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.

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## 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*

**Click Here to down the Polynomials Class 9 PDF**

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.