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 2 years ago
Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function:
f(x)= 5x^3+X^2X+5
 2 years ago
Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function: f(x)= 5x^3+X^2X+5

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mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1How many times did the expression changed sign?

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1There is then a MAXIMUM of 3 positive roots, but it could also be 1.

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1Now flip the sign of the coefficients of the odd powers. How many times does the new expression change sign?

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1The number of sign changes is the maximum number of positive roots, including complex. If there are two complex roots (as is the case), then there is only one (real) positive root.

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0so the negative right converted right?

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, the number of sign changes of the new expression (i.e. replace x by x) gives the MAXIMUM number of negative roots.

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0So for this one there is none for the negative and 1 for the positive

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0so how do you know there is one and not 3 for the positive?

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1Using the Descartes rule of signs, we do know that there is no negative root. But we do not know if there is ONE or THREE positive roots.

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0ok so the answer for positive is 1 and 3 and negative is 0

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1You can find that out by factoring the expression. A cubic has at least one real root, and we know that it is positive in this case.

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function: f(x)=3X^2+2X^2+x+3

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0how does this work then?

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0the positive looks like there is none and same with the negative?

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1Now, let's see. (1) The expression as is has no change in sign, so what do you conclude? (2) flip the sign of the odd powers to give 3X^3+2X^2x+3, sign changed 3 times, what do you conclude?

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0wait why is 3X^2 flipped?

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1I supposed you have a typo! It's 3x^3, ... or not?

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function: f(x)=3X^3+2X^2+x+3

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0ok so the positive is none then the negative is 3X^3+3X^2X+3, so 2 and 1

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1No positive is correct. Can you repeat the conclusion for negative?

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0of so negative is 3 and 1

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1The number of possible roots always differ by an even number because complex roots, if any, come in pairs. 1 and 2 will not be correct. 1 and 3 would be.

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1Perhaps more exactly, 1 OR 3.

HelpMe94
 2 years ago
Best ResponseYou've already chosen the best response.0ok thanks that makes sense
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