Polynomials Class 9 in One Shot 🔥 | Class 9 Maths Chapter 2 Complete Lecture | Shobhit Nirwan

Shobhit Nirwan - 9th・2 minutes read

Shaub Darban introduces Chapter 2 of class ninth, emphasizing teaching with intensity and conceptual understanding, utilizing NCERT, RD Sharma, and RS Aggarwal, and urging active engagement from viewers with notes and questions during the lecture. The chapter explains variables as unknown or changing values denoted by letters, distinguishes them from constants, introduces algebraic expressions and polynotes, emphasizes whole numbers in variable powers for polynotes, and details factorization, using methods like Splitting the Middle Term and the Factor Theorem.

Insights

Shaub Darban introduces Chapter 2 of class ninth, focusing on Paulino Mills, promising to teach without pen.
Emphasizes teaching with intensity and conceptual understanding, encouraging questions from NCERT, RD Sharma, and RS Aggarwal.
Explains the concept of variables as unknown values or values that change, denoted by alphabetic letters.
Defines constants as fixed values and distinguishes them from variables.
Introduces algebraic expressions as combinations of terms using plus and minus.
Details the special form of algebraic expression called polynote, where variables have whole numbers as powers.
The concept of Polynote involves identifying variables with specific powers.
Different types of Polynote are classified based on the number of terms and the degree of the expression.
The degree of Polynote is determined by the highest power of the variable present in the expression.
The process of determining if a given expression is Polynote involves pausing to analyze the presence of whole numbers in variable powers.
The Reminder Theorem simplifies the process of finding reminders in division operations.
Factorization involves writing a number as a product of its factors within brackets.
Practice is emphasized as the key to mastering the formulas.

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Recent questions

What is the concept of variables in algebra?
Variables in algebra represent unknown or changing values.

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Summary

00:00

Teaching Algebra: Variables, Constants, and Polynotes

Shaub Darban introduces Chapter 2 of class ninth, focusing on Paulino Mills, promising to teach without pen.
Emphasizes teaching with intensity and conceptual understanding, encouraging questions from NCERT, RD Sharma, and RS Aggarwal.
Urges viewers to engage actively by taking notes and solving questions during the lecture.
Explains the concept of variables as unknown values or values that change, denoted by alphabetic letters.
Defines constants as fixed values and distinguishes them from variables.
Introduces algebraic expressions as combinations of terms using plus and minus.
Details the special form of algebraic expression called polynote, where variables have whole numbers as powers.
Clarifies that polynotes must have whole numbers as powers of variables to qualify as such.
Provides examples to illustrate the concept of polynotes and the importance of whole numbers in variable powers.
Concludes by emphasizing the significance of whole numbers in variable powers for an expression to be classified as a polynote.

13:23

Understanding Polynote: Variables, Degrees, and Examples

The concept of Polynote involves identifying variables with specific powers.
Polynote is distinguished by the presence or absence of a whole number in the power of the variable.
The process of determining if a given expression is Polynote involves pausing to analyze the presence of whole numbers in variable powers.
The temperature in Delhi during recording reaches 52.3 degrees, affecting the need for frequent water breaks.
The equation 4^x - 3^x is discussed, emphasizing the absence of whole numbers in variable powers for it to be Polynote.
The importance of simplifying expressions to understand Polynote is highlighted.
The General Form of Polynote is explained, focusing on the notation and variables involved.
The degree of Polynote is determined by the highest power of the variable present in the expression.
Different types of Polynote are classified based on the number of terms and the degree of the expression.
Examples of Monomial, Binomial, and Trinomial Polynote are provided to illustrate the classification based on terms and degrees.

28:08

Polynomials: Terms, Degrees, and Value Calculation

Monomials have one term, binomials have two terms, and trinomials have three terms.
Degree one is linear, degree two is quadratic, and degree three is cubic.
Classify polynomials based on the number of terms and the degree.
The first example is 4 or 3x k + 4x s + 7, which has three terms.
The degree of a polynomial is determined by the highest power of the variable.
Constant polynomials have a degree of zero.
The degree of a zero polynomial is not defined.
The value of a polynomial at a specific value of x is found by substituting that value for x.
For example, if a polynomial is 3x^4, its value at x = 2 would be 16.
Polynomials can have different variables, and their values are calculated by replacing the variable with the given value.

40:51

Substitute to Simplify Polynomial Equations

Denoting a given equation as 3y k – 4y + √11, if p equals 2, substitute 2 for y to simplify the equation.
Substituting 2 for y in the equation results in 8 - 8√11.
The value of the polynomial equation 4t 4+ 5 t k - t s + 6 is determined by replacing t with a.
By substituting a for t in the equation, the result is 4 - 5 * a to the power of 4a - a s + 6.
To find the zero of a polynomial, replace x with a value that makes the polynomial equal to zero.
Substituting x with 1 in the polynomial 3x - 3 results in the zero being x = 1.
The process of finding the zero of a polynomial involves substituting different values for x until the polynomial equals zero.
To determine the zero of a polynomial, substitute the given values into the equation and check if the result is zero.
By substituting - m / l for x in the equation, the zero of the polynomial is found to be x = - m / l.
The long division method is used to divide a polynomial by another and find the remainder.

54:03

Factor Theorem Simplifies Division Reminder Calculation

Multiplying p by 12 results in 12x, and if 2 is subtracted from 12x, it becomes 24p.
The leftover value after performing operations is referred to as the reminder, which signifies what remains.
Dividend and divisor are terms used to describe the numbers involved in a division operation.
Long division is utilized to calculate the reminder when dividing a large number by another.
The Reminder Theorem simplifies the process of finding reminders in division operations.
The Factor Theorem explains that if a number is a factor of another, the reminder will be zero when dividing them.
Understanding the Factor Theorem involves recognizing that if x - a is a factor of p, then p equals zero.
The converse of the Factor Theorem states that if p equals zero, then x - a is a factor of p.
Applying the Factor Theorem aids in determining the reminder in division operations.
The Factor Theorem provides a straightforward method to calculate reminders in mathematical operations.

01:07:53

Factorization and the Factor Theorem Simplified

4 is a factor of 36, meaning when 36 is divided by 4, the remainder is 0.
If x - a is a factor of p, then when px1 is divided by p, the remainder is 0.
The Factor Theorem states that if x - a is a factor of px1, and p of a is 0, then x - a is a factor of p.
To check if x + 2 is a factor of a polynomial, substitute -2 for x and see if the result is 0.
The Factor Theorem simplifies to checking if the remainder is 0 when dividing the polynomial by the potential factor.
Factorization involves writing a number as a product of its factors within brackets.
The Splitting the Middle Term method is crucial for factorization in class 10.
Factorization entails writing a number as a product of its factors within brackets.
The Splitting the Middle Term method is essential for factorization in class 10.
Factorization involves writing a number as a product of its factors within brackets.

01:21:20

"Middle term split for effective factorization"

The process involves splitting the middle term in an equation.
The middle term is identified as -7x.
The term is then broken down into two numbers, p and q.
The goal is to find two numbers that sum up to -7 and multiply to 12.
The numbers identified are 4 and 3.
The equation is rearranged in standard form.
Common factors are identified between the terms.
The final factorization is achieved by taking out the common factors.
The process is repeated with another equation involving factorization.
The method of splitting the middle term is emphasized for effective factorization.

01:33:49

Factorizing Polynomials Using Factor Theorem

To factorize a polynomial using the Factor Theorem, start by finding the constant term and then determine its factors.
Identify the factors of the constant term, such as 1, -1, 2, -2, 3, -3, 6, -6, and proceed to divide them.
Test each factor by substituting it into the polynomial to see if it satisfies the equation and becomes zero.
The factors that make the polynomial zero are considered the zeros or roots of the polynomial.
The number of zeros a polynomial can have is equal to its degree, so a polynomial of degree two can have a maximum of two zeros.
Express the polynomial as a product of its factors, such as x - a, x - b, based on the zeros found.
Apply the Factor Theorem to factorize a given polynomial by finding the constant term and its factors.
Use the trial and error method to identify a factor of the polynomial and then divide the polynomial by that factor.
Utilize long division to divide the polynomial by the factor and obtain the quotient and remainder.
Break down the polynomial further by splitting the middle term and finding factors of the constant term to simplify the factorization process.

01:48:16

Factorize Equations Using Factor Theorem and Formulas

Apply the formula: Dividend Equal Divisor into Question Plus Reminder
Dividend equals divisor divisor tha x-1 into ct tha plus Reminder reminder
Use splitting to find two more factors, resulting in three factors
Follow three steps: Remove one factor, Divide by that factor, Set Reminder
Factorize using Factor Theorem: Remove factor, Divide by it, Set Reminder
Factorize using Factor Theorem: Hit and trial, Divide by factor, Set Reminder
Factorize using Factor Theorem: Hit and trial, Divide by factor, Set Reminder
Memorize formulas: a + b whole square, a - b whole square, a + b * a - b, x+a * x + b
Evaluate formulas: x + a * x + b, x + a * x + b, a - b time square
Apply formulas: x + a * x + b, x + a * x + b, a - b time square
Solve equations: x + a * x + b, x + a * x + b, a - b time square
Understand and apply formulas: a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - b, a + b * a - 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02:03:49

Summing Three Terms Formula with Variables

The formula for summing three terms is a + b + c = a^2 + b^2 + 2ab + 2bc + 2ca.
The square of a + b + c is a^2 + b^2 + 2ab + 2bc + 2ca.
The formula involves the variables A, B, and C.
The process includes repeating the formula three times.
The formula for the square of bpc is a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.
The letters PG are mentioned in the text.
The text discusses the importance of comparison in solving equations.
The formula is further expanded to include x, y, and z variables.
The process involves replacing variables with specific values.
The final result is the square of a + b + c, which equals 2x + 3y - 4z.

02:18:28

Mastering Formulas Through Practice and Analysis

The number of questions asked in class is directly related to the formulas being taught.
To practice the formulas effectively, do as many questions as possible.
The fifth formula involves raising questions based on the formulas taught.
The fifth formula remains consistent, focusing on the whole square of a p b+ c.
Practice is emphasized as the key to mastering the formulas.
The sixth formula involves the whole cube of B, focusing on a cubed minus b cubed minus 3ab in A.
Detailed steps are provided on how to expand and compress formulas effectively.
The importance of comparing formulas and identifying patterns is highlighted.
Practical examples are given to demonstrate the application of the formulas.
The process of transitioning from expanded to compressed forms of formulas is explained thoroughly.

02:32:28

Mastering Algebraic Identities Through Practice and Understanding

The formula for a k + b k then a is a k b k, where a is 3y and b is 5z.
To find the period, multiply -3 * 3y * 5z * 3y + 5z.
The cube formulas are discussed, focusing on a p b p b.
When encountering a k - b k, use the formula a - b's cube, which is a k - b k - 3ab.
Practice deriving formulas rather than memorizing them.
The formula a k + b k + c k - 3abc is explained.
The formula a + b + c * a s + b s + c s - a - bc-c is detailed.
Practice is emphasized to understand and apply algebraic identities effectively.
The concept of summing three terms to zero and its relation to the formula a k + b k + c k = 3abc is highlighted.
The importance of practice, understanding, and repetition in mastering algebraic identities is stressed.

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