Descartes\' Rule of Signs a) If c_1, c_2, ellipsis, c_m are any m nonzero real n

Question

Descartes' Rule of Signs a) If c_1, c_2, ellipsis, c_m are any m nonzero real numbers, and if 2 consecutive terms of this sequence have opposite signs, we say that these 2 terms present a variation of sign. With this concept, we may state Descartes' rule of signs, a proof of which may be found in any textbook on the theory of equations, as follows: Let f(x) = 0 be a polynomial equation with real coefficients and arranged in descending powers of x. The number of positive roots of the equation is either equal to the number of variations of signs presented by the coefficients of f(x), or less than this number of variations by a positive even number. The number of negative roots is either equal to the number of variations of signs presented by the coefficients of f(-x), or less than this number of variations by a positive even number. A root of multiplicity m is counted as m roots. Investigate the nature of the roots of the following equations by means of Descartes' rule of signs: x^9 + 3x^8 - 5x^3 + 4x + 6 = 0. 2x^7 - 3x^4 - x^3 - 5 = 0, 3x^4 + 10x^2 + 5x - 4 = 0.

Explanation / Answer

Descartes' rule of sign is used to determine the number of real zeros of a polynomial function.

It tells us that the number of positive real zeroes in a polynomial function f(x) is the same or less than by an even numbers as the number of changes in the sign of the coefficients. The number of negative real zeroes of the f(x) is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number.

1) f(x) = x9+3x8-5x3+4x+6

Our function is arranged in descending powers of the variable, if it were not we would have to do that as a first step. Second we count the number of changes in sign for the coefficients of f(x).

Here are the coefficients of our variable in f(x):

+1   +3   -5    +4   +6

Our variables goes from positive(1) to positive(3) to negative(-5) to positive(4) to positive(6).

Between the first two coefficients there is no change in signs but between our second and third we have our first change, then between our third and fourth we have our second change and between our 4th and 5th coefficients we have no change. Descartes´ rule of signs tells us that the we then have exactly 2 real positive zeros or less but an odd number of zeros. Hence our number of positive zeros must then be either 2, or 0.

In order to find the number of negative zeros we find f(-x) and count the number of changes in sign for the coefficients:

f(-x) = (-x)9+3(-x)8-5(-x)3+4(-x)+6

      = -x9+3x8+5x3-4x+6
-1     +3      +5      -4       +6

Here we can see that we have three changes of signs, hence we have three negative zeros or less but a even number of zeros..

Totally we have 2 or 0 positive zeros or 3 or 1 negative zeros.

(2) g(x) = 2x7-3x4-x3-5

+2     -3    -1     -5

between the first and second term there is one change,

therefore number of positive zeros are 1.

g(-x) = 2(-x)7-3(-x)4-(-x)3-5

           -2x7-3x4+x3-5

between first and second terms no change in the sign where as for second and third there is one change, and third and fourth another change in the sign,

therefore number of negative zeros are 2 0r 0.

Totally we have 1 positive real zero or 2 or 0 negative real zero.

(3) h(x) = 3x4+10x2+5x-4

                 +3     +10     +5    -4

we have exactly one positive real zero.

h(-x) = 3(-x)4+10(-x)2+5(-x)-4

             3x4+10x2-5x-4

    +3     +10     -5     -4

we have exactly one negative real zero.

Totally we have one positive real zero or one negative real zero.

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