In this section we discuss some bounds for the number of positive roots, number of negative roots and number of nonreal complex roots for a polynomial over. These bounds can be computed using a powerful tool called “Descartes Rule”.

**Descartes Rule**

In this section we discuss some bounds for the number of positive
roots, number of negative roots and number of nonreal complex roots for a
polynomial over ¡ . These bounds can be
computed using a powerful tool called “Descartes Rule”.

To discuss the rule we first introduce the concept of **change of sign in the
coefficients **of a polynomial.

Consider the polynomial.

2*x*^{7} - 3*x*^{6} - 4*x*^{5} + 5*x*^{4} + 6*x*^{3} - 7*x *+ 8

For this polynomial, let us denote the sign of the coefficients
using the symbols ‘ + ’ and ‘ − ’as

+, -, -, +, +, -, +

Note that we have not put any symbol corresponding to *x*^{2}. We further note that
4 changes of sign occurred (at *x*^{6} , *x*^{4}, *x*^{1} and *x*^{0} ).

A **change of sign in the coefficients is said to occur **at
the *j *^{th} power of *x *in a polynomial *P*(*x*)
, if the coefficient of *x *^{j }^{+}^{1} and the
coefficient of *x*^{j}* *(or) also coefficient of *x *^{j }^{−}^{1} coefficient of *x
*^{j}* *are of different
signs. (For zero coefficient we take the sign of the immediately preceding
nonzero coefficient.)

From the number of sign changes, we get some information about the
roots of the polynomial using

**Descartes Rule**. As the proof is beyond the scope of the
book, we state the theorem without proof.

If *p *is the number of positive zeros of a polynomial *P*(*x*)
with real coefficients and *s *is the number of sign changes in
coefficients of *P*(*x*), then *s *− *p *is a
nonnegative even integer.

The theorem states that the number of positive roots of a
polynomial *P*(*x*) cannot be more than the number of sign changes in
coefficients of *P*(*x*) . Further it says that the difference
between the number of sign changes in coefficients of *P*(*x*) and
the number of positive roots of the polynomial *P*(*x*) is even.

As a negative zero of *P*(*x*) is a positive zero of *P*(−*x*) we may use the theorem
and conclude that
the number of negative
zeros of the polynomial *P*( *x*) cannot be more than the number of
sign changes in coefficients of *P*(− *x*) and the difference between the number of sign changes
in coefficients of *P*(− *x*) and the number of negative zeros of the polynomial *P*(
*x*) is even.

As the multiplication of a polynomial by *x** ^{k}*, for some positive
integer

We note that nothing is stated about 0 as a root, in Descartes
rule. But from the very sight of the polynomial written in the customary form,
one can say whether 0 is a root of the polynomial or not. Now let us verify
Descartes rule by means of certain polynomials.

The polynomial *P *(
*x*) = (*x *+1)(*x *−1)(*x *− 2)(*x *+ *i*)(*x *− *i*) has the zeros
−1, 1, 2, − *i*, *i *.
The polynomial, in the customary form is *x*^{5} − 2*x*^{4} − *x *+ 2 .This polynomial *P*(*x*)
has 2 sign changes, namely at fourth and zeroth powers. Moreover,

*P*(−*x*) = −*x*^{5} − 2*x*^{4} + *x *+ 2

has one sign change. By our Descartes rule, the number of
positive zeros of the polynomial *P*(*x*) cannot be more than 2; the
number of negative zeros of the polynomial *P*(*x*) cannot be more
than 1. Clearly 1 and 2 are positive zeros, and −1 is the negative zero for the polynomial, *x*^{5} − 2*x*^{4} − *x *+ 2 , and hence the
bounds 2 for positive zeros and the bound 1 for negative zeros are attained. We
note that *i *and −*i *are neither positive nor negative.

We know (*x *+
2)(*x *+ 3)(*x *+ *i*)(*x *− *i*) is a
polynomial with roots −2,
−3, −*i*, *i *. The
polynomial, say *P*(*x*) , in the customary form is *x*^{4} + 5*x*^{3} + 7*x*^{2} + 5*x *+ 6 .

This polynomial *P*(*x*) has no sign change and *P*(−*x*) = *x*^{4} − 5*x*^{3} + 7*x*^{2} − 5*x *+ 6 has 4 sign changes.
By Descartes rule, the polynomial *P*(*x*) cannot have more than 0
positive zeros and the number of negative zeros of the polynomial *P*(*x*)
cannot be more than 4 .

As another example, we consider the polynomial.

*x*^{n}* *−^{n}*C*_{1} *x*^{n}^{−}^{1} +^{n}*C*_{2} *x*^{n}^{−}^{2} −^{n}*C*_{3} *x*^{n}^{−}^{3} +….+ (−1)^{n}^{−}^{1} *n C*_{(n}_{−}_{1)} *x *+ (−1)^{n}* *.

This is the expansion of (*x *−1)^{n}* *. This polynomial has *n
*changes in coefficients and *P*(−*x*) has no change of sign in coefficients. This shows that the
number of positive zeros of the polynomial cannot be more than *n *and the
number of negative zeros of the polynomial cannot be more than 0. The statement
on negative zeros gives a very useful information that the polynomial has no
negative zeros. But the statement on positive zeros gives no good information about
the positive zeros, though there are exactly *n *positive zeros; in fact,
it is well-known that for a polynomial of degree *n *, the number of zeros
cannot be more than *n *and hence the number of positive zeros cannot be
more than *n *.

Using the Descartes rule, we can compute a lower bound for the
number of imaginary roots. Let *m *denote the number of sign changes in
coefficients of *P*(*x*) of degree *n*; let *k *denote the
number of sign changes in coefficients of *P*(−*x*) . Then there are at
least *n *−
(*m *+ *k *) imaginary
roots for the polynomial *P*(*x*) . Using the other conclusion of the
rule, namely, the difference between the number of roots and the corresponding sign changes is
even, we can sharpen the bounds in particular cases.

Show that the polynomial 9*x*^{9} + 2*x*^{5} − *x*^{4} − 7*x*^{2} + 2 has at least six
imaginary roots.

Clearly there are 2 sign changes for the given polynomial *P*(*x*)
and hence number of positive roots of *P *( *x*) cannot be more than two. Further,
as *P*(-*x*) = -9*x*^{9} - 2*x*^{5} - *x*^{4} - 7*x*^{2} + 2, there is one sign change for *P*(-*x*)
and hence the number of negative roots cannot be more than one. Clearly 0 is
not a root. So maximum number of real roots is 3 and hence there are atleast
six imaginary roots.

**Remark: **From the above discussion we note that the Descartes rule gives
only upper bounds for the number of positive roots and number of negative
roots; the Descartes rule neither gives the exact number of positive roots nor
the exact number of negative roots. But we can find the exact number of
positive, negative and nonreal roots in certain cases. Also, it does not give
any method to find the roots.

Discuss the nature of the roots of the following polynomials:

(i) *x*^{2018} +1947*x*^{1950} +15*x*^{8} + 26*x*^{6} + 2019

(ii) *x*^{5} -19*x*^{4} + 2*x*^{3} + 5*x*^{2} +11

Let *P*(*x*) be the polynomial under consideration.

(i) The number of sign changes for *P*(*x*) and *P*(-*x*)
are zero and hence it has no positive roots and no negative roots. Clearly zero
is not a root. Thus the polynomial has no real roots and hence all roots of the
polynomial are imaginary roots.

(ii) The number of sign changes for *P*(*x*) and *P*(-*x*)
are 2 and 1 respectively. Hence it has at most two positive roots and at most
one negative root.Since the difference between number of sign changes in
coefficients of *P*(-*x*) and the number of negative roots is even, we cannot have zero
negative roots. So the number of negative roots is 1. Since the difference between number of sign
changes in coefficient of *P*(*x*) and the number of positive roots
must be even, we must have either zero or two positive roots. But as the sum of
the coefficients is zero, 1 is a root. Thus we must have two and only two
positive roots. Obviously the other two roots are imaginary numbers.

Tags : Definition, Theorem, Solved Example Problems | Theory of Equations , 12th Mathematics : UNIT 3 : Theory of Equations

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