My Math Forum Descartes Rule not verifying

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 August 23rd, 2018, 07:35 AM #1 Member   Joined: Nov 2016 From: USA Posts: 34 Thanks: 1 Descartes Rule not verifying The roots are -1, -2, 5, i, -i for: P(x)= x^5-2X^4-12x^3-12x^2-13x-10 using Descartes Rule to verify: The rule says the number of positive real zeros of P(x) is the same as the number of sign changes in the sign of the coefficients or is less than this by an even number.. I see only one sign change in P(x). For P(-x), I get: -x^5-2X^4+12x^3-12x^2+13x-10 I see four sign changes Shouldn't there be 1 positive and 4 negative roots? The answer is showing 2 positive (5 and i) and 3 negative (-1, -2, and -i). Thank you for feedback.
 August 23rd, 2018, 08:24 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,767 Thanks: 1422 click on link ... https://www.emathhelp.net/calculator...3x-10&steps=on Thanks from topsquark
August 23rd, 2018, 09:27 AM   #3
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Quote:
 Originally Posted by Seventy7 The roots are -1, -2, 5, i, -i for: P(x)= x^5-2X^4-12x^3-12x^2-13x-10 using Descartes Rule to verify: The rule says the number of positive real zeros of P(x) is the same as the number of sign changes in the sign of the coefficients or is less than this by an even number.. I see only one sign change in P(x). For P(-x), I get: -x^5-2X^4+12x^3-12x^2+13x-10 I see four sign changes Shouldn't there be 1 positive and 4 negative roots? The answer is showing 2 positive (5 and i) and 3 negative (-1, -2, and -i). Thank you for feedback.
You are ignoring that the rule applies to REAL roots.

Positive real roots. One change of sign implies one positive real root. That is 5. The number i = 0 + 1i is a complex number rather than a real one. (Furthermore, complex numbers do not have signs in the normal sense.)

You are also ignoring the aspect of the rule that says the number of real roots is specified by the number of sign changes OR by a number less than that by some multiple of 2.

Negative real roots. Four changes of sign inplies four, two, or zero negative real roots. There are two, namely - 1 and - 2. Again, the number - i does not count because it is neither real nor negative.

This was in my opinion a good question because it showed an effort to understand a rule rather than memorize it. Like many rules, this one requires close attention to what it exactly says.

EDIT: Notice that the rule does not tell you the number of roots that are zero.

$f(x) = +\ x^2 - 2x \implies 1 \text { sign change } \implies 1 \text { positive real root.}$

$f(-\ x) = +\ x^2 + 2x \implies \text { no sign change } \implies \text { no negative real root.}$

1 + 0 = 1, but there are two real roots, x = 0 and x = 2.

Also the rule will lead you astray if you think it tells you the number of distinct real roots.

$f(x) = x^2 - 2x + 1 \implies 2 \text { sign changes } \implies 2 \text { or } 0 \text { positive real roots.}$

$f(-\ x) = x^2 + 2x + 1 \implies \text { no sign change } \implies \text { no negative real roots.}$

But the only distinct root is 1. We say that it has multiplicity of 2. The rule would be clearer if it noted that the rule does not cover roots equal to zero at all and counts duplicate roots as equal to the number of times that root is duplicated.

Last edited by JeffM1; August 23rd, 2018 at 09:55 AM.

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