Descartes Rule of signs

This rule says that in a polynomial f(x), the maximum number of positive real roots can be ascertained by counting the number of sign changes in its coefficients. The actual

Number of positive real roots may the maximum number or a number reduced by a multiple of two

For example in f(x) = 3x⁶+x⁵-x⁴+3x³+2x² - x +1, there are 4 changes of sign +, -, + , -, +

First sign change is + to –

Second is – to +

Third is + to –

Fourth is - to +

Hence it can have a maximum of 4 positive real roots or 2 or none

Similarly for negative real roots, we count the sign changes in f(-x)

Here f(-x)= 3x⁶ – x⁵ – x⁴- 3x³+ 2x² +x +1. There are 2 sign changes +, -, +,

Hence it can have a maximum number of 2 negative real roots or none

[Because the Complex roots occur in pairs, that is why the number of real roots ascertained by counting sign changes may be this number or a number reduced by a multiple of 2]

Fundamental theorem of Algebra

First of all complex numbers.

These are of type a+bi, where a is the real part and bi is the imaginary part. A complex number may consist of real part only, imaginary part only, or both real and imaginary. Zeros of a polynomial with imaginary part always occur in conjugate pairs. For example a+bi and a-bi, or i and -i

Fundamental theorem can be stated in the form that every polynomial of degree n , as n zeros, which may be real or complex

Or, alternatively that every polynomial can be factorised as a product of linear and irreducible quadratic factors

Irreducible factors arise, if the polynomial has complex zeros with imaginary part. For example the polynomial x³-1 can be factorised as (x-1)(x²+x+1). Here x-1 is a linear factor and x²+x+1 is an irreducible quadratic factor, that is it cannot be further factorised into linear factors.

Hence for predicting the complex roots, we factorise a polynomial. If it has irreducible quadratic factor, then we will have complex roots,

As another example of irreducible quadratic factor, we can consider the polynomial x⁴-1.

If we factorise it, we will have its factors (x+1)(x-1)(x²+1). Here x²+1 is an irreducible quadratic factor, suggesting that the polynomial has a pair of complex roots +i and –i.

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