**1. The zeros of the polynomial**

**x**^{3} - 33x^{2} + 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

x^{3} + (-3a)x^{2} + (3a^{2}-b^{2})x + (b^{2}a-a^{3})

Equating coefficients, we get

-3a = -33

3a^{2}-b^{2} = 354

k = b^{2}a-a^{3}

Right away we see from the first equation that a=11

From the second equation, 3a^{2}=363, so we see b = 3 or -3

Then from the third equation, we see that k=99-1331=-1232

**2. Let f(x) = px**^{5}+qx^{4}+rx^{3}+sx^{2}+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 px^{5}+qx^{4}+rx^{3}+sx^{2}+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.