## Introduction

- Books Name
- Kaysons Academy Maths Foundation Book

- Publication
- Kaysons Publication

- Course
- JEE

- Subject
- Maths

**Chapter -2**

**Polynomials**

**Introduction**

In this chapter, we shall start our study with a particular type of algebraic expression, called polynomial, and the terminology related to it. We shall also study the Remainder Theorem and Factor Theorem and their use in the factorisation of polynomials. In addition to the above, we shall study some more algebraic identities and their use in factorisation and in evaluating some given expressions.

**Polynomials in One Variable**

Let us begin by recalling that a variable is denoted by a symbol that can take any real value. We use the letters x, y, z, etc. to denote variables. Notice that 2*x*, 3*x*, – *x*, are algebraic expressions. All these expressions are of the form (a constant) × *x*. Now suppose we want to write an expression which is (a constant) × (a variable) and we do not know what the constant is. In such case, we write the constant as a, b, c, etc. So the expression will be a*x*, say.

However, there is a difference between a letter denoting a constant and a letter denoting a variable. The values of the constants remain the same throughout a particular situation, that is, the values of the constants do not change in a given problem, but the value of a variable can keep changing.

Now, consider a square of side 3 units What is its perimeter? You know that the perimeter of a square is the sum of the lengths of its four sides. Here, each side is 3 units. So, its perimeter is 4 × 3, i.e., 12 units. What will be the perimeter if each side of the square is 10 units? The perimeter is 4 × 10, i.e., 40 units. In case the length of each side is x units, the perimeter is given by 4x units. So, as the length of the side varies, the perimeter varies.

Expressions of this form are called polynomials in one variable. In the examples above, the variable is *x.* For instance, *x*^{3} – *x*^{2} + 4*x* + 7is a polynomial in *x*. For instance, *x*^{3} – *x*^{2} + 4*x* + 7 is a polynomial in *x*. Similarly, 3y^{2} + 5y is a polynomial in the variable y and t^{2} + 4 is a polynomial in the variable t.

In the polynomial *x*^{2 }+ 2*x*, the expressions *x*^{2} and 2*x* are called the terms of the polynomial. Similarly, the polynomial 3y^{2} + 5y + 7 has there terms, namely, 3y^{2}, 5y and 7. Can you write the terms of the polynomial –*x*^{3} + 4*x*^{2} + 7*x* – 2 ? This polynomial has 4 terms, namely, –*x*^{3}, 4*x*^{2}, 7*x* and –2.

Each term of a polynomial has a coefficient. So, in –*x*^{3} + 4*x*^{2} + 7*x* – 2, the coefficient of *x*^{3} is –1, the coefficient of *x*^{2} is 4, the coefficient of *x *is 7 and –2 is the coefficient of *x*^{o} (Remember, *x*^{o} = 1). Do you know the coefficient of *x* in *x*^{2} – *x* + 7? It is –1.

2 is also a polynomial. In fact, 2, –5, 7 etc. are example of constant polynomials. The constant polynomial 0 is called the zero polynomial. This plays a very important role in the collection of all polynomials, as you will see in the higher classes.

Now, look at the polynomial p(x) = 3x^{7} – 4x^{6} + x + 9. What is the term with the highest power of x ? It is 3*x*^{7}. The exponent of x in this term is 7. Similarly, in the polynomial q(y) = 5y^{6} – 4y^{2} – 6, the term with the highest power of y is 5y^{6} and the exponent of y in this term is 6. We call the highest power of the variable in a polynomial as the degree of the polynomial. So, the degree of the polynomial 3x^{7} – 4 x^{6} + x + 9 is 7 and the degree of the polynomial 5y^{6} – 4y^{2} – 6 is 6. The degree of a non-zero constant polynomial is zero.

Now, that you have seen what a polynomial of degree 1, degree 2, or degree 3 looks like, can you write down a polynomial in one variable of degree n for any natural number n? A polynomial in one variable x of degree n is an expression of the form

a_{n}*x*^{n} + a_{n – 1}*x ^{x}*

^{ – 1 }+ … + a

_{1}

*x*+ a

_{o}

Where a_{o}, a_{1}, a_{2}, …, a_{n} are constants and a_{n} ≠ 0.

In particular, if a_{o }= a_{1} = a_{2} = a_{3} = … = a_{n} = 0 (all the constants are zero), we get the zero polynomial, which is denoted by 0. What is the degree of the zero polynomial? The degree of the zero polynomial is not defined.

## Remainder Theorem

- Books Name
- Kaysons Academy Maths Foundation Book

- Publication
- Kaysons Publication

- Course
- JEE

- Subject
- Maths

**Remainder Theorem**

Let us consider two numbers 15 and 6. You know that when we divide 15 by 6, we get the quotient 2 and remainder 3. Do you remember how this fact is expressed? We write 15 as

15 = (6 × 2) + 3

We observe that the remainder 3 is less than the divisor 6. Similarly, if we divide12 by 6, we get

12 = (6 × 2) + 0

What is the remainder here? Here the remainder is 0, and we say that 6 is a factor of 12 or 12 is a multiple of 6.

Now, the question is: can we divide one polynomial by another? To start with, let us try and do this when the divisor is a monomial. So, let us divide the polynomial 2x^{3} + x^{2} + x by the monomial x.

We have (2*x*^{3} + *x*^{2} + *x*) ÷ *x* =

= 2*x*^{2} + *x* + 1

In fact, you may have noticed that x is common to each term of 2x^{3} + x^{2} + x. So we can write

(2x^{3} + x^{2 }+ x as x(2x^{2} + x + 1).

We say that *x* and 2*x*^{2} + *x* + 1 are factors of 2*x*^{3} + *x*^{2} + *x*, and 2*x*^{3} + *x*^{2} + *x *is a multiple of *x *as well as a multiple of 2*x*^{2} + *x *+ 1.

Consider another pair of polynomials 3*x*^{2} + *x* + 1 and *x*.

Here, (3*x*^{2} + *x* + 1) ÷ *x* = (3*x*^{2} ÷ *x*) + (*x* ÷ *x*) + (1 + *x*).

We see that we cannot divide 1 by x to get a polynomial term. So in this case we stop here, and note that 1 is the remainder. Therefore, we have

3*x*^{2} + *x* + 1 = {*x *× (3*x* + 1) } + 1

In this case, 3x + 1 is the quotient and 1 is the remainder. Do you think that x is a factor of 3x^{2} + x + 1? Since the remainder is not zero, it is not a factor.

Now let us consider an example to see how we can divide a polynomial by any non-zero polynomial.

i.e., Dividend = (Divisor × Quotient) + Remainder

In general, if p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x) and g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that:

p(x) = g(x)q(x) + r(x),

Where r (x) = 0 or degree of r(x) < degree of g(x). Here we say that p(x) divided by g(x), gives q(x) as quotient and r(x) as remainder.

In the example above, the divisor was a linear polynomial. In such a situation, let us see if there is any link between the remainder and certain values of the dividend.

**Remainder Theorem**: Let p(x ) be any polynomial of degree greater than or equal to one and let a be any real number. If p (x) is divided by the linear polynomial x – a, then the remainder is p (a).

**Proof:** Let p(x) be any polynomial with degree greater than or equal to 1. Suppose that when p(x) is divided by x – a, the quotient is q(x) and the remainder is r (x), i.e.,

p(x) = (x – a) q(x) + r (x)

Since the degree of x – a is 1 and the degree of r(x) is less than the degree of x – a, the degree of r(x) = 0. This means that r(x) is a constant, say r.

So, for every value of x, r(x) = r.

Therefore, p(x) = (x – a) q(x) + r

In particular, if x = a, this equation gives us

p(a) = (a – a) q(a) + r = r,

Which proves the theorem?

Let us use this result in another example.

## Factorization of Polynomials

- Books Name
- Kaysons Academy Maths Foundation Book

- Publication
- Kaysons Publication

- Course
- JEE

- Subject
- Maths

**Factorization of Polynomials**

Let us now look at the situation of Example 10 above more closely. It tells us that since the remainder, (2t + 1) is a factor of q(t), i.e., q(t) = (2t + 1) g(t) for some polynomial g(t). This is a particular case of the following theorem.

**Factor Theorem:** If p(x) is a polynomial of degree n > 1 and a is any real number, then (i)

x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x).

**Proof:** By the Remainder Theorem, p(x)=(x – a) q(x) + p(a).

(i) If p(a) = 0, then p(x) = (x – a) q(x), which shows that x – a is a factor of p(x).

(ii) Since x – a is a factor of p(x), p(x) = (x – a) g(x) for same polynomial g(x). In this case,

p(a) = (a – a) g(a) = 0.

## Algebraic Identities

- Books Name
- Kaysons Academy Maths Foundation Book

- Publication
- Kaysons Publication

- Course
- JEE

- Subject
- Maths

**Algebraic Identities**

From your earlier classes, you may recall that an algebraic identity is an algebraic equation that is true for all values of the variables occurring in it. You have studied the following algebraic identities in earlier classes:

**Identity I**: (x + y)^{2} = x^{2} + 2xy + y^{2}

**Identity II**: (x – y)^{2} = x^{2} – 2xy + y^{2}

**Identity III**: x^{2} – y^{2} = (x + y) (x – y)

**Identity IV**: (x + a) (x + b) = x^{2} + (a + b)x + ab

You must have also used some of these algebraic identities to factorise the algebraic expressions. You can also see their utility in computations.