Finding the roots of a 4th degree polynomial. Reddit precalculus r/HomeworkHelp
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- čas přidán 30. 06. 2024
- Learn how to find all the real and complex solutions to this quartic (i.e. 4th degree) polynomial equation 2z^4-3z^3+2z^2=6z+4. We will first have to use the rational zero theorem to find an easy solution, and then we will have to use either polynomial long division or synthetic division to factor the polynomial. This is not an easy equation especially if you are learning this for the first time, usually in a precalculus class or an algebra 2 class. This problem is from Reddit r/Homeworkhelp.
See the original post here: / zi3o69ozwf
0:00 Where do you even start with something like this?
2:43 The video is out synced
3:43 The video is normal again! : ) yay!
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For some reason, the audio is out synced from 2:43 to 3:43 but the rest is okay. Please let me know if that’s the case. Thanks.
It is the case. Rest is okay thankfully. Isn't "z" a bit of information that there will be complex roots BTW?
Thank you. And yes, z is usually for complex roots : )@@jannegrey593
I didn't even notice it until I read your comment 😂
I had thought there was a problem with my Bluetooth. But then, yes, it fixed itself.
Was probably from the timelapse and the audio didn't get sped up at the same rate or something, fixing itself at the next cut
bprp, I can't help but laugh every time you say x but the problem contains z. We're so used to finding x. 😂😂
LOL, yes. Maybe the first step should have been a substitution!
That’s because we are using the complex domain
I am never paying taxes
How are you 6days ahead?
Based
How??
Wtf is this dude
based
I just found you randomly and your videos are now increasing my math knowledge day by day
Thanks
Congrats on 100k, huge milestone! Love the math vids!
Thanks so much!
We studied synthetic division in university tho we called Horner's method, and mostly that was the only method taught to us to solve polynomials. Well, we also learned about the Bézout lemma and long division, but we almost never used them, they were just stepping stones for learning Horner's method. It can be tedious to use it, and I'm certain there are cases where it doesn't work, but if you limit the possible roots with rational zeroes and just iterate over and over again you'll eventually find the roots
Well there's really no way that Horner's algorithm "can't" work. It will always output the value of the polynomial in a point you give it
@@methatis3013 our teacher gave us polynomials that were unsolveable by horner's method tho
I tried multiple times to solve them but I wasn't able to
In Italy we use Ruffini's method
If you didn’t happen to spot the simple factorization of the cubic, you could continue trying out candidates generated by the rational root theorem. Decartes’ rule of signs tells us that we can stop looking for positive roots, but there is still a negative root to be found. -4 is no longer a possibility and -1 is easily eliminated, so that still leaves -2 and -1/2 to check. The remaining two roots must be complex.
-2 is also easily eliminated as any real number less than -1 is clearly not going to work based on the coefficients m
I love your videos! They are so informative and fun to watch!
Solve x^3-2x+1=0
Answer here
czcams.com/video/mZFrPKirbak/video.html
This dude is awesome
x1 = 1; x2 = -phi; x3 = 1/phi
That's an easy one. if Sum of coefficients=0 then 1 is one of the roots after that we can do synthetic division and then factorisation or grouping will give other roots.
I have question, which is more correct to say -
(i) Adding a number both sides or
(ii) Bring a number to the other side?
@@Killer_Queen_310 Both are correct, depends a bit on the context.
Synthetic division and noticing the third degree polynomial factoring out (to be quicker than testing all Synthetic division possible oermutations) were two important things to learn in the quartic polynomial root finding or "factors" of the quartic polynomial. Finding all the roots would have not found the 2x + 1 factor but instead found the x + 1/2 factor instead by normalizing the first coefficient of the resulting cubic polynomial taking x - 2 out. The factored out 2 in x + 1/2 factor would then be divided out into 0 on the right and the remaining x^2 + 2 would have remained the same forming: (x - 2)(x + 1/2)(x^2 + 2) factored result only matching your answer if 2 was brought back into the middle factor 2(x + 1/2) or 2x + 1 👍
Congrats for 100k :D
Sounds like a job for the quartic formula. You can even preprogram it in your TI84.
You know what they say: Maths is only fun if you know how to math. Pretty sure this applies to everyone here.
Congratulations for 100K❤!!
This is a great refresher on synthetic division. I had completely forgotten how it works.
It was relatively easy to get real roots by subtracting 8 in both sides and moving -3z^3 and 2z^2 to the other side and factoring by grouping
Polynomial long division is simple, I don't know why everyone avoids it.
This is really written out version of Horner's method. I think everyone avoids the other one because it's simpler and usually faster than long division
@@bill_the_butcherNah, it really is very simple. It literally is doing multiplications and then additions or subtractions. The problem is that most people that find it hard either have crappy teachers that don't know how to explain it in simple terms or they don't have the desire to learn it in the first place.
Sure, long division works, but you're ultimately doing the same thing while having to write a lot more.
It is simple, but it takes a lot of time. People don't avoid it because they don't know how to do it, they avoid it because they don't want to do it, there are better ways
Because it's writing a lot of things fo rno good reason? Even his box thing is overkill for me, you can just factorize on the fly, it's way faster (but I've never seen an American do it, it's probably not taught that way over there).
I didn’t remember how to do this at all which is completely annoying. Usually it comes back to me. I’ve been really enjoying your videos. Now I need more practice on this. 🤓
I sometimes try to use "short division", where I split the original expression into several terms, and then rewrite the terms in ways so that they allow me to factor out a common factor from them.
people should call this “peter’s method” if you discovered it first
After the little fast forward when checking z=2 is a root at 2:40, the sound becomes desynced... ahh, it's fixed after the cut at 3:43, nice...
Congratulations on 100k Subs
5:00 or so - I like to do this instead. Since (z-2) is a factor, I want all the stuff in the expression to look like multiples. So 2z^4 has to be followed by - 4z^3. But we don't have - 4z^3, we just have - 3z^3. So I add in z^3, and this has to be followed by - 2z^2. I need to add back 4z^2, which must be followed by - 8 z. I add back in 2z, which must be followed by the - 4 we have. So now it's simple to factor into (z-2)(2z^3+z^2+4z+2) = 0. Then when I find - 1/2 is another root, I do the same thing with (2z+1). In this case, the expression is already in the form I want, so we now have (z-2)(2z+1)(z^2+2) = 0. And the quadratic is a piece of cake, +/- root2*i.
You can also divide it by z ^2.
congrats on 100k
This completely fried my brain circuits 😮
Thanks my synthetic devision concept was unclear you helped me out
You can also divide the entire equation by z^2, make a quadratic in z-(2/z), and then solve this quadratic. These biquadratic quartic polynomials are a common occurrence in the JEE, it was interesting to see this at a 10th grade level...
You can also find the -1/2 by elimination: Since all parts of the cubic term are positiv (no minus), z *has* to be negativ. If you try out -1, the result is -3, so -1 is to big. Therefore only -1/2 remains of the possible roots from the beginning.
or u just use ur brain and don’t waste ur time
the first part is right . but the second part where u say you tried -1 and the answer is -3 so -1 is big is wrong . If the coefficient of z^2 is greater the result will lesser
@@SaiPrasannaS yes, that is why it is too big (as in the absolute value is too big): -3 is already negative, so if x < -1, the result gets even more negative...
What is this synthetic division magic 😮. Never had seen that before.
It's usually called Horner's algorithm. It can be used to calculate a value of a polynomial in any point
could you do a video explaining how synthetic division works
best method is to take z common then get 2 zeroes from quadratic and multiply tht common z with tht , u will get 4
i tried tho
Would you please do a video about principal roots? For example:
What is the square root of 9?
I know that the principal root is the positive one, 3.
But, what is the cubic root of -8?
I’m not sure, but I think the principal root is -2, because this is the largest (and only) real root.
I think that would be a nice subject for a video. 😉
poly long division between this polynomial and z-2. factor the quotient.
My brain jumped a little when the blue "pen" came out. "Whoa!"
Polynomial equations always start scary as hell, but are pretty easy to resolve. Just a bit tiring.
My personal favorite way of resolving them is by dividing the equations. Takes a bit longer than the process in the video, but feels much more intuitive for me.
grouping
(2z² - 5z + 2) could have kept going into (2z-1)(z-2)
2z⁴ - 3z³ + 2z² = 6z + 4
(z²)(2z² - 3z + 2) = 6z + 4
that's as far as i can see... without trying to look at your spoiler.
i didn't think it would be that easy
Im curious, is it possible to find all the roots using binomial theorem?
Ohhh.... Congratulations on 100K subs!
Thanks! 😃
Quartic Formular go brrr
My method is 1. Assume your teacher is nice like he won’t give you the 17th root of unity + sqrt5i as a root. Then try to factor out using newton raphsom approximation because it saves a ton of time. Hope you can somehow find the complex numbers. They always come in pair and their algebraic conjugate is their complex conjugate.
Ahh algebra , my mortal enamy in school. I like to think that if somebody bothered explaining to me that the process rather than the nombers and simbols were important i won't hate it as much today.
I wish I'd have mastered the move we see at 2:41 - that would have saved me a lot of time in school!
Sad that teachers don't teach the basics. That was the first thing we learned when we started talking about roots
@@rennepenneYou mean, they used to teach us that "ultra fast writing technique" move as the basic?
@@loveblue7 and then they ask if anyone has any questions and if you do the teacher let’s out a big sigh and gets angry so eventually everyone is too scared to ask anything.
@@loveblue7 Oh wtf. never thought about that the comment was about the fast forward. I was probably a bit high
it's always weird seeing people use i for root negative 1 since I'm used to j from my EE courses
Idk use the quartic equation
I remember this in school
Just completely skipped it got 0 and continued my life passed my year
Ah yes, solving 4th degree polynomials, the most widely applicable method in daily life
Can we solve without the given root (z-2)
you know its serious when he uses a blue pen
Huh, that was not how I was taught to do polynomial devision, but I suppose it works.
I point out some slips and skips, but: you are an extremely smart and knowledgeable man! Some of your videos that CZcams suggests - I'm scared to open so not to feel dumb... (limits, exponents with e, logs, trig...)
But what should we do if we don't know 2 is a root?
The rational roots theorem.
Given any polynomial in the form:
a*x^n + b*x^(n - 1) + other terms + k = 0
The possible rational roots will all fit the form of:
factors of k / factors of a
Including both negatives and positives
The roots will add to -b/a, and will multiply to +k/a for even n, or -k/a for odd n.
Descartes's rule of signs, can also help you predict whether the roots are positive or negative. If you have uniform positive signs on all coefficients, there are no positive roots. If you have sign swaps, the number of sign swaps you have tells you the number of positive roots, or less by an even number (because the other "positive" roots are a pair of complex conjugates with positive real parts). This assumes all real coefficients.
This allows you to take educated guesses for the roots, which narrows down your search.
There is a cubic and a quartic formula, but they are complicated and seldom taught outside a math degree. For quintics and beyond, Galois (pronounced GAL-wah) proved there can be no such formula, just using elementary functions.
@@carultch And Descartes's Rule of Signs helps narrow down the candidates.
After doing the first step, I split 2z^4 into z^4 and z^4. Then we have terms z^4 - 3z^3 + 2z^2 - 6z + z^4 - 4. Which, after taking common, becomes, z^3(z - 3) + 2z(z - 3) + (z+2)(z-2). Which ultimately becomes, (z^3 + 2z)(z - 3) + (z+2)(z - 2) = 0. Since both the terms are positive, both of them have to be equal to zero. We get 4 zeroes;
root of -2
3
2 and -2.
But after checking, I found that 3 and -2 do not satisfy the equation so I finally got
root of -2 and 2 as roots. But I didn't get -1/2 as one. Can anyone please check why.
I did something different.
I divided the expression with (z-2), I got a three degree equation. Every odd degree function has range of all real numbers. That means, function will have atleast one more root ( expression is defined for every real number). I put z=0 in the three degree quotient, I got 2, then I put z=-1, I got -3. That means one root lies between z=0 and z=-1. So I thought let's hunt down the root by narrowing the possible range of root. I put z=-1/2. Luckily that was the root .Again I divided my degree expression with z+1/2. I got a two degree equation. And after factorising it. I got z=√2 and z=-√2 . This solved??
Sorry I fucked up my iota😅
Thank for the video 👑👑
How are you 6 days ahead?
@@samarjitdas6378 maybe a patreon
Nice
@@samarjitdas6378 im the one who posted the original question so he sent me the video early lmfao
@@Spicychicken_ Oow Now I get it
Thanks man
But what about the other guy who was 6 days ahead also? Do you know him?
6:34 you said "z minus 2 is the root". you meant "z minus 2 is a factor" or "z equals 2 is a root". I think...
I was never taught the rational root theorem in school, even with my applied math minor, granted that was a couple decades ago. Are people being taught this in high school now?
Same, that's news to me. All we did back then was "just try small easy numbers : 1, -1, 2…"
Just change the z to an x. It doesn't matter. If you must, change it back at the end.
i made something complicated for the procedure but the results
Im in grade 10th
But Ive done this since i was in grade 9th. Things are getting harder and harder now
Idk if 10th grade has complex numbers I don't think so tbh so it would just be 2 awnsers not 4
what's the difference between a root and a factor?
In this example a factor would be (z - 2) , and the corresponding root is z = 2 .
For a given polynomial,
A factor is a term that can be divided out , so that you can rewrite the polynomial as the product of two terms: the factor (which is itself technically a polynomial) and a new, smaller polynomial. So you go from
( Big polynomial ) = 0 , to
(Factor)(smaller poly) = 0.
A root is a value where if you plug it into the polynomial, the entire polynomial equals 0. If you have the function f(z) = (polynomial), then a root is a value where if you plug it in for z you get f(root) = 0.
Roots and factors are directly related, because if you have a value of z such that a factor = 0, then it will cause the entire polynomial to go to 0. For example when you plug z=2 into the factor (z - 2) , then that factor becomes (0). So if the equation f(z) = (polynomial) can be factored into f(z) = (z - 2)(smaller polynomial), then when you plug z=2 into f(z), the entire equation will go to zero.
Factoring is just a regrouping of the terms of a large polynomial into smaller terms which are multiplied together, but the factored version is still the same expression as the unfactored version. So whatever values of z cause the factored version to = 0 (by causing one of the factors to =0, and then in turn multiplying the rest of the expression by 0) will also cause the unfactored version to go to zero, because they are the same expression.
Easier method
Divide entire polynomial by z²
We will get a quadratic equation and then just plug in the quadratic formula
Edit:just realised the flaw with my method
Dividing z² would make a problem
2(z²-( 2/z²))-3(z+ (2/z))+2=0
And yes this method is still pretty effective and you CAN evaluate something like 4/x or 4/x² in quadratic formula
But can you use the quadratic formula if you have 6/z and 4/z²?
What do you do with z/6and 4/z^2 then
No, that won't work. But you can use Ferrari's method to turn this equation into
(4z² − 3z + 2)² = (3z + 6)²
and then you only need to solve two quadratics to get all four roots of the equation.
@@user-hg2sh5dq5hjust cry😂 he is nub bro dont believe
@@debikk4204 pretty much yeah
In this case we would have a problem
I would have liked to have seen how to go about it without trial and error of z=2 root.
There is a cubic and a quartic formula. I've used the cubic formula before, and I've attempted to use the quartic formula, but not successfully.
A much easier way to get the root with an educated guess, is to use the rational roots theorem.
In general, consider a polynomial in the following form:
a*x^n + b*x^(n - 1) + other terms + k = 0
The rational solutions for x will all have the following form:
factors of k / factors of a, including both positives and negatives.
The roots also will add to -b/a, and multiply to +k/a for even n, or multiply to -k/a for odd n.
This means you can make a list of all factors of k, and factors of a, and then try the ratios thereof. Descartes's rule of signs is another rule to guide you for guessing roots.
For the given example:
2*z^4 - 3*z^3 + 2*z^2 - 6*z - 4 = 0
There are 4 solutions due to 4 as the highest power. Because there are 3 sign swaps, Descartes's rule of signs tells us that there are either 1 or 3 positive real roots.
Factors of final constant = 1, 2, and 4
Factors of leading coefficient = 1 & 2
This means, our candidates for rational roots are:
+1/2, +1, +2, and +4, as well as negatives
Roots will add to 3/2, and the complex roots (if they exist) will come in conjugate pairs that add up to a real number that contributes to this sum.
The following sets of candidates for roots meet these criteria:
Set 1: -2, +1/2, +1, +2
Set 2: -1/2, +2, and an unknown complex conjugate pair of roots that add up to zero.
Both possible sets contain +2, which means +2 is our strongest candidate to guess as a root.
Of these candidates, z = -1/2 and z=2 are confirmed as the two real roots.
I would recommend you to use Newton Raphson method if using trial and error method may be time consuming 1:48
That’s Algebra 2 and pre calculus
Not that I can make sense of any of this, but why Z instead of X? What does X and Z intend to represent? Why not D or K or any of the other letters in the english alphabet?
Nothing stops you from using whatever you want. But there are conventions. x is a general variable. z tends to represent complex numbers. k is usually an index. d might represent distance or diameter. s sometimes means arc length. r means radius or the roots of a characteristic equation in a differential equation.
The unknown variable we use really depends on the question (In this case the question uses Z.). Any alphabet is fine but we mostly use X and Y because they both represents the coordinate system. Or greek alphabets but that's another story.
Descartes set a precedent to generally use the beginning of the alphabet for constants, and the end of the alphabet for variables. That's why x, y, and z tend to be the default choices for variables. They are the trio that is also least likely to stand for anything in particular.
You of course can use any letters you want, but the letter may also represent something in the original application. Perhaps z represents vertical position. One application of this, might be solving for the equilibrium elevation of a boat floating in water, where the shape of its hull is defined by a polynomial. You may want to use z for this, if x and y are spoken for, for representing horizontal position.
@@major__kong Well, all those makes sense, R for radius and such. I find myself using X and Y frequently, but this video just made me think, "why not other letters?"
@@carultch Ohh, Thank you.
My favourite way which works for every quartic equation as long as we know how to solve cubic looks as follows
We are trying to rewrite quartic on the LHS of equation as difference of two squares
2z^4-3z+2z^2-6z-4=0
Lets multiply both sides by 8 to avoid square roots and fractions
16z^4-24z^3+16z^2-48z-32 = 0
Lets group the terms
(16z^4 - 24z^3) - (-16z^2 + 48z + 32) = 0
Lets complete the square in the leftmost brackets
Anything we add to the expression in leftmost brackets we must add to the expoession in other brackets
(16z^4 - 24z^3 + 9z^2) - (9z^2-16z^2 + 48z + 32) = 0
(16z^4 - 24z^3 + 9z^2) - (-7z^2 + 48z + 32) = 0
(4z^2 - 3z)^2 - (-7z^2 + 48z + 32) = 0
Lets observe that expression inside second brackets (from the left)
is quadratic trinomial so it will be a perfect square when discriminant is equal to zero
If we calculate discriminant it may happen that it is not equal to zero
so we must make it dependent from parameter
We introduce parameter in such way that expression inside leftmost brackets is still
a perfect square keeping inmind that anything we add to the one expression we have to add to the other
(4z^2 - 3z + y/2)^2 - ((4y-7)z^2 + (-3y+ 48)z+y^2/4 + 32) = 0
Now we are ready to calculate discriminant of this quadratic expression
4(y^2/4 + 32)(4y-7) - (-3y+ 48)^2 = 0
(y^2+128)(4y-7) -9(y-16)^2 = 0
(4y^3-7y^2+512y-896)-9(y^2-32y+256) = 0
(4y^3-7y^2+512y-896) - (9y^2-288y+2304) = 0
4y^3 - 16y^2+800y - 3200 = 0
y^3 - 4y^2 + 200y - 800 = 0
y^2(y-4)+200(y-4) = 0
(y-4)(y^2+200) = 0
y = 4
(4z^2 - 3z + y/2)^2 - ((4y-7)z^2 + (-3y+ 48)z+y^2/4 + 32) = 0
(4z^2 - 3z + 2)^2 - (9z^2 + 36z+ 36) = 0
(4z^2 - 3z + 2)^2 - (3z+6)^2 = 0
((4z^2 - 3z + 2) - (3z+6))((4z^2 - 3z + 2) + (3z+6)) = 0
(4z^2-6z-4)(4z^2+8) = 0
(z^2 + 2)(2z^2-3z-2) = 0
(z^2 + 2)(z - 2)(2z+1) = 0
Exactly. This is Ferrari's method for solving quartic equations. Note that in many cases, especially with quartics from math contests and so on which often have a nice factorization into two quadratics with _integer_ coefficients, you can save yourself the trouble of solving the cubic resolvent by first inspecting the coefficients of the quadratic which is to become a perfect square. Here we have
(4y − 7)z² + (−3y + 48)z + (y²/4 + 32)
If this is to be the square (px + q)² of a linear polynomial (px + q) with _integer_ coefficients p and q then both 4y − 7 = p² and y²/4 + 32 = q² will need to be squares of integers and therefore nonnegative, so y needs to satisfy 4y − 7 ≥ 0 i.e. y ≥ 7/4. Also, y needs to be an even integer otherwise y²/4 + 32 will not be an integer and _eo ipso_ not the square of an integer. So, we only need to test positive even integers for y to see if our quadratic in z can be the square of a linear polynomial in z with integer coefficients. With y = 2 we have 4y − 7 = 1 = 1² which is the square of an integer but y²/4 + 32 = 33 is not the square of an integer. But with y = 4 both 4y − 7 = 9 = 3² and y²/4 + 32 = 36 = 6² are squares of integers and indeed with y = 4 we have 9z² + 36z + 36 = (3z + 6)² as required.
you should have used the quartic formula :)
What is the quartic formula?
@@emilyesnyman czcams.com/video/gDCu2pZd_LY/video.htmlsi=yE4W2Y0BW4gs9zZN this video explains the quartic formula in a concise and understandable manner
It’s a huge formula! Instead of using the quartic formula, it is easier to find the roots of the given quartic equation by trial and error,long division, and factoring.
Just do polynomial long division it is very easy and similar to the regular long division
@@RishankPratti Well there are systematic methods to solve quartic equations which do _not_ require you to memorize or use hugh formulas and which are easy to learn, like Ferrari's method which is still taught in India (there are Indian CZcams videos which show how to do this).
In fact if you first multiply both sides of the equation by 8 (to avoid fractions later on) you can use Ferrari's method to transform the equation from this video into
(4z² − 3z + 2)² = (3z + 6)²
which gives
4z² − 3z + 2 = 3z + 6 ⋁ 4z² − 3z + 2 = −3z − 6
and so
4z² − 6z − 4 = 0 ⋁ 4z² + 8 = 0
and then you only need to solve these two quadratic equations to get all four roots of the original quartic equation. Of course both sides of the first quadratic can be divided by 2 to give 2z² − 3z − 2 = 0 which factors as (2z + 1)(z − 2) = 0 giving the roots z = −½ and z = 2 and both sides of the second quadratic can be divided by 4 to get z² + 2 = 0 which gives the nonreal roots z = √2·i and z = −√2·i.
How's the dude in the comments 6days ahead??? Can somebody make sense of this
The video was released early to channel members or something like that.
The date the video is made available to all viewers doesn't have to be the same as when the video was uploaded or made available to private groups.
ohhhh Yes! Solved it in under 3 minutes, while being in Grade 9 😄
is it just me or is the audio not in sync?
Im too bz to do this question
jst do synthetic division (im im the 10th grade)
Luckily, this was not a grade 10 polynomial!
How does the other guy be 8 days ahead?💀
They found my unlisted video in the playlist or on Reddit.
Ruffini
you learn bi-quadratic eqn in 10th grade in us!!! Here, in india we only learn how to solve quadratic eqn!! if I am wrong then pls correct me as I have no knowledge about u.s. educational system. would anyone take the efforts to tell me about educational system in U.S.
That might have meant to be solvable by other methods(that could come from experience), finding two roots and therefore able to simply the equation into a quadratic equation
~7:20 I know and you know, but does everybody know? z^1=z and z^0=1 (not "no z")...
Using z when you can use x is criminal
a + b + c = 6
a² + b²+ c² = 14
a³ + b³ + c³ = 36
ab + bc + ac = 11
abc =6
Values of a , b and c are ?
Help me, solve this
I can. Is it too late? Should I bother?
Hi
Hey
Your way of writing "z" hurts my eyes and brain. Also i guess this is 7th class questions thats when i did these but i don't know about other countries' education system.
(I am in India)
Is it just me or your voice doesn't sync well with the video?
It’s fine but that does happens to me randomly on other people’s videos. No clue why but it’s super annoying!!
Not quite what I x-pected.
en.wikipedia.org/wiki/Factorization#Factoring_by_grouping
For example, for P ( x ) = x³ − 3 x + 2 , one may easily see that the sum of its coefficients is 0, so r = 1 is a root. As r + 0 = 1, and r² + 0 r − 3 = − 2 , one has
i see
x³ + x² -2x -x² -x +2
merging
x³ -3x + 2
okay makes sense now... how do i do this backwards now?
x³ − 3 x + 2 = ( x − 1 ) ( x² + x − 2 ) .
just randomly guess, no? pull out sh1t tons of x and then pull out sh1t tons of negative 1s?
how the fu _ck is this even practical in the real world?
If I had this problem in a practical application today, I'd visit Geogebra, and graph it to find the solutions. The reason for doing it manually, is to understand what people would've had to do in the past. Another reason for knowing how to do it manually, is that it allows you the insight into how the coefficients relate to roots, so you can know how to tune the parameters (that become coefficients) of a system modeled with a polynomial, to get desired results.
An application of higher degree polynomials, is beam theory. Given a beam carrying a uniform load, the slope of the beam at any point along its length will be a cubic function of the x-position, and the deflection curve will be a quartic function of the x-position. Examples with symmetry are easy, but for examples without symmetry, you'd solve a cubic equation to find a location of maximum deflection, and you'd solve a quartic equation to find locations of zero deflection (that don't coincide with a support).
@@carultch
what is the quartic function of the x-position if the affluent or luxurious misplace my support on a disadvantaged slope?
where y of x and y is the value of the disadvantaged versus the affluent or luxurious?
where does the deflection curve reflect a change in potential disenfranchisement in favor of universal sufferage?
where does the slope place
who controls it? how do i find out how this works now?
@@HydraulicsMe I don't understand your question. It sounds like you just copied a bunch of keywords from my post, and made a word salad from a bunch of other words that have nothing to do with it.
How's the dude in the comments 6days ahead??? Can somebody make sense of this