A polynomial f with integer coefficients is written on the blackboard. The teacher is a mathematician who has 3 kids named Andrew, Beth and Charles. Andrew, who is 7, is the youngest, and Charles is the oldest. When evaluating the polynomial on his kids' ages he obtains:
f(7) = 77
f(b) = 85, where b is Beth's age,
f(c) = 0, where c is Charles' age.
How old is each child?
(In reply to
question by Steven Lord)
If there is some polynomial that passes through a given finite set of points then there's an infinite number of polynomials that pass through those of points.
I will illustrate using the problem.
The given points are (7,77), (9,85) and (14,0). By polynomial fitting f(x)=-3x^2+52x-140 is the polynomial of lowest degree which passes through those points.
Construct g(x) = (x-7)*(x-9)*(x-14)+1 = x^3-30x^2+287x-881. In its semi-factored form it is obvious that g(7)=g(9)=g(14)=1.
Now take h(x) = f(x)*g(x) = -3x^5+142x^4-2561x^3-21767x^2-85992x+123340.
Because of how g is constructed then h(7)=f(7)=77, h(9)=f(9)=85, and h(14)=f(14)=0. h(x) is then a fifth degree polynomial passing through the points and h has integer coefficients.
g(x) can be generalized into G(x)=(x-7)^i*(x-9)^j*(x-14)^k+1, where i, j, and k are nonnegative integers. My example was i=j=k=1 and the trivial case h(x)=f(x) occurs when i=j=k=0, making G(x) the constant polynomial G(x)=1.
Of course then it follows that there are an infinite number of G(x), and thus an infinite number of h(x).
re: question
