Solve for x
Вставка
- Опубліковано 29 чер 2023
- The zeros of a function are the x values at which the value of the function is zero. This is equivalent to solving for x by setting the function equal to 0.
The video shows the following
* How to find the possible zeros (rational)
* How to do synthetic division
* How to check and see which of the possible zeros are actual zeros
* What to do after finding one of the zeros to find the rest of the zeros
That was beautiful!!! (75 retired Electrical engineer here) Never heard of synthetic division. Wait, in hindsight, I remember polynomial division looking like long division, but it was demonstrated quickly, and rarely used in developing other math. Just needed to know it could be done, but again rarely used.. YOU ARE a prime tutor. I value your lessions!
Although I am an engineer now I struggled with maths in secondary school and in University. I had a sucession of very bad maths teachers. No, I'm not a bad workman blaming my tools, but for maths in particlar, one needs a good methodical teacher. I learned more here in these videos that in 8 years of maths study and must admit that I had never heard of synthetic division? I did matrices in UNI, but nobody ever said they could be applied to solve equations? I'm 67 and semi-retired but always fascinated by maths! Thanks.
I am happy that you are frequently posting lecture videos❤
More to come!
Maybe a video on "synthetic division" would be good?
That's a great idea 💡.
I'll post one soon.
I was going to say I needed a refresher 😀
I see all of your tutoring shorts. I have searched UA-cam for about a week finally discovered your channel and I had to subscribe.
Thank you so much!
I simply used factoring by grouping.
(3x^4-9x^2-12) -10x^3+40x=
3(x^4-3x^2-4) -10x(x^2-4)= 3(x^2)^2-3(x^2)-4 -10x(x^2-4)=
X^2-4(3(x^2+1)-10x)
x=+/-2,3,1/3
Note: i did have to use the quadratic formula. Hence, im not sure if it was less steps.
@forcelifeforce you're correct. Once I saw the coefficient, I defaulted to wanting to use the quadratic formula instead of factoring directly.
expand out second bracket: 3x^2 -10x+3 then ( 3x-1)(x-3 )
x=+-2, 3,1/3
It great that you are posting these videos!!
To teach it's the best thing a man can do! 👍
I appreciate that!
Thanks prof, could you please kindly increase your upload and concentrate one specific playlist like precalculus, Trig, Calculus, and linear algebra. You are greatly needed in your expertise. Thanks again
Will try
You are amazing teacher!
Thanks Prof T. I learnt a new cool way of factoring polynomials from your video, I taught it to my learners and they enjoyed it. Please do a video on synthetic multiplication.
Excellent!
Thank you, I love this VDO clip.
I watch his videos religiously.
But This one went over my head. 😢😢
After getting 12 values for p/q, why pick 2.🧐??
I could've picked any number from the list of p/q. However, I picked 2 because I knew it would work since I knew the answer ahead of time. This was done to save time for the video.
@@mrhtutoringso how we also know the Answer ahead of Time.
@@torque7742 I tried dividing by other p/q before making the video.
@@mrhtutoring could you make a video how we get Quick answers
In less time.
You can also use Descartes' rule of signs to narrow down the list of possible rational zeros. If you determine that there is only 1 positive zero, once you find it you can focus on the negative zeros.
Sorry, I had an Amazon delivery, could you repeat that? Just kidding, you are the best. Thank you!
i found this pretty helpful
Thank u teacher
Great going..
좋은강의입니다
댓글 감사합니다 ~
Nice solution!
Thank you
I remember before taking limits, derivatives and integrals there was a whole ass unit about partial fractions, the remainder theory and many more.
Ooh nice it is !
Outstanding
ขอบคุณครับ
sir upload all algebra and arthematic concepts viodes and shortcuts
have you ever heard a method called "double cross"?
1 -5 6
3 5 -2
1*3=3, 6*(-2)=-12, 1*5+3*(-5)=-10,(-5)*(-2)+5*6=40, 1*(-2)+3*6+(-5)*5=-9
(x^2-5x+6)(3x^2+5x-2)=0
(x-2)(x-3)(3x-1)(x+2)=0
easy peacy
How did he factorize the equation of degree 2?
It's mostly guessing and checking.
@@mrhtutoring and years of experience 😀
You can also use the quadratic equation to figure it out if you're not confident in factoring: ax2 + bx + c
Cant we just test the all the factors of 12 by substituting it into p(x) then take the one that gives zero ? Then do synthetic division from there?
That's possible only when q=1.
If not, you have do p/q for possible zeros
👍
Does Rational Zeros Theorem work on all degrees of Polynomials quintic and above ?
Yes.
@@mrhtutoring Thanks !
I tried it on my TI-84 CE+.
Gotta start doing more Matlab & Python
Thank You so much sir, I just started learning math from zero I stuck on this topic for over 5 hours now, I was asking why I'm dividing those terms in synthetic division I tried chat gpt, bard but nothing and now you showed from start to end how to use the method and it's really beautiful method 🤍, I will still try to figure out how the method came into existence .
Professor could you please bring chapters of class 11th math
Long video.
Basic concepts.
Is the possible rational roots method only applicable for polynomials of 4?
No, it works for all
@@mrhtutoring then if we get a binomial like 4x^2 + 3x + 2 = 0, then the roots willbe two conjugate complex numbers... How can you derive a possible rational root which is complex, otherwise we are not getting a solution...
@@md.hossain4974 The rational roots theorem in general, works for a polynomial, regardless of the degree of the polynomial. It works for quadratics, cubics, quartics, and anything beyond.
The reason it doesn't work for your specific example, is that the rational roots theorem isn't a guaranteed way to solve for polynomial roots. It's just a way to take educated guesses to solve for polynomial roots in the special case that you have at least one root. Once you find a rational root, you can use it to reduce the degree of the polynomial, and simplify your work to find the rest. For instance, if you find one rational root to a cubic, you can reduce it to a quadratic, and use the quadratic formula to find the two remaining roots.
There are formulas that work as master keys for general solutions to the quadratic, the cubic, and the quartic. But the quartic is the end of the line; Galois proved there is no quintic formula, or anything beyond.
Is there another way of doing these types of questions? 😅 I thought it was quite time consuming😊 thx
Not without a graphing a calculator.
@@mrhtutoring Could we not use an obvious root and then use a, b constants for the x² and x terms, then develop the equation and finally solve the two unknown system by equating the constant terms to the initial equation? Seems it would be faster.
@@pureffm Without a graphing calculator, there is no "obvious" root in most of these problems. There is a guess and check method unless you try factoring by grouping but I'm not confident in that.
There is a quartic formula, that is the master key for solving any quartic equation with a 4th power as the highest power. The quartic is as far as we can go, if we want a formula that just uses elementary functions to find the solution to a polynomial. Quintics and beyond, have no such formula.
I am a little lost. but will come back to it later.
but what if there is a quartic without any rational zeros? like this one x^4-8x^3+6x^2+7x-3=0? How would You solve it?
Sad to say, but the method he used only works if P(x) contained at least rational real roots-the definition of rational root or rational zero theorem itself. It means u need to use the general formula for P(x) with degree of n.
@@eccentrico5965 Yeah i know, I was only wondering how one would solve a quartic without any real rational roots.
@@HH_Vision There is a quartic formula that can solve ANY quartic, just like there is a quadratic formula, but it is very difficult and computationally intensive to use, that it is rarely used in practice. You are generally better off using Newton's method to extract the decimal expansion of the root of a quartic, which is what I'd usually do if I had to solve one in a practical application. I know how to solve cubics with Cardano's formula, and I've only been able to figure out how to use Ferrari's quartic formula in special cases of quartics.
Interestingly enough, the quartic is the end of the line, when it comes to a formula as the master key for finding the roots of a polynomial. There is no quintic formula, or anything beyond. Not that we haven't discovered one yet, but Galois (pronounced Gal-wah) proved that it is impossible for one to exist, using only arithmetic, powers and roots, and even elementary transcendental functions like trig.
Professor, one those or any number must have a singular verb. For example, one of those numbers is the correct one. You use 'is' because the verb agrees with 'one' not these numbers or numbers. That's because the verb in the sentence agrees with the subject noun or pronoun that precedes the preposition in a a sentence or phrase. Google verb and preposition agreement. However, things for showing us your brilliant techniques of computations.
Thanks for pointing out the mistake.
But when I'm teaching or making the videos, grammar is the last thing on my mind. There's so much going on in my head when I'm making these videos by myself~
This would have been harder if you don’t state that the zeros are rational.
Moses!
Horner's rule does it much faster.
The guessing part is so annoying/random. Yet that’s what you have to do, right?
Yes, but doing it systemically helps
@@mrhtutoring How so?
If you could see my face at the door of that 😮 I’m going to need a minute or 2 🤤🙃👀
Door is nothing like End is Ai going crazy
All i know is that 0 doesn't excist....