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164 ALGEBRA

2 . A constant K is called an upper bound for the real roots of equation (5.1.5.1) or the
◦
polynomial P n (x) if equation (5.1.5.1) has no real roots greater than or equal to K;in a
similar way, one defines a lower and an upper bound for positive and negative roots of an
equation or the corresponding polynomial.
Let
K 1 be an upper bound for the positive roots of the polynomial P n (x),
K 2 be an upper bound for the positive roots of the polynomial P n (–x),
n
K 3 > 0 be an upper bound for the positive roots of the polynomial x P n (1/x),
n
K 4 > 0 be an upper bound for the positive roots of the polynomial x P n (–1/x).
Then all nonzero real roots of the polynomial P n (x) (if they exist) belong to the intervals
(–K 2 ,–1/K 4 )and (1/K 3 , K 1 ).
Next, we describe three methods for finding upper bounds for positive roots of a
polynomial.
Maclaurin method. Suppose that the first m leading coefficients of the polynomial
(5.1.5.2) are nonnegative, i.e., a n > 0, a n–1 ≥ 0, ... , a n–m+1 ≥ 0, and the next coefficient is
negative, a n–m < 0.Then
B
1/m
K = 1 + (5.1.5.5)
a n
is an upper bound for the positive roots of this polynomial, where B is the largest of the
absolute values of negative coefficients of P n (x).
Example 3. Consider the fourth-degree equation from Example 2. In this case, m = 2, B = 36 and
1/2 4 2
formula (5.1.5.5) yields K = K 1 = 1+(36/9) = 3. Now, consider the polynomial P 4(–x)= 9x –9x +36x+1.
4
4
3
2
Its positive roots has the upper bound K 2 = 1+(9/9) 1/2 = 2. For the polynomial x P 4(1/x)= x –36x –9x +9,
4
4
2
3
we have m = 1, K 3 = 1+36 = 37. Finally, for the polynomial x P 4(–1/x)= x +36x –9x +9,we have m = 2,
k 4 = 1 + 9 1/2 = 4. Thus if P 4(x) has real roots, they must belong to the intervals (–2,–1/4)and (1/37, 3).
Newton method. Suppose that for x = c, the polynomial P n (x) and all its derivatives

P (x), ... , P n (n) (x) take positive values. Then c is an upper bound for the positive roots
n
of P n (x).
Example 4. Consider the polynomial from Example 2 and calculate the derivatives
4 2
P 4(x)= 9x – 9x – 36x + 1,
3
P 4 (x)= 36x – 18x – 36,

2
P 4 (x)= 108x – 18,

P 4 (x)= 216x,

P 4 (x)= 216.
It is easy to check that for x = 2 this polynomial and all its derivatives take positive values, and therefore c = 2
is an upper bound for its positive roots.
A method based on the representation of a polynomial as a sum of polynomials. As-
suming a n > 0, let us represent the polynomial (5.1.5.4) (without rearranging its terms) as
the sum P n (x)= f 1 (x)+ ... + f m (x), where each polynomial f k (x)(k = 1, 2, ... , m)has a
positive leading coefficient and the sequence of its coefficients does not change sign more
than once. Suppose that for c > 0 all these polynomials are positive, f 1 (c)> 0, ... , f m (c)> 0.
Then c is an upper bound for the positive roots of P n (x).
Example 5. The polynomial
7 6 5 4 3 2
P 7(x)= x + 2x – 4x – 7x + 2x – 3x + ax + b (a > 0, b > 0)
can be represented as a sum of three polynomials
2
4
5
4
3
2
3
2
7
6
f 1(x)= x + 2x – 4x – 7x = x (x + 2x – 4x – 7), f 2(x)= 2x – 3x = x (2x – 3), f 3(x)= ax + b