The Art and Craft of Problem Solving

5.4 POLYNOMIALS^167

has eight zeros, but only five distinct zeros. The zero 7 appears with multiplicity 3 and

the zero - 6 has multiplicity 2.

Here is an example that uses an analysis of zeros combined with clever substitution

and the define a function tool that you encountered in Example 3.3.8 on page 89.

Example 5.4.2 (USAMO 1975) If P(x) denotes a polynomial of degree n such that

P(k) = k/(k+ 1) for k = 0, 1,2, ... ,n, determine P(n+ 1).

Solution: Go back to the Factor Theorem. A reinterpretation of this theorem from
a problem solver's perspective is

To know the zeros of a polynomial is to know the polynomial.

In other words, if you don't know the zeros of the polynomial under consideration,

either expend some effort to find them, or shift your focus to a new polynomial whose

zeros are apparent. In our case, knowing that P(k) = k/ (k + 1) does not tell us anything

about the zeros of P(x), since k/(k+ 1) is neither zero nor a polynomial. We eliminate

both difficulties simultaneously by mUltiplying by (k + 1) and subtracting:

(k+ I) P(k) - k = 0.

We have information about the zeros of another polynomial, namely the (n + 1 )-degree

polynomial

Q(x) := (x+ I)P(x) -x.

Clearly the zeros of Q(x) are just 0, (^1) ,2, ... , n, so we can write

(x+ 1 )P(x) -x = Cx(x - I)(x - 2)··· (x - n),

where C is a constant that must be determined. Since the above equation is an identity,

true for all x values, we can plug in any convenient value. The values x = 0, 1, ... , n

don't work, since they make the right-hand side equal to zero. The left-hand side

contains the troublesome term (x + 1 )P(x), so clearly our choice should be x = -1.

Plugging this in yields

1 =C(-I)(-2)(-3)···(-(n+I)),

so

( _1)n+!

C = -'::--'--:-:--:-

(n+I)!

Finally, we can plug in x = n + 1, and we have

so

( 1 )n+! ( 1 )n+! (n+ 2)P(n+ 1) -n-I = (

: + I)!

(n+ I)n··· 1 = (