Division of polynomial with cubic divisor.?

Question

A polynomial represented by f has a constant term 5, and f(3) = f(-3) = 6. What is the remainder when f is divided by x^3 - 9x?

I've managed to factorise the divisor to x(x-3)(x+3) which seems to match up with the values above. After that I have no idea.

I managed to get (1/9)x^2 + 5, but my friend asked me - why must the divisor be quadratic?

Answer 1

Let d(x) = x^3-9x. Then f(x) = d(x)q(x) + r(x), where q is the quotient polynomial and r(x) is a quadratic polynomial. If you substitute 0, -3, or 3 for x, d(x) becomes 0, because of your factorization. So f(x) = r(x) for these choices of x. This means that r(x) is that quadratic polynomial that gives 6 when x = -3, 5 when it gives 0, and 6 when it gives 3. This tells you that f(x) = ax^2+bx+5. Plug in x = 3. 9a+3b+5 = 6. Plug in x = -3. You get 9a-3b+5 = 6. This is a system of two equations in two unknowns which will give you your desired remainder. Hope this helps.

Answer 2

The remainder is a polynomial of degree 2 (since all higher powers are divided out by the x^3), so it has the form

R = ax^2 + bx + c

Since x^3 - 9x = x(x-3)(x+3) is zero when x = -3, 0 or 3, the remainder must have the same values for these x as the original polynomial, namely 6, 5 and 6.

It follows that
R(-3) = 9a - 3b + c = 6
R(0) = 0a + 0b + c = 5
R(3) = 9a + 3b + c = 6

from which it is clear that c = 5, b = 0 and a = 1/9. Therefore the remainder is

R = 1/9 x^2 + 5


NOTE: why is R quadratic? *By definition*, the remainder of P / Q given by R such that

A Q + R = P

with A a polynomial and R a polynomial of degree less than that of Q. Since in your case Q is cubic, R must be quadratic (or lower degree).