A (univariate) polynomial is an expression of the form
where the aᵢ s are constants (aₙ ≠ 0) and x is a variable. The aᵢ s are called the coefficients of the polynomial. n is the degree of the polynomial. aₙ is the leading coefficient and a₀ is the constant term. If the leading coefficient is 1, then the polynomial is a monic polynomial.
A root of the polynomial is a value for x such that P(x) = 0.
The Fundamental Theorem of Algebra states that every polynomial of degree n will have n roots. Note that the roots need not be real and there may be repeated roots. If r₁, r₂, …, rₙ are the roots of P(x), then
Expanding the product on the right hand side and comparing with the coefficients of P(x) gives
The expressions on the left hand sides are called the elementary symmetric polynomials in the rᵢ s
(SMO 2011) Let P(x) be a polynomial of degree 2010. Suppose P(n) = n/(1 + n) for n = 0, 1, 2, …., 2010. Find P(2012)
Solution: n/(1 + n) isn’t a polynomial, which makes it harder to understand. Consider the expression (1 + n)P(n) – n, which is a degree 2011 polynomial. It has 2011 distinct roots n = 0, 1, 2, …, 2010, hence it must be of the form an(n – 1)(n – 2)…(n – 2010) for some constant a. Substituting n = -1, we find that a = 1/2011! . Substituting n = 2012, we find that P(n) = 0.
(SMO 2012 Open) The product of two of the four roots of the quartic equation x⁴-18x³+kx²+200x-1984=0 is -32. Determine the value of k.
Let a, b, c, d be the four roots of the quartic, with ab = -32. Then
Hence cd = -1984/-32 = 62. -200 = cd(a + b) + ab(c + d) = 62(a + b) – 32(c + d). Solving this together with 18 = (a + b) + (c + d) gives a + b = 4 and c + d = 14. Then k = -32 + ac + ad + bc + bd + 62 = (a + b)(c + d) + 30 = 86.