Complex numbers and polynomials - University College Cork

Example 6. Suppose x 1 , x 2 , x 3 are three distinct real numbers and y 1 , y 2 , y 3 are anygiven set of real numbers, determine the equation of the poly that passes through thepoints(x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ).Let the desired poly be p(x) = ax 2 + bx + c. Then p(x i ) = y i , i = 1, 2, 3, i.e.,ax 2 1 + bx 1 + c = y 1 ,ax 2 2 + bx 2 + c = y 2 ,ax 2 3 + bx 3 + c = y 3 ,three equations for three unknowns a, b, c, which can be solved by the process ofelimination. □Another approach is to solve three similar, but easier problems by letting two of they’s be 0 and the other 1. Then, say, p 1 (x 2 ) = p 1 (x 3 ) = 0, p 1 (x 1 ) = 1. The first twoconditions say that x 2 , x 3 are roots of the quadratic p 1 . Sop 1 (x) = a(x − x 2 )(x − x 3 ).The third condition now fixes a and so p 1 :a =1(x 1 − x 2 )(x 1 − x 3 ) , p 1(x) = (x − x 2)(x − x 3 )(x 1 − x 2 )(x 1 − x 3 ) .Cycling the points x 1 , x 2 , x 3 we determine p 2 , p 3 asp 2 (x) = (x − x 1)(x − x 3 )(x 2 − x 1 )(x 2 − x 3 ) , p 3(x) = (x − x 1)(x − x 2 )(x 3 − x 1 )(x 3 − x 2 ) .Thenp(x) = y 1 p 1 (x) + y 2 p 2 (x) + y 3 p 3 (x)solves the problem.This can be extended to deal with the general problem of finding a poly of degree nthat passes through n + 1 points (x i , y i ), i = 1, 2, . . . , n with distinct x-coordinates.7.4 Some Olympiad style problems1. The value of the polynomial x 2 +x+41 is a prime number for x = 0, 1, 2, . . . , 39.The value of the polynomial x 2 − 81x + 1681 is a prime number for x =0, 1, 2, . . . , 80. These examples might suggest that there is a poly p, withinteger coeffs, such that p(n) is a prime for every integer n ≥ 0. This is false.No such poly exists. [Hint: If such a poly f exists, p = f(0) is a prime andp|f(mp), m = 0, 1, . . ., i.e., Hence f(mp) = p, m = 0, 1, 2, . . ., which is absurd.]22