, then a = b which is impossible since a =/= b. Thus, N = 0 as needed to show.

Done!