Like this video but want to see more examples? Check out this other video I did on Descartes Rule of Signs here: th-cam.com/video/et4yALjAwYM/w-d-xo.html
I wish I had you as my maths teacher when I was a teenager, but I am grateful even now in my older age to clarify all the maths and regain my confidence from your excellent teaching.
Good question...you want to remember that if all the coefficients of the polynomial are real that the imaginary zeros come in conjugate pairs (2) so that is why you either decrease from the maximum number of negative zeros by 2 and the number of imaginary zeros increase by 2 or when you decrease from the maximum number of positive zeros by 2 that the number of imaginary numbers increases by 2. Hope that helps some.
Oh my gosh! Thank you so much!!!! Your way of teaching is so understandable unlike my professor 😩 I understand like 98% online and self teach while only 2% in my professor's class.
Thank you so much!! I was having a hard time understanding this concept, and I don't really have a teacher to ask for help since I'm learning via a computer program administered from my school.
Great explanation on Descartes Rule of Signs. THANK YOU you cleared my confusions. However I noticed that on your Synthetic Division THE REMAINDER is not 8, It should be -24.
+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.
Gam, here is a more recent video I did on the topic that hopefully will clarify it: Descartes Rule of Signs Examplesth-cam.com/video/et4yALjAwYM/w-d-xo.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.
I am confused! The positive candidates for a possible solution for the first example are 1,2,4,and 8. However none of them would work! Please tell me the reason! Because the idea of Descarte's sign rule is to help with Rational root theorem ! Mario any suggestion as to why?! Also you are an amazing teacher. Thank you.
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..
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 th-cam.com/video/et4yALjAwYM/w-d-xo.html
amazing video, this helped me a lot but could you please elaborate on the part on the table about the conjugate? how did you get the second row on the table? How did you get 0 on the negative column?
I need help with a problem, why does x^2 + x not violate Descartes rule of signs? The zeros are -1 and 0, and when I do the table it gives me one negative and one imaginary. I really need help
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 😊😊
I’m a private math tutor. I do have a degree in math education to teach middle school and high school. I’m glad you like my videos...here’s a more organized list if you want to help you through your class: www.mariosmathtutoring.com/free-math-videos.html -Mario
+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.
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 th-cam.com/video/et4yALjAwYM/w-d-xo.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.)
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
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.
+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.
Like this video but want to see more examples? Check out this other video I did on Descartes Rule of Signs here: th-cam.com/video/et4yALjAwYM/w-d-xo.html
You taught me more in 4 minutes than my teacher could in an hour
You are an amazing teacher. your intonation and speed and way to present the information is really great. May God bless you abundantly
+Felicidad Completa thank you. I'm glad you liked the video.
Thank u sir. My teacher didn't teach this rule in the class but it's actually in our syllabus..
Best lecture on Fundermental counting principle, thank you Mario
I wish I had you as my maths teacher when I was a teenager, but I am grateful even now in my older age to clarify all the maths and regain my confidence from your excellent teaching.
Happy to help!
thanks for the help! getting ready for my exam tommorrow
+William Ordas your welcome! Good luck on your exam!
tnk u
Same here
Good question...you want to remember that if all the coefficients of the polynomial are real that the imaginary zeros come in conjugate pairs (2) so that is why you either decrease from the maximum number of negative zeros by 2 and the number of imaginary zeros increase by 2 or when you decrease from the maximum number of positive zeros by 2 that the number of imaginary numbers increases by 2. Hope that helps some.
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
You’re welcome! Glad my videos are helping you!
Just started the new school year and you are already helping me. Thanks!
You are the Math Monk that enlighten the mathdumbs
Oh my gosh! Thank you so much!!!! Your way of teaching is so understandable unlike my professor 😩 I understand like 98% online and self teach while only 2% in my professor's class.
This was so helpful, thanks for the crystal clear explanation!
I was completely stuck on how the multiples of 2 part of it worked thank god I found this
Thank you so much!! I was having a hard time understanding this concept, and I don't really have a teacher to ask for help since I'm learning via a computer program administered from my school.
Awesome! Glad my video helped you Sophia!
I love your way of teaching... Great video, thank you so much!
Thanks Rebecca! Glad you like my teaching style and video!
Hey thanks.. from india..noce explanation.
You’re welcome Mukul!
you are more helpful than my teacher. THX!!!
You’re welcome Kayla! Glad my video helped you!
Great explanation on Descartes Rule of Signs. THANK YOU you cleared my confusions. However I noticed that on your Synthetic Division THE REMAINDER is not 8, It should be -24.
+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.
Thank you , I'm in grade 10 and I'm having a hard time doing graph with the p ( X ) , do you perhaps have a tutorial for it?
Really clever technique, they should seriously teach us some of these tricks in maths in the UK
Glad you liked the video Shrimat!
We are going to have a test about descartes rule of signs and I really had a hard time understanding my teacher. Your video helped a lot :)
Glad my video helped you Carlos. Report back if you want and let us all know how you did on your test.
Thanks alot.....This help me alot .....😄
You’re welcome Shreya! Glad my video helped you!
that was the best explanation i heard, thank you very much.
+Mayuna Kyouko your welcome...glad my video helped
This totally helps😭❤️
Thank you so much! You made this easy for me now!!
Your very welcome! Glad my video helped you!
Simple and concise. Thanks a lot ☺️
Super ah solli kuduthada sotta thalaya
Put the speed to 1.25x (in TH-cam settings)
Thank me later
paasitive to paaaasitive
@@sauravkarmakar3704 Lol
Glad my video helped you Benjamin!
thanks for you simplest explanation
+Akshay Kumar your welcome! Glad it helped you!
Just a clarification at 5:11 minutes by descates rule you subtract by 2 . Isn't that 0+,0-,4 imaginary?I'm a bit confused.
Gam, here is a more recent video I did on the topic that hopefully will clarify it:
Descartes Rule of Signs Examplesth-cam.com/video/et4yALjAwYM/w-d-xo.html
Thank you for helping me!!
Your welcome Sarah. Glad the video helped!
Great video better than my math teacher geez
Thanks chief. Very cool
Glad you liked the video Yeeet!
Nyc explanation....😎
Glad you liked the video Manisha!
Hell yeah classic Mario! My main man
Glad you like my videos James!
Sir, at 5:30 , why don't you choose 0 4 0 as another possibility solutions.? or maybe 4 0 0 at 5:19.
+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?
I wished you go a bit more into, why this is so?, rather than just giving a rule that frankly is not easy to remember . Thanks.
Good Video. Thank you Mario!
Great video!
Glad you liked the video Iffat!
This was explained so well!
Glad you liked my explanation!
So if we have missing terms, we just skip them? Or no?
Yes just skip them Azura.
I am confused! The positive candidates for a possible solution for the first example are 1,2,4,and 8. However none of them would work! Please tell me the reason! Because the idea of Descarte's sign rule is to help with Rational root theorem ! Mario any suggestion as to why?! Also you are an amazing teacher. Thank you.
Excellent succinct video!
Thanks, you made it easy to understand
+m0ng0l0yd your welcome. Glad it helped.
Thank you sir....god job sir..
You’re welcome Surinder.
Thank you very much 😊
You're welcome 😊
I thought to find the number of maximum negative roots was to replace x with -1 in the equation. Any clarification?
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..
For the first example, can't you also have one negative and two imaginary?
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 th-cam.com/video/et4yALjAwYM/w-d-xo.html
Allah Razi olsun senden :)
Learning is all about "Understanding how it works", but most teachers only tell you to "remember the rules".
amazing video, this helped me a lot but could you please elaborate on the part on the table about the conjugate? how did you get the second row on the table? How did you get 0 on the negative column?
I am the 100,001st viewer1!!!!!
Thank you so much! I have a really great teacher, but he went very fast this lesson! I was the 420th like btw 😂
Glad my video helped you Ayden!
4:20th damn
excellent explanation sir
+rahul puri thanks!
What will you say about roots of f(x) =x3 + x2 + x+ 1
this was sooo helpful!!!
Thank you so much.
Thanks a lot! Very nice video!
+Science Man your welcome. Glad you liked it.
Good explanation
Glad you liked the explanation Roberta!
I need help with a problem, why does x^2 + x not violate Descartes rule of signs? The zeros are -1 and 0, and when I do the table it gives me one negative and one imaginary. I really need help
And also why it does not violate complex conjugate, please I do not understand that at all
Nvm I figured it out
Sir in cardon method if x^2 term is not missing but x term is missing than how to solve. ..
You put in a zero when doing the division
thank you sir
+Ankit Singh your welcome.
Sir what if in a cubic eqn the square term is missing so can we still use this rule??
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 😊😊
Karan, you can check out the SAT playlists on my channel to give you some help...
Can it be used for quadratic equations?
Yes
How to know the other possibilities for the positives and negatives?
I need some help with these when the degree is 5
Are u a professor at some university ?? I mean ur way of teaching is great
I’m a private math tutor. I do have a degree in math education to teach middle school and high school. I’m glad you like my videos...here’s a more organized list if you want to help you through your class:
www.mariosmathtutoring.com/free-math-videos.html
-Mario
Mario's Math Tutoring i would like to have a good conversation with u .. How can i contact u
I’m pretty busy helping students ...Best way to contact me is through TH-cam or my website...
Mario's Math Tutoring then please stay tuned
what if the roots are odd numbers? how you will decrease it by two?
+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.
Mario's Math Tutoring thank you sir...
Michael Rooney thanks😉
very helpful, thank you!
You’re welcome Jack!
Thankyou so much!
does descartes rule of signs is only applicable on just complete equation or it can be applied on missing equation like (x^3-1)
+tariq pathan there can be missing terms just make sure the polynomial is written in descending order from the highest degree to the lowest.
Thank you!!!
You're welcome!
if root of any equation occurs zero then it will be considered as positive or negative root like (x^2-x)
+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.
At 7:08 we find that -16 becomes 16 even though the root is 1 and not -1. Just wanted to point out that error. Great video though!
Why in the second example we can not have 1 positive root,1 negative root and 2 complex roots
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 th-cam.com/video/et4yALjAwYM/w-d-xo.html
@@MariosMathTutoring Thank You certainly cleared my doubts and Keep Up the good work
Thanks again
You’re welcome Rohitashwa.
Thank you for explaining everything you do and going slower so my dumb ass can follow.
In the first example, u could also have 0 positive, 1 negative and 2 imaginary roots right ?
Thank you so much sir
+Alejandro Lopez your welcome. Glad my videos have helped you.
Hi sir, thanks for teaching this great theorem, but I wonder can't the first polynomial have one negative root and two imaginary roots?
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 Examplesth-cam.com/video/et4yALjAwYM/w-d-xo.html
In the first example, 012 is also possible
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
He explained it better than my math teacher lol
Glad you liked my explanation!
Sir, at 1:53 did you mean "Rational Coefficients" instead of "Real Coefficients"?? Bit confused in this part. Very nice video though, thank you:)
Thank you again
Thank you so much
Your welcome Anshul!
I need deep explanation of Descartes rule
At the very end, negative 16! Still gives us -24 so the 1 still isn't a zero but still. My professor did the same thing!
How to find complex roots by this method
Explain roots of x^2+x+1=0 using Descartes rule of signs
ty how do I give you credit for your simple explanation.
If you find my videos and channel helpful if you could give me a shoutout to friends on your social media accounts that would be a great help!
I don't mean to pry, but how do you earn money if you put these lessons online? Through the ads?
Good question Hans....through the ads, through the sale of my own video courses, and some people support me on Patreon...
Thank u so muchhh
In the first problem why can't we have one -ve root and 2 imaginary roots
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.
Youre awesome
Glad you like my videos Carlos!
under 2nd example under 8 you wrote positive 16. It should be negative (-16)
finally someone said it, i thought i was missing something
-2x4 + 2x2 = 0
X = 0, 1 , -1
why it is not applicable in above example
Or How it is applicable
Positive excludes 0
Let's start a subwar between Mario's and Brian McLogan's
👍🏻👍🏻👍🏻👍🏻
Why can't there be 4 positives, 0 negatives, and 0 imaginary, or 0 positives, 4 negatives, and 0 imaginary?
+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.
Can you solve the equation in your thumbnail
It must be complex rather than imaginary