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.