Can You Factor This Interesting Expression? | Factorise x^10+x^5+1 | Aman Malik Sir
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- Опубліковано 25 лют 2024
- In today's video, we are going to factorize an interesting expression.
We have to factorize the following expression - x^10+x^5+1
If you want to excel in Algebra, you should learn the skill to factorize algebraic expressions.
This expression x^10+x^5+1 is an interesting expression to make you understand the skill of factorization.
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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At first glance.: EASY ❤❤❤❤❤
After sometime : ☠️
Use my method of letting x^5 as t then t will become w and w^2 then find easily😊😊
He is not telling to find roots he is telling to factorize
@@vishalfgm oops
@@LearnerAryan bro if u would find the roots then u can write it in (x-x1)(x-x2)....(x-x10) dont u know this
@@ParthAG12 are u crazy bro
1+x^5+x^10 =( x^15-1)/(x^5-1) by sum of gp formula....then factorize by n-nth roots of unity concept
Taking x⁵ as t² and hence x¹⁰ will be t⁴, so the expression x¹⁰+x⁵+1 = t⁴+t²+1 which can be factorised as (t²+t+1)(t²-t+1). Now substitute 't' in terms of x and we will get the required factors of the expression.
Same to you❤
bro powers integer nhi hogi
Sir let x^5=t. Then we have t^2+t+1. Complex ki equation. Hence x^5=omega
**only if x^10+x^5+1=0
nice
Factor bata sidha...
@@user-ls1qq6px8d
Factors hai [x-(omega)⅕] and [x-(omega)⅖]
@user please be polite he was just giving an approach which you couldn't develop.
My 1st approach par 0 nhi Diya lol
taking x⁵ as t and than after finding x⁵= w or w² we can use 5th root of unity to get all 10 factors
Did same good
amazing explanation sir aaj maza aagya👍👍👍
Nice explanation sir ji 🙏
Sir we can also do the same using synthetic division 😊
Love you Bhannat sir and your mathematics ❤
I substituted omega and (omega) ^2 and i found that it is the root for the given equation. Then i divide the given expression by x^2+x+1 and obtain the expression x^8-x^7+x^5-x^4+x^3-x+1
Beautiful way sir
I need playlist or lecture of Pair of Straight line (math-1) 12th .
Give pls
Sir factors mai Descartes laga sakte hai
Sir it's very loving expression ❤❤❤
Thanks
Sir could you make a practice series on ioqm sir
Sir by synthetic division method we can find factor
Amazing ❤
Sir am your new fan math teacher
Sir Ji ,
STEP-1== x^10 and x^5 mai sai x^5 comman laingai , and we will get x^5(X^5+1)+1
STEP-2== we will write 1 as 1/x^5+1multiply by x^5+1[it will cancel out and we will get 1 again]
STEP-3== now we can see that we have two factors (X^5+1) and (X^5+1/x^5+1)...... solving the second factor we get omega and omega sqaure as the solution of that equation
x¹⁰+x⁵+1=(x¹⁵-1)/x⁵-1 from this we can conclude that the value of x can be x=e^i2kpi/15, except for the values k=0,3,6,9,12 for the denominator becomes zer at tthese values of x.
My answer came X=Cis(2π/5(k+1/3) where k=0,1,2,3,4
Is it correct?
I didn't get. Please explain brother.
How did u manage to insert trigo in this please explain
ek hi bari me chamak gaya masterbki omega ayega 👌👍
Sir please solve this
Integration of ln(1+a^2x^2) /1+b^2x^2 limits 0 to infinite
sir factorised it using de moiver's theorem by considering x^5=t then t^2+t+1=0 here x^5=cos(2pi/3)+isin(2pi/3) so now we can use de moiver's theorem
Who said that t²+t+1=0?Wrong method.
Olympiad questions pa video banaya sir
God tussi great ho awesome 👍👍 Jai hind sir Deepak Lucknow
Sir aap Bhopal se hain kya
Take x^5 as t, u get t^2+t+1, add and subtract t, get perfect square in t+1, and - t
As
(t+1)^2 - t
Take t as root t^2
Get a^2-b^2 whos factors are a+b a-b
(T+root t + 1) * (t - root t + 1)
Replace t with x^5 and u get 2 factors, although weird factors but what was asked was done.
Bhai yese hi hum bhi kiye..but uska koi real value nhi aa rha
Is it x+ 1? It’s 0: 32 and I solved by trial method it was easy
Sir, By the way how did you know that we have to do all that modification..
What were the requirements that forced us to do that stuff..
@BHANNATMATHS
okay I learned this nice little trick from a number theory textbook that an expression like x^(a_1)+x^(a_2)+...x^(a_p) where p is a prime and the set {a_1, a_2, a_3,...a_p} on division by p leaves remainders {0, 1, 2,...., p-1} (not in order) is divisible by 1+x+x^2+...+x^(p-1) which basically generalises all these types of expressions
Which textbook?
Trick dhang se bata satisfy hi nahi kar rahi😂😂😂....
Constant term na lu phir bhi satisfy nahi karega..... 😂😂😂
Bhai x^5 + x^3 + 1 mein nahi bethega... Bithake dekhke.... 😂😂😂
@@user-ls1qq6px8d bro read the comment carefully, it says that a_n's must have all the remainders mod p which yours doesnt x^3 and 1
Which book?
Sir jee advanced 2015 me aaya integration ke questions ka video banaye.❤❤❤❤
Me 8th me hu kuch bhi nahi aata lekin video dekhne me maja aata hai concept understanding Aman Malik Sir Thank you ❤❤
Divided by borders and United by Knowledge..
Love from Pakistan 🇵🇰
Love FROM India
completely unnecessary@@Abhi_83669
@@Abhi_83669hindustan
Oh its new to see pakistanis are able to learn till 12th
@@ParthAG12 Is this the 12th Class??😂😂
What about perfect square
Complete the square method.
x¹⁰ + x⁵ + 1 = (x⁵ + 1)² - x⁵ = (x⁵ - x² ½ +1)(x⁵ + x ² ½ + 1)
I used long division
👍
Sir, I saw your video on the factorization of x^10 + x^5 + 1. Before, I saw your video, I tried this in the following way:
x^10 + x^5 + 1
(x^5)^2 + x^5 +1
Let x^5 = a
a^2 + a + 1
a^2+ 1^2 + a [ here m^2 + n^2 = (m+n)^2 - 2a))
(a+1)^2 - 2a + a
(a+1)^2 - a
(x^5 + 1)^2 - x^5
(x^5 + 1)^2 - (x^(5/2))^2
= (x^5 + 1 + x^(5/2))( x^5 + 1 - x^(5/2))
Sir, I approached the problem in this way. Is this solution correct?
Bro i did same to same😮
Omega wala method sir?
Sir a factories main ne akele Kiya tha and a bohut asan tha and main class 9 me parthahu।
Sir just from seeing i thought what if we take x^5 as t then t woule be omega and omega swuare fir aaram se factors aajayenge??
Same bro
Sir ka jo answer aaya hai vo aana chahiye.... W and W2 se vo nahi aayega...
Sir ke answer mein x^0 wala term real number hai....
@@user-ls1qq6px8d sir ne jo kiya he woh byi further simply karsaktw he
Sir my factors were (X⁵+X^5/2 +1)(X⁵-X^5/2+1)
When we expand them we get X^10+X⁵+1 that means they are factors it is just like factor of X⁴+X²+1 factors where one is plus other is minus and power is halved is this solution acceptable?
I did by the same method
Taking x⁵ as t² and hence x¹⁰ will be t⁴, so the expression x¹⁰+x⁵+1 = t⁴+t²+1 which can be factorised as (t²+t+1)(t²-t+1). Now substitute 't' in terms of x and we will get the required factors of the expression
@@chandraprakash227 yes this method only I used
Coordinate geometry shortcut bataa do sirr please Varna fail ho jaunga
Add and subtract x power 5
At first glance mein hi Pata karliya jab khud solve kiya sir ki uske roots w,w² hai
Maine pahle X ke power 5 se divide kiya then aage procude kiya......
Sir atlast Can we also apply "DESCARTE RULE" ❓
I mean there can be 6 more real factors because in the bracket (x⁸-x⁷.....)sign changes 6 times (i.e.+ then - then + ........)
OR SIR TRY TO MAKE IT LAMBERT W FUNCTION
this method must be out of the jee syllabus..right?
@@aishvarydwivedi2531 Right Jee advanced syllabus
Ohh I think Lambert W FUNCTION works , but it doesn't works So sorry for commenting
@@aishvarydwivedi2531friend LAMBERT W FUNCTION is in Basic maths
Let x⁵=t t²+t+1=0 then t= omega and omega ² so x⁵= omega or omega ² so x= omega^1/5 and x=omega^2/5
x^11+x^9+1 wala sawal jaisa h ye to
{x^5+1+x^(5/2)}{x^5+1-x^(5/2)}
This took me 30 min 😅
So at first I tried x⁹-x⁹ + x⁸-x⁸....... So on and tried factoring but too lazy so I noticed that a common factor is x²+x+1 so I did long division method and divided x¹⁰+x⁵+1/x²+x+1 and i got x⁸-x⁷+x⁵-x⁴+x³-x+1 the calculation took me almost 20 min yes I am very slow idk the answer yet
Edit : the ans is right but the process is very time consuming
If we put ( x⁵ = t² ) and then proceed
What a method yaar!!!!!
This factorize is easy
Hello genius teacher and students give me remainder without devision (x^100-x56+1)/x^4-x^3+x^2-x+1
Sir my answer: (x⁵ + 1 - √x⁵) (x⁵ + 1 + √x⁵)
Is it right
Mene bhi yahi Kiya@@MS_306
आपने जो कहा था omega वाला मैंने सबसे पहहले उसी से किया
(x^5+x^(5/2)+1)(x^5-x^(5/2)+1)
Same
Taking x⁵ as t² and hence x¹⁰ will be t⁴, so the expression x¹⁰+x⁵+1 = t⁴+t²+1 which can be factorised as (t²+t+1)(t²-t+1). Now substitute 't' in terms of x and we will get the required factors of the expression
Can we divide entire expression by x^5 and repeat process?
No
@@abhiboss4662 bruh we can
6:07
Sir mera aa gaya
Who agrees that sir is best ( like 👇)
x⁵ = t
=> t² + t + 1
=> (t-1)(t²+t+1)/(t-1)
=> (t³-1)/(t-1)
=> (x¹⁵ - 1)/(x⁵-1)
Abb bas isse jada me isko nahi jhel skta 🗿
This can be more factorised
5 ka square das hota ky
@@user-bz4hp9sf5l you really need education
@@kavyajain_64 karde bro fir...aur batana.merko kaise kiya
@@mr.unknown1070 x¹⁵ can be written as x^(15/2)^2 and similarly x⁵ can be written as x^(5/2)^2 and we can go further using the identity {a²-b²=(a+b)(a-b)}
Sir binomial
maine bhi cube roots of unity socha
Aur wah sae t = omega and omega²
W is the answer
Maine to quadratic lga ke, complex m ghusa diya😂
x5 = t let krke Quadratic bna li
First comment 😗😗😗😗
Sir i am class 10 student jitna mene padha mera roots -1+√3/2 and -1 -√3/2 araha he jab x^5=y let kare agar koi bade bhaiya comment padh rhe he to please galti kha hoi he bata dijiyega 😅
Actually roots are not real
-1/2+√3/2(i ) likh dono mein because i = √(-1)
Dekte hi pata chalna chahiye ki iske 10 ke 10 roots imaginary hai... Bata kaise..?
Dekte hi pata chalna chahiye ki iske 10 ke 10 roots imaginary hai... Bata kaise..?
Lallu question but tricky😂😊
Bro thinks he's smart
Are aapne kaha ki x²+x+1 ek factor dikh raha hai, to maine to fir direct divide krke nikal liya dusra factor 😅
Bro exam main ese pta nhi chalta ki x²+x+1 factor hoga.
Isi 2023 ke exam me aya tha ye aur isi 2020 ke aaspas bhi aya hua h same question
@@sakshamsingh1778kya phek raha hai
Sir nhi huya
Sir mera to
x⁵(x⁵+1+(1/x⁵))
Aur
x⁴(x⁶+x+(1/x⁴))
A raha tha 😅😅😅
😂😂
Aapne acha samajhaya mere sir ne mujhe gali di aur bola ghar nikalo khaccharo
Sir my full processs 🥱🥱:-
Take x⁵ = y,
Then x¹⁰ = y²,
Thus, the new eq. will be,
y² + y + 1
y² + 2y + 1 - y
y² + y + y + 1 - y
y(y+1) + 1(y+1) -y
(y+1)² - (√y)²
(y + 1 + √y) (y + 1 - √y)
Then final solution:
(x⁵ + 1 + √x⁵) (x⁵ + 1 - √x⁵)
This was mine one too
Sidha sidha bolo ki x^4 + x^2 + 1 mein bithaya hai😂😂
Same I did 😮
Mine ans was the same but the process different
This is wrong both the factors must be polynomial i.e. their power must be whole number.
Before watching your full video i tried once then I came to result as
(X^5 + 1 - x^2√x) (X^5 + 1 + x^2√x)
Pls comment on this
I am in class 10th unknown of omega
@@RajeshYadav-ju5mu then u should be patient until class 11th or learn advance polynomial problem solving skills
Your multiplication is wrong
Both the factors must have powers as whole number otherwise the are not considered polynomials
I don't think its a relevant method....i think using demovire theorem to get all the roots will be an easy approach.
Sir jee advance 2026 ke liye course nikalo mein apse padhne chahta hu online
Total time w.r.t JEE
Please help me😢 :
5 = (3+2cot² theta )/ cot theta
5-3 =( 2cot² theta) / cot theta
2/2 = cot² theta / cot theta
Cot theta = 1
Is these all steps are correct? All
Because my maths sir said that you can't take -3 you have to cross multiply first 😢
Wrong hai bhai tera 😭
Phle cot theeta multiple bad me 3 minus karna
@@abhi-nashpandit1966 bhai reason toh bata please 🥺
@@notgamingpro1804 bhai reason toh batao Aisa kyu mein samjha nhi please 🥺
@@MathematicalAnalysesu can’t subtract it on both sides since left hand side still has a division
So when u subtract -3 on both sides ur subtracting to the entire fraction rather than just numerator
1 like = 2 push up
Please don’t hesitate to break my arm 😂😂
Sorry of bad English 😢😅
Me first to Comment 😂😂
Juth pakda gaya sale... 😂😂😂upar 1 hr ago likha hai aur vedio 2hr ago hai😂😂
As 9th class student , I was able to solve it..
.
Because ek orr que sir ne isi method se solve kiya thaa.....😅😅😁
Agree to like krnaa..