A (univariate) polynomial is an expression of the form

where the *a*ᵢ s are constants (*a*ₙ ≠ 0) and *x* is a variable. The *a*ᵢ s are called the **coefficients** of the polynomial. *n* is the **degree** of the polynomial. *a*ₙ is the **leading coefficient** and *a*₀ 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 *r*₁, *r*₂, …, *r*ₙ 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 *r*ᵢ 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*-18*x³*+*kx²*+200*x*-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.