Vieta’s Formula and Newton’s sum
Prajit Adhikari
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Introduction to Vieta’s Formula
Consider a polynomial P (x) of degree n,
P (x) = an xn + an−1 xn−1 + .... + a1 x + a0
with complex coefficients, and roots r1 , r2 , r3 , ....rn , and let Qj denote the j t h
elementary symmetric polynomial of the roots. The examples of elementary
symmetric polynomial can be seen below:
For n = 1 :
e1 (X1 ) = X1
For n = 2 :
e1 (X1 , X2 ) = X1 + X2 ,
e2 (X1 , X2 ) = X1 X2 .
For n = 3 :
e1 (X1 , X2 , X3 ) = X1 + X2 + X3 ,
e2 (X1 , X2 , X3 ) = X1 X2 + X1 X3 + X2 X3 ,
e3 (X1 , X2 , X3 ) = X1 X2 X3 .
For n = 4 :
e1 (X1 , X2 , X3 , X4 ) = X1 + X2 + X3 + X4 ,
e2 (X1 , X2 , X3 , X4 ) = X1 X2 + X1 X3 + X1 X4 + X2 X3 + X2 X4 + X3 X4 ,
e3 (X1 , X2 , X3 , X4 ) = X1 X2 X3 + X1 X2 X4 + X1 X3 X4 + X2 X3 X4 ,
e4 (X1 , X2 , X3 , X4 ) = X1 X2 X3 X4 .
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So, from Vieta’s formula, we have,
an−1
an
an−2
Q2 = r1 r2 + r1 r3 + · · · + rn−1 rn =
an
..
.
a0
Qn = r1 r2 r3 · · · rn = (−1)n .
an
Q1 = r1 + r2 + · · · + rn = −
For n=3, we have,
The roots r1 , r2 , r3 of the cubic polynomial P (x) = ax3 + bx2 + cx + d satisfy
b
r1 + r2 + r3 = − ,
a
2
r1 r2 + r1 r3 + r2 r3 =
c
,
a
d
r1 r2 r3 = − .
a
Problems
1. Find a quadratic whose roots are 5 and 8.
2. Find all polynomials
P (x) = xn + an−1 xn−1 + · · · + a0
such that ai = ±1 for all 0 ≤ i ≤ n − 1 satisfying the condition that all
roots of P (x) are real.
3. Let x1 , x2 , ..., x10 be the roots of the polynomial x10 + x9 + · · · + x + 1.
Find the value of
10
X
1
.
1
−
xn
n=1
4. The polynomial x3 −ax2 +bx−2010 has three positive integer roots. What
is the smallest possible value of a?
(A) 78
(B) 88
(C) 98
(D) 108
(E) 118
5. Let r, s, and t be the three roots of the equation
8x3 + 1001x + 2008 = 0.
Find (r + s)3 + (s + t)3 + (t + r)3 .
6. For certain real numbers a, b, and c, the polynomial
g(x) = x3 + ax2 + x + 10
has three distinct roots, and each root of g(x) is also a root of the polynomial
f (x) = x4 + x3 + bx2 + 100x + c.
What is f (1)?
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Newton’s sum
Newton’s identities, also known as Newton-Girard formulae, is an efficient way
to find the power sum of roots of polynomials without actually finding the roots.
If x1 , x2 , . . . , xn are the roots of a polynomial equation, then Newton’s identities
are used to find the summations like
n
X
xki = xk1 + xk2 + · · · + xkn .
i=1
It can be derived from the Vieta’s formula playing with the elementary
symmetric polynomial.
Consider a polynomial P (x) of degree n,
P (x) = an xn +an−1 xn−1 +· · ·+a1 x+a0 Let P (x) = 0 have roots x1 , x2 , . . . , xn .
Define the sum:
Pk = xk1 + xk2 + · · · + xkn .
Newton’s sums tell us that,
an P1 + an−1 = 0
an P2 + an−1 P1 + 2an−2 = 0
an P3 + an−1 P2 + an−2 P1 + 3an−3 = 0
..
.
(Define aj = 0 for j < 0.)
We also can write:
P1 = S1
P2 = S1 P1 − 2S2
P3 = S1 P2 − S2 P1 + 3S3
P4 = S1 P3 − S2 P2 + S3 P1 − 4S4
P5 = S1 P4 − S2 P3 + S3 P2 − S4 P1 + 5S5
..
.
where Sn denotes the n-th elementary symmetric sum.
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Problems:
1. consider the polynomial P (x) = x3 + 3x2 + 4x − 8. Let the roots of P (x)
be r, s and t. Find r2 + s2 + t2 and r4 + s4 + t4 .
2. Consider the polynomials P (x) = x6 − x5 − x3 − x2 − x and Q(x) =
x4 − x3 − x2 − 1. Given that z1 , z2 , z3 , and z4 are the roots of Q(x) = 0,
find P (z1 ) + P (z2 ) + P (z3 ) + P (z4 ).
3. Let sk denote the sum of the kth powers of the roots of the polynomial
x3 − 5x2 + 8x − 13. In particular, s0 = 3, s1 = 5, and s2 = 9. Let a, b,
and c be real numbers such that sk+1 = a sk + b sk−1 + c sk−2 for k = 2,
3, .... What is a + b + c?
4. Given the following system of equations:
x+y+z =1
2
x + y2 + z2 = 2
x3 + y 3 + z 3 = 3,
find the smallest positive integer value of n (¿ 3)n (¿3) such that xn +
y n + z n is an integer.
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References:
1. Brilliant
2. Art of Problem Solving
3. Wikipedia
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