Can You Factor This Wonderful Expression? | Factorise x^13+x^11+1 | Aman Malik Sir

In today's video, we are going to factorize a very amazing expression.
We have to factorize the following expression - x^13+x^11+1
If you want to excel in Algebra, you should learn the skill to factorize algebraic expressions.
This expression x^13+x^11+1 is a wonderful expression to make you understand the skill of factorization.
This will help you in your basic mathematics exams, olympiads, and JEE Mains and Advanced examinations.
Let's see how we can factorise this expression and learn a lot of core mathematics skills.
Stay tuned with @BHANNATMATHS
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[@XYZ-21]

Here's the one using the cube roots of unity, for the equation x³=1, we've by the x³-y³=(x-y)(x²+xy+y²), it as (x-1)(x²+x+1)=0. Now, for the polynomial x¹³+x¹¹+1, manipulating can be rewritten as: → x¹¹(x²+1)+1 Add and subtract 'x' to produce the factor of the cube root of unity equation as → x¹¹{(x²+x+1)-x}+1 Expand: → x¹¹(x²+x+1)+(1-x¹²) Use a²-b²=(a+b)(a-b) → x¹¹(x²+x+1)+(1-x⁶)(1+x⁶) Again expand as: → x¹¹(x²+x+1)+(1-x³)(1+x³)(1+x⁶) Take a minus sign common out of the second term to make one of the factors of the term, x³-1 as; → x¹¹(x²+x+1)-(x³-1)(1+x³)(1+x⁶) Rewrite as: → x¹¹(x²+x+1)-(x-1)(x²+x+1)(1+x³)(1+x⁶) Take (x²+x+1) common; → (x²+x+1)[x¹¹-{(x-1)(x³+1)(x⁶+1)}] Simplifying; → (x²+x+1){x¹¹-(x¹⁰-x⁹+x⁷-x⁶+x⁴-x³+x-1)} Finally we've: →→→ (x²+x+1)(x¹¹-x¹⁰+x⁹-x⁷+x⁶-x⁴+x³-x+1)

[@pritideval3727]

Really great way of thinking, thnx dude ❤

[@MyOnlyMaster]

Thank you

[@nikhildubey5054]

Good job brother 👍

[@Arkaprabha_saha]

Very good process❤

[@chipixe9326]

Same bhai

[@SatyamKumar-wg2kg]

I've another elegant solution using COMPLEX NUMBERS.. It can be seen that the function contains the roots only in (-1,0). So we can consider a root of this equation as z=cosӨ + isinӨ. By de Moivre's :- cos13Ө+isin13Ө+cos11Ө+isin11Ө+1=0 This yields two equations as cos11Ө+cos13Ө+1=0 & sin11Ө+ sin 13Ө=0 2nd equation gives cos11Ө=cos13Ө Putting this into (1) we get cosӨ=-1/2 ..thus sinӨ=sqrt(3)/2 or -sqrt(3)/2 Thus the roots are (-1±i√3)/2 which are the roots of x^2+x+1!!!!!!! We've got a factor of this monster equation. Now long division method proceeds--- Thank u🤘👍

[@BruhGamer05]

Maine to equation dekhne par notice kiya ki ye cube roots of unity hain , to omega aur omega² factors aa gaye waise.

[@SatShriakal97]

@@BruhGamer05 yaa

[@The.Sigma.]

nice brother

[@Naitik-pyarlawar]

बहोत ही बडा solution दे दिया मेरे भाई.

[@sagnikbiswas3268]

Slick

[@aljbbethsrikanth2794]

The only man in KZbin who teaches us to explore maths Rather than tell us to prepare for JEE This difference makes me love you sir 💝💝💝

[@smrutiranjanparida8517]

👍👍 Yes

[@gyaneshpathak6709]

You didn't explore KZbin yet

[@aljbbethsrikanth2794]

@@gyaneshpathak6709 You maybe correct and that's ok

[@focus_on_what_you_want.]

😂😂😂 Bhai use yt properly

[@An_ony_mous]

@@focus_on_what_you_want. He is using it properly itself. I go to coaching and that's enough for me. I explore the things which I'm interested on YT.

[@prajnaparamitapatra9722]

Bohut pyara Sawal tha sir. Hats off

[@Pie_670]

Are sir aap itna padhate ho ki m kuch saal me expert ban jaunga 😅😅😅😅 Edit: i am so famous

[@alisharehankhan5586]

😂😂😂😂😂😂😂😂

[@Sanatan_saarthi_1729]

Galat fehmi ☕☕☕☕☕

[@focus_on_what_you_want.]

Women 😂😂😂

[@somamandal338]

Really 😂😂

[@g.d.243]

Positive attitude is the best part about sirji

[@joshuajoseph9236]

2:36 start

[@jaythakkar4298]

complex number se hue yeh question x^13 + x^11 + 1 = 0 ...(1) as we know from complex numbers w^2 + w + 1 = 0.... (2) the second equation can be written as w^2.(1) + w.(1) + 1 = 0 w^2.((1)^9) + w.((1)^12) + 1 = 0 ( because 1 to the power any number is 1 itself ) w^2.((w^3)^9) + w.((w^3)^12) + 1= 0 ( since w is the cube root of unity ) w^11 + w^13 + 1 = 0 .... (3) comparing equatins (1) and (3) :- we get x = w and similarly we can get x = w^2 too as the answer

[@sardaar_sanveer_singh]

Great!

[@habibahmad1050]

Sir ek baar reimman hypothesis ko samjao plz sir

[@shiwoshiwoismyactualname]

Ye reimer aur teimann ko combine krke bna h kya

[@ganeshpatwal6555]

@@shiwoshiwoismyactualname 😅🤣

[@quantumphilosopher1729]

​​​@@shiwoshiwoismyactualname ye Reimann Hypothesis ek elegant problem hai jiska mazak banana matlab isse solve karne ki ability rakhna kya tere paas wo ability hai to bol varna kai mathematicians ne apni life di hai is hypothesis pe to mazak mat uda.

[@shiwoshiwoismyactualname]

@@quantumphilosopher1729 i didn't demean it by joking on it. I appreciated the hypothesis while making a light joke on it.

[@BossAbhay212]

gajab explanation sir ji❤❤❤

[@Leviackerman-hm4vl]

other method add and subtract x^12 we get x^11(x^2+x+1)+(1+X^6)(1+x^3)(1-x^3) now take x^2+x+1 common from 1-X^3 and other term that its one line answer no need complex number and other methods

[@narayanmodi11]

1:41 Sir thank you hamare bare me sochne ke liye .

[@amarjeetahlawat4945]

Jabardast guruvar

[@aadisingh8742]

Sir aise question videos kaafi acche lagte hai. Keep making these types of videos. 😊😊😊

[@aadijaintkg]

Sir In this question this same factorization can be take place by only adding X^9 and subtraction x^9 in between of x^11 and 1

[@PabitraT]

Dear sir, one video on FERMAT'S LAST THEOREM ❤

[@ujjualgamers9865]

Yes, let's factor x^13 + x^11 + 1. To factor this expression, we can first notice that it does not appear to be easily factorable using traditional methods. However, we can rewrite it using a substitution to make factoring easier. Let's substitute y = x^11: x^13 + x^11 + 1 = (x^11)^1 * x^2 + (x^11) + 1 Substituting y into the equation: y(x^2 + 1) + 1 = y(x^2 + 1) + 1(x^2 + 1) Now, we can see that it has a common binomial factor of (x^2 + 1), so we can factor it out: (x^2 + 1)(y + 1) = (x^2 + 1)(x^11 + 1) Therefore, we have factored x^13 + x^11 + 1 as (x^2 + 1)(x^11 + 1).

[@vask5500]

Where did you get y(x2+1) + 1 = y(x2+1) + 1(x2+1) When the answer is expanded, you get x13 + x11 + x2 + 1

[@ajajjsbsjsbkbqpbpqnpa9549]

Chatgpt

[@Godgaming-re3wm]

Sir isse aise kar sakta hai X¹³+x¹¹+1=0 , then multiply both side by x . Now eq is x¹⁴+x¹²+x =0 Now take x common. X(x¹³+x¹¹+1) =0 Aise kuch kar ka answer a sakta hai kya Main abhi 10th paper Diya hai. Uss hisab se

[@Mohit_Choudhary._]

If you multiply x both side then in this particular question should not be equal to zero Because jab x ko RHS me transfer kroge then it is 0/x and x means denominator can not be zero, so basically your question again becomes same

[@ssf4915]

Multiply karke common lia 😭

[@Mohit_Choudhary._]

@@ssf4915 😂😂, bro he is in 10th class , don't cry 😂😂

[@Godgaming-re3wm]

@@ssf4915 acha samjhe gaya question wahi ho gya Jo tha

[@prabhagupta6871]

1:49 correct sir

[@s.sgameing654]

Sir samjh nhi aya jab apna x⁹ common liya uska baad aya kaisa baki ka ???

[@ІгорСапунов]

Let x^13+x^11+1=0. Then x^13+x^11=-1. If x0 is root of 1, then x^13 and x^11 are equal to -1/2+/-i*sqrt(3). So x^2+x+1 factorizes x^13+x^11+1

[@meenakshisinghal6126]

Sir aap isko complex no ka use krke bhi factor Krna sikha do

[@bbmathematics224]

Sir Riemann hypothesis samjha dan.

[@lalikantkumar-vz9cm]

Sir Aap Bsc ki maths prasandhaiye please

[@sumitbhawanani253]

Love the question ❤

[@kabirhossain8603]

X and x^2 are not needed. Because we can divide in factor -(x^3-1) and then so on.....

[@Lifechangingmotivational-tk5go]

Agar aspke jaisa mhan guru mil jaye to Kisi ki bhi Jeet pakka ho sakti 🙏 love you guru ji 😘

[@avyakthaachar2.718]

Amazing question sir.❤ Simpler method :- Notice that omega and omega² (complex cube roots of 1) are roots of x¹³+x¹¹+1. Thus x²+x+1 must be a factor, and the other factor can be found by simple polynomial long division.

[@jaythakkar4298]

bhai long division se iske factors nahi aate ...

[@avyakthaachar2.718]

Try again, I was able to get the second factor when I divided x¹³+x¹¹+1 by x²+x+1.

[@7k_Satyam]

@@avyakthaachar2.718 I got (x¹¹+x⁹+x⁸-x⁶-x⁴-x²-x+1)

[@xeindercage1666]

Sir please request hai aapse jee ki series continue krdo vector3d ka baad

[@honestadministrator]

Factors of x^11 - x^10 + x^9 - x^7 + x^6 - x^4 + x^3 - x + 1 ???

[@debaprasadhalder7982]

অসাধারণ

[@sarthakkhawle]

Can anyone explain how to factorise using Complex numbers??

[@Hariprasadb7]

somehow using polar form, but then there will be 2 variables theta and radius, idk.

[@trueblood541]

Take x=omega Then equation becomes w2+w+1 which is a hint that x2+x+1 is a factor of the given polynomial You can find the other factor by dividing the two equations

[@Dharmarajan-ct5ld]

Don't know why you missed simple method of factorisation and went for reverse construction

[@amnkumar8432]

How did you cancel out x square against -1 Please explain me

[@rameshn282-u7q]

integral of f(x) = f(x) then f(x)=? f(x) is not e power x. please tell sir

[@Dikku716]

e^cx is the only solution

[@bhanusharma233]

amazing

[@DebarthoGuptaXIIA]

Sir since w and w² are roots then x²+x+1 must divide the expression so I divided it by x²+x+1 and got the ans

[@arnavjindal2857]

Sir please batado is type ki factorisation problems kaha se karu practice ke liye??

[@niteshsingh4772]

Sir ek bar comlex number s factroise Karen

[@Subham_Kumar_chaudhary]

Sir please complex number se bhi samjha dijiye.

[@YuezhiTribe]

Bohot mzedaar ques tha sir.....thanks

[@iusepc2317]

video 2:30 se start hoti hai..

[@RohitKumar-ou5qz]

Sir please ek video remainder problem pr banaiye please 😢

[@AjayDas-yx9kn]

can you please factor 4xto the power 4+8x+15

[@NotSoSmart740]

Sir make a video on how to factorise it using cube root of unity.... Please 🙏

[@Arushdixit6546]

Iske factors (x⁹)(x⁴+x²+1) bhi to ho skte h shyd

[@super40educationalinstitut29]

Sir apki paid class kaise milegi

[@lkcreations594]

Dekhkar smjha aa gya tha imaginary cube roots of unity satisfy kr rhi to puri eqn ko x^2+x+1 se divide karke do min me a jayega ans

[@dhritishmankumar881]

Love you sir 🔥

[@jumblefumble]

Itna add or subtract karne ke bajiye Khali x⁹ add or subtract Karo X¹³+x¹¹+1+x⁹-x⁹ = x⁹(x⁴+x²+1) -(x⁹-1) Then factorizing x⁹(x²+x+1)(x²-x+1)-(x-1)(x²+x+1)(x⁶+x³+1) I multiples the brackets also = (x²+x+1)[x¹¹-x¹⁰+x⁹-x⁷+x⁶-x⁴+x³-x²+1] Btw I am in 9th 😊

[@050138]

Awesome! 👏

[@raghvendrasingh1289]

Your solution is awesome I am a iit teacher have u heard about formula for a^3+.. j^3 when a+ .. +j=o

[@jumblefumble]

@@raghvendrasingh1289 thanks , but no I haven't heard about that formula, i ve even learned a lot of non routine maths for exams like nmtc and ioqm but i haven't heard about that formula. I have seen the formula for a³+b³+c³ when a+b+c is zero but not for more than three terms

[@rooikriti6466]

I’m also in 9th. Do you mind explaining this sum in just a bit more detail please?

[@jumblefumble]

@@rooikriti6466 sure x¹³ +x¹¹+1 =0 Now you have to add x⁹ both sides as we can see that x⁹ can be taken common from first two terms and with 1 it can be factorized to x³-1 which has the common factor same as the first three terms (x¹³+x¹¹+x⁹) Regrouping the terms: (x¹³+x¹¹+x⁹) -(x⁹-1)= x⁹(x⁴+x²+1) -(x³-1)(x⁶+x³+1) Now you gotta know that x⁴+x²+1 can be factorized to x²+x+1 and x²-x+1 =x⁹(x²+x+1)(x²-x+1) - (x-1)(x²+x+1)(x⁶+x³+1) Here we see that there is a common factor so we finally factorize it by taking it common =(x²+x+1)[x⁹(x²-x+1)-(x-1)(x⁶+x³+1)] Then multiply the brackets = (x²+x+1)(x¹¹-x¹⁰+x⁹-x⁷+x⁶-x⁴+x³-x²+1)

[@PrayasianRamesh]

First comment 🙂🙂 love you sir❤

[@vikashkumar96314]

Sir 1 bechare ko itna Durr kyu likh diya

[@username_taken_se_pareshan_boi]

Sb maths padhate h.. Aman sir maths feel krwa dete h 🙌

[@AnirudhSahu]

Sir aap samjhate ho to aajata h pr khud se nhi ho rha kya kru

[@nikhilgone-ey3yf]

Sir one shot lectures lao please jee ke liye

[@lakhi22794]

Math is so simple 0 to 9 numbers and other different operations like stories

[@proman9297]

Can we solve this by using the idea of Geometric Progression

[@kakashiamv7837]

Ya same doubt but i believe it will not help us ti facotorize it rather it would have been helpful to find sol

[@MultiGurmukh]

I don't know what type of questions comes in IIT, but factoring an equation like does not help me in anyway studying this equation, the equation does not have any postive factors and there is only one one negative factors and rest all factors are imaginary. After factoring equation like this what paper setter is trying to achieve. Can you give an example where this octic equation are use and how do you every time come up with random method of solving an equation.

[@sagnikbiswas3268]

Algebraic manipulation is definitely a tool in the JEE Advanced, and even mains. Will this exact question come, probably not but who knows. That is not to say it is not a good use of getting exposed to algebraic manipulation. Also others in the comments have alluded to an approach with Roots of Unity which is definitely an important topic.

[@joydebtantubayedit]

Sir apka math ka tarikha alag hi hai

[@samirbhattacharya4145]

Good .

[@meetpatil5736]

w aur w^2 roots dikh rahe the toh polynomial ko x^2+x+1 se divide kar diya

[@RandhirKumar-kt8lz]

Root8 ka power root2 devided by root 2 ka power root8 plzz sir solve it

[@Arushdixit6546]

Answer kya h broo

[@Heisenberg_blackteee]

Sir physics ke bhi numerical ko bhi layi pls😅

[@arnab7681]

Sir ji ajj maja agaya

[@kinshuksinghania4289]

x^3 = 1

[@vipinpandey3461]

👍👍👍

[@deepshinde5703]

X=omega

[@gauravaggarwal]

any polynomial with exponents of the form 3k , 3k-1, 3k+1 in equal qty is always divisible by x^2+x+1 example x^23 +5 x^19 +4 x^11 + 2 x^3 + 3 now 23 and 11 are of the form 3k-1, and are in total 5in number 19 is of the form 3k+1 so we have exponents each of the form 3k-1,3k,3k+1 so need not check, it is divisible by x^2 + x + 1

[@darkside17790]

Literally isne bande ne rap ga ke maths samjhaye

[@manishsharmamanishsharma8979]

Sir apne wrong likha hai

[@swrshayariwitharrehman7141]

What is your Qualification sir?

[@pratulsingh7823]

Sir isko x ki power 11 ko t mana lenge gaya aur phir 13 ko t2 lekin denge gya aur phir d nikala denga gya aur -b+√d/2a kar ka nikal denge gya

[@joydebtantubayedit]

Bsc padao na app

[@sankalp5649]

x¹³+x¹¹+1

[@rajkumarvermasir5465]

Why you make fooooool other

[@amarnaths3014]

Can we try x^11 = y. Therefore x^13 = y^3. The equation becomes y^3 + y + 1. Perhaps we can proceed from there.

[@amarnaths3014]

@Target IPhO sorry, my bad. Thanks for correcting.

[@050138]

😂

[@naitik_singh_]

Iske to only two factor bane

[@johnpeter6779]

koi bdi baat nhi RH criterion se aaram se ho jayega kitne bhi degree ka polynomial ho😂

[@tarunpal8648]

The factors are (X^0) (X^13 + X^11 + 1) simple 😂😂😂😂

[@rakeshrajak5801]

Naman sir

[@Gyanendra47]

Mai 11-1=10 me ane wala hu

[@PanchalSahib-lh2op]

Any 9 students here

[@Gyanpath77]

Yes

[@Amol-Bhāratvāsi]

Yes

[@prmotivation5641]

Yeah

[@Sumit77661]

Yeah

[@arpan8584]

Lambi kahani hai

[@biranchisarangi4011]

E sala ko math k bare mein kuchh nehin pata..... Go for arvind kalia sir .... See the difference....

[@lakhi22794]

Math is so simple 0 to 9 numbers and other different operations like stories