Descartes Rule of Signs 

+Felicidad Completa thank you. I'm glad you liked the video.

+William Ordas your welcome! Good luck on your exam!

tnk u

Same here

Happy to help!

try routh herwitz criteria...lot easier

What if we decrease positive and negative each by 1

According to Descartes theorem you can only decrease the possible positive zeros by 2,4,6,... and the negative by 2,4,6,...

@@MariosMathTutoring thanks. Appreciated replying to my doubt

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+Ernesto Cortez glad you liked the video. When you do synthetic division you want to ADD straight down NOT SUBTRACT like when you do long divison. Hope that helps clear up any confusion.

+Ernesto Cortez Yes you are absolutely correct. I took a second look at that - good catch! It sounds like you are understanding the process I was demonstrating so that's fantastic. Good job.

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paasitive to paaaasitive

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4:20th damn

Yes just skip them Azura.

+tariq pathan there can be missing terms just make sure the polynomial is written in descending order from the highest degree to the lowest.

Good question Chetan... the way i understand it we replace x with -x in the polynomial then simplify it....we then count the number of sign changes to find the maximum number of negative zeros...but substituting -1 and then counting the sign changes should give you the same ...hope that helps you..

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Glad you liked my explanation!

Gam, here is a more recent video I did on the topic that hopefully will clarify it: Descartes Rule of Signs Examplesru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-et4yALjAwYM.html

+Ankit Singh your welcome.

Music God here is another video I did on the topic that will hopefully help you better understand the topic: Descartes Rule of Signs to Determine Number of Positive & Negative Zeros ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-et4yALjAwYM.html

According to Descartes rule of signs when we counted the sign changes we found the maximum number of positive and maximum number of negative real zeros. Since the maximum number of negative real zeros was 2...according to Descartes...there can either be 2 or 0 negative zeros...they go down by an even number (2, 4, 6, etc.)

Thanks for the explanation, maybe there are some points I have missed, I will look over it!

Here is a more recent video I did on the topic that may help you.... Descartes Rule of Signs Examplesru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-et4yALjAwYM.html

+Onny Hoo there was a maximum of 2 positive and 2 negative zeros so it has to be lower than 2. Hope that helps. We found the maximum number by looking at the sign changes.

ok ok Mario, I will watch the video again.. many thanks..

and what if all givens are positive and or negatives?

Good question Hans....through the ads, through the sale of my own video courses, and some people support me on Patreon...

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And also why it does not violate complex conjugate, please I do not understand that at all

Nvm I figured it out

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Yes

You put in a zero when doing the division

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Yes you can Karan.

Thanks a lot Sir for your help. To be frank I didn't expect you to reply . I have my SAT subject test exam this Saturday . Would you like to advice me something that can come out handy in my exam 😊😊

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+tariq pathan you will be able to factor out x if that is the case and identify it that way. Then proceed with Descartes rule of signs.

No, it's not because there were 2 sign changes when he did f(-x).

i don't understand here either. why is 0 1 2 not possible?

ah it's because it decreases by even number. as Mario said 1-2 would be negative you can't have a negative number of real zeros... so it was explained in the video

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Just remember according to Descartes rule of signs that when we find the maximum number of positive zeros and maximum number of negative zeros...we then subtract 2, 4, 6, etc. from the positives and/or the negatives but not one from each...hope that helps Rohitashwa.

Here is another video I did that can help you understand Descartes rule of signs: Descartes Rule of Signs to Determine Number of Positive & Negative Zeros ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-et4yALjAwYM.html

@@MariosMathTutoring Thank You certainly cleared my doubts and Keep Up the good work Thanks again

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+Juan Gabriel say there are a maximum of 3 negative zeros...you could then decrease it to 1 negative and then increase the imaginary zeros by 2 more. You would not then be able to decrease the negatives by 2 more because you can't have a negative number of zeros. Again Descartes rule is to narrow down the possible number of positive, negative and imaginary roots. Hope that helps.

"Decreasing by two" simply means that you take the number of possible real roots that you found and subtract two until you hit zero. So if you found 5 possible negative real roots for an equation, you could have either 5 roots, (5 - 2) roots, or (5 - 2(2)) roots. This is because, as mentioned in the video, if you don't subtract by two, you can end up having an odd number of non-real complex roots. This is impossible for polynomials with real coefficients as complex conjugates come in pairs, something that can't be achieved if you have an odd number of them.

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Michael Rooney thanks😉

finally someone said it, i thought i was missing something

According to Descartes rule of signs the number of possible positive or negative zeros can only decrease by 2 or 4 or 6 etc....you saw how we found the maximum number of possible positive zeros and the maximum number of negative zeros...so we subtract multiples of 2 from those numbers and increase the possible complex zeros by 2 (since they come in conjugate pairs). We can’t subtract 1 from the maximum positive and 1 from the maximum negative by the way. Hope that helps.

Positive excludes 0

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+Kevin Ma you want to calculate the number of sign changes in the original equation - this will give you the maximum number of positive zeros. Then when you substitute -x in for x in your function and simplify - you can count the number of sign changes and this gives you the maximum number of negative zeros. Then if all the coefficients are real numbers then you want to recognize that any imaginary zeros come as conjugate pairs. Also, you can only decrease the number of positive zeros or negative zeros by an even number ...2, 4, 6 etc. as you increase the number of imaginary zeros. Also remember the total number of zeros will always add up to the highest degree of your function.