# Class 9 Polynomials Study Notes

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The basic understanding of calculus starts at the school level and continues to be present in a student’s life until higher education. Calculus is an essential mathematical division that solves a range of problems in the STEM field. Its foundation is laid in class 9 with chapter Polynomials. This chapter familiarizes you with the use of variables along with numbers and arithmetic operations. Preparing the class 9 Polynomials chapter thoroughly will not only assist you in scoring well in exams but enable you to easily grasp a plethora of upcoming concepts for class 10th.

## What are Polynomials?

Whenever we combine the use of a particular type of algebraic expression which consists of variables and coefficients with the use of operations like addition, subtraction, multiplication. For example: 2x + 1, 5y + 6, etc.

## Polynomials in One Variable

In a polynomial, a variable can be expressed by a symbol that can assume any real value. For example – 3x, 4y, 7z, -3/4x, etc. are algebraic expressions. If the constant value is unknown then the same expression can be written as ax or by.

The exponential power cannot be negative like 5x-2 as per the class 9 polynomials chapter. Expressions that are written in only one variable like 8+2×2, x2-4, etc. can be classified into polynomials with one variable. A coefficient is primarily a number used to multiply a variable so, in the above examples 3x+6, 3 is the coefficient of x.

## Zeros of a Polynomial

The value or values of x for a given polynomial is equal to zero, then it is the root or a zero of a polynomial. Let us understand this using an example. Take any expression like p(x) = 3x + 1 , now p(x) = 0 i.e. 3x + 1 = 0, this gives us x = -1/3. Therefore, -1/3 is the zero of the polynomial 3x+1.

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## Remainder Theorem

Remainder Theorem states that when a polynomial p(x), where p(x) means any polynomial p whose variable is x, is divided by a linear factor say x-b, where b is an arbitrary number, then the result of this division is equal to p(b). This theorem only works when the function is divided by a linear polynomial of the form x+ or x- any number.

To better understand this, consider a pair of polynomials 2×2+ x + 1 and x.
Here,  (2×2 + x + 1) ÷ x = (2×2 ÷ x) + (x ÷ x) + ( 1÷ x)
Since 1 cannot be divided by x to get a polynomial term, so we stop the division here. The remainder, in this case, is 1 while the quotient is 2x+1
It can be written as 2×2  + x + 1 = { x *(2x + 1)} +1, this gives us a very useful formula i.e.

Dividend = (Divisor × Quotient) + Remainder

## Factorisation of Polynomials

Factorisation of Polynomials is a method of expressing a polynomial into the products of its factors. From our class 9 polynomial chapter, we will consider an expression say p(x) = 4×3 + 3×2 – 4x -3, now let us check whether x-1 is a factor of the above expression or not.

Equate x-1 with 0
i.e., x – 1 = 0 which gives us x = 1
Now if we substitute the value of x in p(x) we get,
p(1) = 4(1)^3 + 3(1)^2 – 4(1) -3
= 4 + 3 – 4 -3 = 0

Thus, we can conclude that x – 1 is a factor of the given expression (as the remainder is equal to 0)
We can even factorise a polynomial equation ax2  + bx + c by splitting the middle term as

ax^2  + bx + c = (px + q) (rx + s) = pr x^2 + (ps + qr)x + as
So, by comparing the coefficients of x^2, we get a = or
Similarly, comparing the coefficients of x, we get b = ps + qr
and, c = qs

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To get clarity for this method, let us have a look at a solved example.

For example:
We have to factorise x3  – 2×2  – x + 2 by splitting the middle term.

Solution:
This can be achieved by
x^3 – 2x^2 – x + 2 = x^2 (x – 2) – 1 ( x -2 )
= (x – 2) (x^2 -1) as the answer.

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## Algebraic Identities

If an algebraic equation is true for all the values of the variables occurring in it, then it is called an algebraic identity. There are four different identities which are there in the class 9 polynomials mentioned below :

Identity 1 : (x + y) ^2 = x^2 + 2xy+ y^2

Identity 2 : (x – y) ^2 =x^2 – 2xy +y^2

Identity 3: x^2 -y^2 = (x+y) (x-y)

Identity 4 : ( x + a) (x + b) = x^2 + (a+b)x + ab

We hope that through this article we could clear your concepts and assist you in understanding the class 9 polynomials chapter. Get in touch with our Leverage Edu experts to get your career direct towards the right path by simple and effective decisions.

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