Polynomials II: Vegetables
Any field. Roots and coefficients in any field:
1. Given polynomials P and D (if D isn’t the zero polynomial) we can write
P = QD + S
where Q is some polynomial (“quotient”) and S (“remainder”) has degree less
than D.
2. r is a root of P  (x-r) divides P(x).
3. root r has multiplicity k  (x-r)k divides P(x), but (x-r)k+1 doesn’t
 r is a root of P, DP, D2P, …, Dk-1P, but not of DkP
4. P is “irreducible” if we can’t factor P(x) = Q(x) D(x) with Q and D both having
smaller degree than P.
5. P factors uniquely as a product of irreducibles (except for multiplication by constants)
6. If P has degree n, then P has at most n roots (even counting by multiplicity)
Complex numbers. Roots and coefficients in :
1. Any polynomial with degree ≥ 1 has a root in
.
2. Irreducible = linear
3. Every polynomial splits into linear factors.
4. We can count the roots using winding numbers.
5. If the coefficients are real, then the non-real roots come in conjugate pairs.
Any polynomial that splits into linear factors:
1. Can write P(x) as an ( x  r1 )( x  r2 ) ( x  rn ) . (Another way to write polynomials!)
2. Coefficients are symmetric functions of roots.
Real numbers. Roots and coefficients in
:
1. Irreducible = All linear, some quadratics (the ones without real roots)
2. P factors into linear and quadratic factors, nothing worse
3. Between any two real roots of P there is a real root of DP
4. If P splits into linear factors (“has all real roots”) then so does DP.
(Harder complex analog: If all the roots of P are in a convex polygon, so are all
roots of DP)
5. Rule of signs and better root-counting tricks
Rationals. Roots and coefficients in
or :
1. There are irreducibles of all degrees.
2. Monic, coefficients in  all the rational roots are really integers.
3. If P and D have integer coefficients, and if D is monic, we can write
P = QD + S
where Q and S also have integer coefficients, and S has lower degree than D.
(We need for D to be monic to make up for the fact that
is not a field.)
4. If P has integer coefficients, then r is a root  P(x) = (x-r) Q(x), where Q also has
integer coefficients.
Quadratic Equations:
1. When a is small, the usual formula for the roots of ax2+bx+c is numerically
dangerous. So, try:
2c
.
x
b  b2  4ac
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Polynomial Problems
Problem A. Show that if P and D have coefficients in , and if D is monic, then P can be
written as P = QD + S where S has smaller degree than D and both Q and S have
coefficients in . Conclude that if P has coefficients in
and an integer n is a
root of P, then
P(x) = (x-n) Q(x)
where Q also has coefficients in .
Problem B. Suppose that P has coefficients in , and
P(a) = P(b) = P(c) = P(d) = 12
for four distinct integers a, b, c, d. Is it possible that P(k) = 25 for some integer k?
Problem C (USAMO 1983). Suppose that
P( x)  x5  ax 4  bx3  cx 2  dx  e
and that
2a 2  5b.
Show that P does not have 5 real roots.
Problem D (USAMO 1984). Suppose that
P( x)  x 4  18x3  kx 2  200 x  1984
and that the product of two of the roots of P is -32. What is k?
Hint for D and E: If the roots are a, b, c, d, then note the special roles of ab, cd, a+b, and c+d.
Give them names; say s=ab, t=cd, p=a+b, q=c+d.
Problem E (USAMO?). If a and b are roots of x 4  x3  1  0, show that ab is a root of
x 6  x 4  x3  x 2  1  0.
Problem F. Left and Right play a game. In the polynomial equation,
x3 
x2 
x
0
first Left fills in a blank with a real number, then Right fills in a blank with a real
number, then Left fills in a blank with a real number. Left wins if the equation has
three distinct real roots. Can Left always win?
Problem G (Bogdan).
(A) Can one obtain the polynomial x by starting with f(x) = x2+x,
g(x) = x2-2, and combining them using only additions, subtractions, and multiplications?
(B) What if f(x) = x2+x and g(x) = x2+2 ?
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More Problems !
Problem H. Suppose that three of the roots of
x 4  px3  qx 2  rx  s  0
are tan A, tan B, and tan C where A, B, C are the angles of a plane triangle. Let’s
just call the fourth root x. Write x in terms of p, q, r, s.
Problem I. Solve exactly:
2 x 4  5 x3  x 2  5 x  2  0 .
Problem J. Show that
x 6  x 4  x3  x 2  1  0
can be solved exactly in radicals.
Problem K (1994 Putnam Exam). For which real numbers c is there a straight line that
intersects the curve
y  x 4  9 x3  cx 2  9 x  4
in four distinct points?
Problem L (MOSP). Show that none of the equations
x6  ax 4  bx 2  c  y 3
with a 3, 4,5 , b 4,5, ,12 , c 1, 2, ,8 have solutions in integers x, y.
Problem M (MOSP). Find all polynomials p(x) such that for all x,
( x  16) p(2 x)  16( x  1) p( x) .
Problem N (Galperin). If x, y, z are positive real numbers and
x 2  xy  y 2  16
3 y 2  z 2  27
3x 2  z 2  3xz  75,
then what is
xz  2 yz  3xy ?
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