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5.1. POLYNOMIALS AND ALGEBRAIC EQUATIONS 165

(in the first two polynomials the sign of the sequence of coefficients changes once, and in the last polynomial
the coefficients do not change sign). It is easy to see that all these polynomials are positive for x = 2. Therefore,
c = 2 is an upper bound for the positive roots of the given polynomial.

5.1.5-5. Theorems on the number of real roots of polynomials.
The number all negative roots of a polynomial P n (x) is equal to the number of all positive
roots of the polynomial P n (–x).
1 . The exact number of positive roots of a polynomial whose coefficients form a sequence
◦
that does not change sign or changes sign only once can be found with the help of the
Descartes theorem.
DESCARTES THEOREM. The number of positive roots (counted according to their mul-
tiplicity) of a polynomial P n (x) with real coefficients is either equal to the number of sign
alterations in the sequence of its coefficients or is by an even number less.
Applying the Descartes theorem to P n (–x), we obtain a similar theorem for the negative
roots of the polynomial P n (x).
Example 6. Consider the cubic polynomial
3
P 3(x)= x – 3x + 4.
Its coefficients have the signs + – +, and therefore we have two alterations of sign. Therefore, the number of
3
positive roots of P 3(x) is equal either to 2 or to 0. Now, consider the polynomial P 3(–x)=–x + 2x + 1.The
sequence of its coefficients changes sign only once. Therefore, the original equation has one negative root.
2 . A stronger version of the Descartes theorem. Suppose that all roots of a polynomial
◦
P n (x) are real ; then the number of positive roots of P n (x) is equal to the number of sign
∗
alterations in the sequence of its coefficients, and the number of its negative roots is equal
to the number of sign alterations in the sequence of coefficients of the polynomial P n (–x).
Example 7. Consider the characteristic polynomial of the symmetric matrix
1
–2 – x 1
3

1 1 – x 3 =–x + 14x + 20,
P 3(x)=
1 3
1 – x
which has only real roots. The sequence of its coefficients changes sign only once, and therefore it has a single
positive root. The number of its negative roots is equal to two, since this polynomial has three nonzero real
roots and only one of them can be positive.
◦
3 . If two neighboring coefficients of a polynomial P n (x) are equal to zero, then the roots
of the polynomial cannot be all real (in this case, the stronger version of the Descartes
theorem cannot be used).
4 . The number of real roots of a polynomial P n (x) greater than a fixed c is either equal to
◦
the number of sign alterations in the sequence P n (c), ... , P n (n) (c)or isbyanevennumber
less. If all roots of P n (x) are real, then the number of its roots greater than c coincides with
the number of sign alterations in the sequence P n (c), ... , P n (n) (c).
Example 8. Consider the polynomial
4 3 2 2 2
P 4(x)= x – 3x + 2x – 2a x + a .
2 2

For x = 1,we have P 4(1)=–a , P 4 (1)= –1 – 2a , P 4 (1)=–2, P 4 (1)= 6, P 4 (1)= 24. Thus, there is a single

sign alteration, and therefore the polynomial has a single real root greater than unity.
This is the case, for instance, if we are dealing with the characteristic polynomial of a symmetric matrix.
∗