1. The zeros of the polynomial

x³ - 33x² + 354x + k

are in arithmetic progression.  What is the value of k?

2. Let f(x) = px⁵+qx⁴+rx³+sx²+tx+1 be a polynomial such that f(1)=4, f(2)=11 and all the coefficients p,q,r,s and t are integers. Prove that the equation f(x)=0 has no integer roots.

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1. The zeros of the polynomial

x³ - 33x² + 354x + k

are in arithmetic progression.  What is the value of k?

-1232.  Here's how to solve it:

Let the zeros be a-b, a, and a+b.  Then the polynomial is

(x-a+b)(x-a)(x-a-b),

which multiplies out to

x³ + (-3a)x² + (3a²-b²)x + (b²a-a³) 

Equating coefficients, we get

-3a = -33
3a²-b² = 354
k = b²a-a³ 

Right away we see from the first equation that a=11
From the second equation, 3a²=363, so we see b = 3 or -3
Then from the third equation, we see that k=99-1331=-1232

2. Let f(x) = px⁵+qx⁴+rx³+sx²+tx+1 be a polynomial such that f(1)=4, f(2)=11 and all the coefficients p,q,r,s and t are integers. Prove that the equation f(x)=0 has no integer roots.

Suppose there is an integer root, r.  Then (x-r) is a factor of px⁵+qx⁴+rx³+sx²+tx+1, which means r must be a factor of 1.  The only integers that divide 1 are 1 and -1, and we know from f(1)=4 that 1 is not a root of f(x).  So r=-1.

Now we know the value of f at four points:

f(-1)=0,
f(0)=1 (which we know by setting x=0),
f(1)=4, and
f(2)=11.

We can find successive differences of f for consecutive integer values of x, to learn something about the roots of this polynomial.  The first differences are:

f(0)-f(-1)=1
f(1)-f(0)=3
f(2)-f(1)=7

The second differences are:

(f(1)-f(0)) - (f(0)-f(-1)) = 2
(f(2)-f(1)) - (f(1)-f(0)) = 4

The third difference -- that is, the difference of the differences of the differences of the values of f -- is:

((f(2)-f(1))-(f(1)-f(0)))-((f(1)-f(0))-(f(0)-f(-1))) = 2

Simplifying the left hand side of this equation,

-f(-1)+3f(0)-3f(1)+f(2)=2

(You might recognize the binomial coefficients here.)  Expressing the left hand side in terms of p, q, r, s, and t, and adding:

 -f(-1)=   p   -q  +r  -s  +t -1
 3f(0) =                       3
-3f(1) = -3p  -3q -3r -3s -3t -3
  f(2) = 32p +16q +8r +4s +2t +1
         -----------------------
         30p +12q +6r            =2 

6(5p+2q+r) = 2, which contradicts the statement in the problem that the coefficients are integers.

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