Polynomials Class 9 Maths Easiest Explanation One-Shot Lecture | Class 9 Maths 2024-25 

Best of luck👍 legends for Tomorrow's EXAM😅🤣😂🥴

46:14 middle term factor 28:38 division of polynomials 37:17 remainder theorem 41:38 factor thorem 52:15 algebraic identities, factorisation of polynomials hope this helps

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This video script is a comprehensive explanation of polynomials in Class 9 Maths, covering various topics and examples in detail. The speaker encourages viewers to engage with the content, subscribe to the channel, and participate in interactive learning activities. Key moments: 00:00 Understanding Mathematical Expressions and Variables -Explaining the concept of variables and constants -Discussing the importance of power in mathematical expressions -Emphasizing the significance of maintaining numerical order in expressions 05:53 Understanding Degree and Polynomials in Mathematics -Writing in Descending Order -Explanation of Degree and Polynomials -Importance of Degree in Polynomials 11:47 Discussion on Degree Certificates and Polynomial Degrees -The importance of degree certificates and polynomial degrees -The significance of different levels of degrees -Activating the next step 17:47 Understanding a Complex Mathematical Concept -A unit to remind him of team matters -Explaining the sequence for pop3 -Solving a mathematical problem involving squares and numbers 23:41 Understanding Mathematical Concepts and Problem Solving -Explaining the implementation of Maithili 10 -Discussing mathematical equations and problem-solving techniques 29:34 Understanding the Concept of Reminder and Division -Explaining the concept of reminder and its significance -Discussing the process of division and its implications -Exploring mathematical calculations and practical applications 35:28 Techniques for Dividing and Sharing in Weight Loss -You can add Aamir -Divide and share the weight loss technique 41:23 Understanding Mathematical Concepts and Operations -Explaining the concept of remainder and division -Discussing the process of division and factors 47:17 Mathematical Problem Solving and Explanation -Product on sensitive issues -Explanation of mathematical calculations -Solving mathematical problems 53:11 Importance of Channel and Personal Assistance in Education -The type of channel is important for education -Personal assistance based on achievement -Efficient study techniques and interactive learning

41:26 question bhaiya forget to put (-) in 2nd term, don't forget to change the sign according to powers , hope this will help 00:01 HELLO! 28:38 DIVISION OF POLYNOMIALS 37:11 NINJA TECHNIQUE 41:38 FACTOR THEOREM 46:14 MIDDLE TERM FACTOR 52:15 ALGEBRIC IDENTITIES, POLYNOMIALS 56:01 BYE BYE!

➡️ Anyone Having doubt in 41:40 here: Term 1: 4*(-1/2)^3 = -4/8 -> (-1/2) Term 2: 4(1/2)^2 = 4/4 -> (+1) Term 3: -(-1/2) -> (+1/2) Term 4: And then there is a -1 ➡️ We got all 4 terms solved: -1 +1 +1/2 -1/2 = 0 Hence it gives us answer as 0 ❤❤

41:25 After solving the remainder is 0 so YES 4t³+4t²-t-1 is a multiple of 2t+1 ( thank you so much for this session 🙏 it was worth it! )

Who is watching in 2024 march

Exam needs preparation, preparation needs notes notes need time aur wo bacha nahi.. so, here come our favorite tutor, which make maths like easy peasy.............. thnk u soooo much

Omg this lecture helped me to be strong in polynomials for tomorrow's maths paper. Wish me luck 🤞

Excellent video! The concept is taught in a nice manner, easy for learners to follow. I am a retired teacher of English. I was trying to help my niece with her mathematics exam and found this video. Though I don't remember to have studied all this in my school days, this video made me learn Polynomials easily.

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00:01 HELLO! 28:38 DIVISION OF POLYNOMIALS 37:11 NINJA TECHNIQUE 41:38 FACTOR THEOREM 46:14 MIDDLE TERM FACTOR 52:15 ALGEBRIC IDENTITIES, POLYNOMIALS 56:01 BYE BYE! For context...

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46:14 middle term factor 28:38 division of polynomials 37:17 remainder theorem 41:38 factor thorem 52:15 algebraic identities, factorisation of polynomials hope this helps like dedena

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Division of polynomial 28:15 Reminder theorem 37:11 Factor theorem 39:27 Factorisation 45:54 Identities 51:42

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Polynomials was a difficult chapter for me but after this video it is a very simple chapter for me Thankyou@Just Padhle

41:35 yes the answer is 0 If we have 4(-1/2)³+4(1/2)²-(-1/2)-1 Then we will apply law of exponent and 4(-1³/2³) + 4(1²/2²)+1/2-1 4(-1/8) + 4(1/4) +1/2 -1 Four is divided by denominator and -1/2 + 1/2 +1-1 And then answer is zero

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41:26 question bhaiya forget to put (-) in 2nd term, don't forget to change the sign according to powers , hope this will help 00:01 HELLO! 28:38 DIVISION OF POLYNOMIALS 37:11 NINJA TECHNIQUE 41:38 FACTOR THEOREM 46:14 MIDDLE TERM FACTOR 52:15 ALGEBRIC IDENTITIES, POLYNOMIALS 56:01 BYE BYE! edited: Sir Maths Ka exam Ko Hai

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41:26 sir yaha pe second term me bhi (-1/2) ana chahiye nahi (1/2) , actually ☝️🤓

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H/w: 4/2+4/2+1/2-1 4 would be cut by 2 and 1-1 become 0 and 2 multiple 0 is 0 So the 2t+1 is factor of 4tcube +4tsq.-t-1

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41:25 Yes it is a multiple of 2t+1, as the remainder is 0. thank you sososo much sir, you're the best teacher. all my doubts have been cleared up now. thank you. 🙏🏻🙏🏻

Bohot cute sa chapter hai, bohot pyaara sa chapter hai polynomials -the nominee of joke of 2024

41:30 Yess, i got 0 as remainder after solving so, 4t³ + 4t² - t - 1 is a multiple of 2t+1

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Thanks You sooooooo much sir It help me a lot Today was my polynomials test Now I can score Good marks 😄😄... It was worth it 😁😁🤩

41:33 ans= -1\2+1+1\2-1 = 0 yes it is multiple bhaiya maja aa gaya 🤩🥰

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Timestamp : 41:32 Given : q(t) = 4t^3 + 4t^2 - t -1 2t + 1 To Prove : Whether 4t^3 + 4t^2 - t - 1 is a multiple of 2t + 1 Proof : Let 2t + 1 = 0 Then, 2t = 0 - 1 t = -1/2 Now, 4 (-1/2)^3 + 4 (-1/2)^2 - (-1/2) - 1 = 4 (-1/8) + 4 (1/4) - (-1/2) - 1 = -1/2 + 1 + 1/2 -1 = 1/2 + (-1/2) = 0/2 = 0 Hence, Remainder = 0 And Therefore, 4t^3 + 4t^2 - t - 1 is a multiple of 2t + 1

1:10 video starts :)

To determine whether the polynomial \( q(t) = 4t^3 + 4t^2 - t - 1 \) is a multiple of \( 2t + 1 \) and to find the remainder when \( q(t) \) is divided by \( 2t + 1 \), we can use polynomial long division or synthetic division. Alternatively, we can use the Remainder Theorem. The Remainder Theorem states that if a polynomial \( f(t) \) is divided by \( t - c \), the remainder is \( f(c) \). For our case, we need to divide by \( 2t + 1 \). First, we convert it to the form \( t - c \): \[ 2t + 1 = 0 \implies t = -\frac{1}{2} \] Thus, we need to evaluate \( q(t) \) at \( t = -\frac{1}{2} \): \[ q\left(-\frac{1}{2} ight) = 4\left(-\frac{1}{2} ight)^3 + 4\left(-\frac{1}{2} ight)^2 - \left(-\frac{1}{2} ight) - 1 \] Now we calculate each term step by step: 1. \(\left(-\frac{1}{2} ight)^3 = -\frac{1}{8}\) 2. \(4 \left(-\frac{1}{2} ight)^3 = 4 \times -\frac{1}{8} = -\frac{4}{8} = -\frac{1}{2}\) 3. \(\left(-\frac{1}{2} ight)^2 = \frac{1}{4}\) 4. \(4 \left(-\frac{1}{2} ight)^2 = 4 \times \frac{1}{4} = 1\) 5. \(-\left(-\frac{1}{2} ight) = \frac{1}{2}\) Combining these results, we get: \[ q\left(-\frac{1}{2} ight) = -\frac{1}{2} + 1 + \frac{1}{2} - 1 \] Simplifying: \[ q\left(-\frac{1}{2} ight) = \left(-\frac{1}{2} + \frac{1}{2} ight) + \left(1 - 1 ight) \] \[ q\left(-\frac{1}{2} ight) = 0 + 0 \] \[ q\left(-\frac{1}{2} ight) = 0 \] Since the remainder is 0, the polynomial \( q(t) \) is a multiple of \( 2t + 1 \).

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41:31 Remainder is 0....so,that would be a multiple of that polynomial.. Thanks bhaiya

Legends trying to complete whole semester day before exam

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43:50 ANS IS To find the remainder when \( x^3 - ax^2 + 6x - a \) is divided by \( x - a \), we can use the Remainder Theorem. The Remainder Theorem states that the remainder of the division of a polynomial \( f(x) \) by \( x - c \) is \( f(c) \). Here, the polynomial \( f(x) = x^3 - ax^2 + 6x - a \) is divided by \( x - a \). According to the Remainder Theorem, the remainder is given by \( f(a) \): \[ f(a) = a^3 - a \cdot a^2 + 6a - a \] Simplifying this: \[ f(a) = a^3 - a^3 + 6a - a \] \[ f(a) = 0 + 5a \] \[ f(a) = 5a \] So, the remainder when \( x^3 - ax^2 + 6x - a \) is divided by \( x - a \) is \( 5a \).

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BHAI EXAM SE 1 DIN PEHLE PADHNA CHOD DO SCHOOL ME UNIT END HONE KE BAAD EK QUICK REVISION JARURI HAI ACCHE GRADES LAANE KE LIYE , JO KAAM MAHINO ME NAHI HUA VO 1 DIN ME KAISA HOGA . I HOPE YOU ALL UNDERSTOOD . 😊😊 ( YOU ARE NOT LEGEND IF YOU ARE WATCHING THIS 1 DAY BEFORE EXAM , YOU KNOW WHO YOU ARE ... )

Ram Ram Sare jana 26 tarik se 9th ke paper h bol (best of luck)