Polynomial and Its Types

By now you are aware of the polynomial equation in one variable and

their degrees. In this article, we will look at the various types of

polynomials to establish a foundation for further studies into them.

Introduction to Polynomial Equation

If p(x) is a polynomial equation in x, then the highest power of x in

p(x) is called the degree of the polynomial p(x). So, p(x) =

● 4x + 2 is a polynomial equation in the variable x of degree 1

● 2y2 – 3y + 4 is a polynomial in the variable y of degree 2

● 5x3 – 4x2 + x – 2 is a polynomial in the variable x of degree 3

● 7u6 – 3u4 + 4u2 – 6 is a polynomial in the variable u of degree

6

Further, it is important to note that the following expressions are NOT

polynomials:

● 1 / (x – 1)

● √x + 2

● 1 / (x2 + 2x + 3)

Types of Polynomials

Let’s look at the different types of polynomials that you will come

across while studying them.

Linear Polynomials

Any polynomial with a variable of degree one is a Linear Polynomial.

Some examples of the linear polynomial equation are as follows:

● 2x – 3

● y + √2

● x √3 + 5

● x + 5/11

● 2/3y – 5

Any polynomial where the degree of the variable is greater than 1 is

not a linear polynomial.

Download NCERT Solutions for Class 10 Maths

Quadratic Polynomials

Any polynomial with a variable of degree two is a Quadratic

Polynomial. The name ‘quadratic’ is derived from the word ‘quadrate’

which means square. Some examples of the quadratic polynomial

equation are as follows:

● 2x2 + 3x – 5

● y2 – 1

● 2 – x2 + x√3

● u/3 – 2u2 + 5

● v2√5 + 2/3v – 6

● 4z2 + 1/7

To generalize, most quadratic polynomials in x are expressed as, ax2 +

bx + c … where a, b, and c are real numbers where a ≠ 0.

Cubic Polynomials

Any polynomial with a variable of degree three is a Cubic Polynomial.

Some examples of the cubic polynomial equation are as follows:

● x3

● 2 – x3

● x3√2

● x3 – x2 + 3

● 3x3 – 2x2 + x – 1

To generalize, most quadratic polynomials in x are expressed as, ax3 +

bx2 + cx + d … where a, b, c, and d are real numbers. Also, a ≠ 0.

Some more concepts

To begin with, let’s look at the following polynomial p(x),

p(x) = x2 – 3x – 4

Next, let’s put x = 2 in p(x). So, we get p(2) = (2)2 – 3(2) – 4 = 4 – 6 –

4 = –6. Note that the value ‘– 6’ is obtained by replacing x with 2 in

the polynomial x2 – 3x – 4. Hence, it is called ‘the value of x2 – 3x – 4

at x = 2’. Similarly, p(0) is the value of x2 – 3x – 4 at x = 0.

Therefore, we can say, If p(x) is a polynomial in x, and if k is any real

number, then the value obtained by replacing x by k in p(x), is called

the value of p(x) at x = k and is denoted by p(k).

Zero of a Polynomial

So, what is the value of p(x) = x2 – 3x – 4 at x = – 1? p(– 1) = (– 1)2 –

{3(– 1)} – 4 = 1 + 3 – 4 = 0. Also, the value of p(x) = x2 – 3x – 4 at x

= 4 is, p(4) = (4)2 – 3(4) – 4 = 16 – 12 – 4 = 0. In this case, since p(-1)

and p(4) is equal to zero, ‘-1’ and ‘4’ are called zeroes of the quadratic

polynomial x2 – 3x – 4.

Therefore, we can say, A real number k is said to be a zero of a

polynomial p(x) if p(k) = 0. In the previous years, you have already

studied how to find zeroes of a polynomial equation. To elaborate on

it a little more, if ‘k’ is a zero of p(x) = 2x + 3, then

p(k) = 0

Or 2k + 3 = 0

i.e. k = – 3/2

Let’s generalize this. If ‘k’ is a zero of p(x) = ax + b, then p(k) = ak +

b = 0. Or, k = – b/a. In other words, the zero of the linear polynomial

(ax + b) is: – (Constant Term) / (Coefficient of x)

Hence, we can conclude that the zero of a linear polynomial equation

is related to its coefficients. You will study more about if this rule is

applicable to all types of polynomials discussed above.

Solved Examples for You

Question: What are the three types of polynomials and how are they

differentiated?

Solution: The three types of polynomials are:

1. Linear

2. Quadratic

3. Cubic

The linear polynomials have a variable of degree one, quadratic

polynomials have a variable with degree two and cubic polynomials

have a variable with degree three.

Value of Polynomial and Division Algorithm

Arithmetic operations like addition, subtraction, multiplication and

division play a huge and most basic rule in Mathematics. Maths is

made by these operations. All other operations go easy with the

polynomials except the division operation, which gets complex when

dealt with polynomials. But this section will explain to you the

division of polynomials and the division algorithm related to it, from

basics.

So, what’s the basic formula we are learning from the day we solved

our first division problem? This is:

Dividend = Quotient × Divisor + Remainder

Example: Divide the polynomial 2x2+3x+1 by polynomial x+2.

Solution: Divisor= x+2

Dividend=2x2 + 3x + 1

Note: Put the dividend under the division sign and divisor outside the

sign.

Steps for Division of Polynomials

● Step 1: Firstly, Arrange the divisor as well as dividend

individually in decreasing order of their degree of terms.

● Step 2: In case of division we seek to find the quotient. To find

the very first term of the quotient, divide the first term of the

dividend by the highest degree term in the divisor. In the

current case,

2x2/x = 2x.

● Step 3: Write 2x in place of the quotient.

● Step 4: Multiply the divisor by the quotient obtained. Put the

product underneath the dividend.

● Step 5: Subtract the product obtained as happens in case of a

division operation.

● Step 6: Write the result obtained after drawing another bar to

separate it from prior operations performed.

● Step 7: Bring down the remaining terms of the dividend.

● Step 8: Again divide the dividend by the highest degree term of

the remaining divisor. Follow the same prior procedure until

either the remainder becomes zero or its degree is less than the

degree of the divisor.

● Step 9: At this stage, the quotient obtained is our answer.

Quotient Obtained = 2x + 1

Note: Division Algorithms for Polynomials is same as the Long

Division Algorithm In Polynomials

Download NCERT Solutions for Class 10 Maths

Division Algorithm For Polynomials

Division algorithm for polynomials states that, suppose f(x) and g(x)

are the two polynomials, where g(x)≠0, we can write:

f(x) = q(x) g(x) + r(x)

which is same as the Dividend = Divisor * Quotient + Remainder and

where r(x) is the remainder polynomial and is equal to 0 and degree

r(x) < degree g(x).

Verification of Division Algorithm

Take the above example and verify it.

Divisor = x+2

Dividend = 2x2 + 3x + 1

Quotient = 2x – 1

Remainder = 0

Applying the Algorithm:

2x2 + 3x + 1 = (x + 2) (2x + 1) + 0

2x2 + 3x + 1 = 2x2 + 3x + 1

Hence verified.

Finding Factors of Polynomials with Division Algorithm

Long division algorithm is used to find out factors of polynomials of

degree greater than equal to two. We’ll be describing the steps to find

out the factors along with an example.

Example: Find roots of cubic polynomial P(x)=3x3 – 5x2 – 11x – 3

Solution

● Step 1: Use the factor theorem to find a factor of the

polynomial.

● Step 2: First divide the whole equation by the coefficient of the

highest degree term of the dividend.

P(x)=3x3 – 5x2 – 11x – 3

On dividing the whole equation by 3,

P(x) =x3 – (5/3)x2 – (11/3)x – 1

● Step 3: Find out factors of the constant term so obtained. In the

present case, factors of the constant term are 1 and -1.

● Step 4: Put the value of x in P(x) = 3x3 – 5x2 – 11x – 3 equal

to 1 and find the remainder. Again put the value of remainder

equal to -1 in and find the remainder using remainder theorem.

Find the value of x for which remainder is zero for the cubic

polynomial.

P (1) = 3(1)3 – 5(1)2 – 11(1) – 3 = -16

P(-1) = 3( -1 )3 – 5( -1 )2 – 11( -1 ) – 3 =0

● Step 5: Remainder is zero for x = -1. So, (x + 1) is a root of the

polynomial.

● Step 6: By Division Algorithm, find out the quotient. It comes

out: 3x2 – 8x – 3

● Step 7: Now, Quotient = 3x2 – 8x – 3

Dividend = (Divisor) * (Quotient) + Remainder

In present case,

3x3 – 5x2 – 11x – 3 = (x + 1) (3x2 – 8x – 3) + 0

By factorizing the quadratic polynomial we shall be able to find out

remaining factors of the cubic polynomial.

● Step 8: Break middle term in terms of a pair of numbers such

that its product is equal to -9 and summation equal to -3.

● Step 9: On factorizing, possible pair of number satisfying both

conditions is (-9, 1). Breaking the middle term,

f(x) = 3x2 – 8x – 3

= 3x2 – 9x + x – 3

● Step 10: Form pairs of terms and factor out GCD of the two

pairs separately. Then again factor out GCD of the remaining

two products.

● Step 11:

f(x) = 3x2 – 8x – 3 = 3x2 – 9x + x – 3

= 3x(x – 3) + 1(x – 3) = (x – 3)(3x + 1)

Now,

3x3 – 5x2 – 11x – 3 = (x + 1) (3x2 – 8x – 3) + 0

= (x + 1) ( x – 3)(3x + 1)

Factors of cubic polynomial are -1, 3 and -1/3.

Degree of Polynomial

Well, before starting, I would like to tell you that this ‘degree’ has

nothing to do with your thermometer’s degree or to your course

completion certification. The term ‘degree’ has come to the important

part of Mathematics, i.e., Polynomials and is adding an essential

meaning to it. So, let’s hit directly to understand the Degree of

Polynomials in details. 

Degree of Polynomials

Source: Wikihow

Polynomial in One Variable

The degree of polynomials in one variable is the highest power of the

variable in the algebraic expression. For example, in the following

equation: x2+2x+4. The degree of the equation is 2 .i.e. the highest

power of variable in the equation.

Multivariable polynomial

For a multivariable polynomial, it the highest sum of powers of

different variables in any of the terms in the expression. Take

following example, x5+3x4y+2xy3+4y2-2y+1. It is a multivariable

polynomial in x and y, and the degree of the polynomial is 5 – as you

can see the degree in the terms x5 is 5, x4y it is also 5 (4+1) and so the

highest degree among these individual terms is 5.

A polynomial of two variable x and y, like axrys is the algebraic sum

of several terms of the prior mentioned form, where r and s are

possible integers. Here, the degree of the polynomial is r+s where r

and s are whole numbers.

Note: Exponents of variables of a polynomial .i.e. degree of

polynomials should be whole numbers.

Download NCERT Solutions for Class 10 Maths

How to find the Degree of a Polynomial?

There are 4 simple steps are present to find the degree of a

polynomial:-

Example: 6x5+8x3+3x5+3x2+4+2x+4

● Step 1: Combine all the like terms that are the terms of the

variable terms.

(6x5+3x5)+8x3+3x2+2x+(4+4)

● Step 2: Ignore all the coefficients

x5+x3+x2+x+x0

● Step 3: Arrange the variable in descending order of their

powers

x5+x3+x2+x+x0

● Step 4: The largest power of the variable is the degree of the

polynomial

deg(x5+x3+x2+x+x0) = 5

Classification Based on the Degree of the Equation

Based on the degree, the equation can be linear, quadratic, cubic, and

bi-quadratic, and the list goes on.

Name of the Equation Degree of the Equation

Linear Equation 1

Quadratic Equation 2

Cubic Equation 3

Bi-Quadratic equation 4

Importance of Degree of polynomial

Case of Homogeneous Polynomial

The degree of terms is a major deciding factor whether an equation is

homogeneous or not. A polynomial of more that one variable is said to

be homogeneous if the degree of each term is the same. For example,

2x7+5x5y2-3x4y3+4x2y5 is a homogeneous polynomial of degree 7 in x

and y.

Relation of Degree of Polynomials with Zeroes of Equation

Theorem 1: A polynomial f(x) of the nth degree cannot vanish for

more than n values of x unless all its coefficients are zero.

Name of the Equation Degree of the Equation Possible Real Solutions

Linear Equation 1 1

Quadratic Equation 2 2

Cubic Equation 3 3

Bi-Quadratic equation 4 4

The above table shows possible real zeros /solutions; actual real

solutions can be less than the degree of the equation.

Note: A constant polynomial is that whose value remains the same. It

contains no variables. The power of the constant polynomial is Zero.

Well, you can write any constant with a variable having an

exponential power of zero. If the constant term = 4, then the

polynomial form is given by f(x)= 4x0

Before going to start other sections of Polynomials, try to solve the

below-given question.

A Question for You

Question: Find the degree of polynomial x3+4x5+5x4+2x2+x+5.

Solution: x3+4x5+5x4+2x2+x+5

=4x5+5x4+x3+2x2+x+5

=x5+x4+x3+x2+5

Degree of equation is the highest power of x in the given equation .i.e.

5.

Factorization of Polynomials

Do you know what factorization of polynomials means? Now that you

have an idea of what polynomials are, let’s learn how to factorize a

polynomial. This is an important step towards solving an equation in

mathematics. Let’s find out.

Before starting, let’s try to solve an example. If you are given 12

chocolates, choose the correct option in which you can divide them,

such that you are left with no chocolates:

● Group of 3 people

● Group of 4 people

● Group of 6 people

Well, the answer is, any one of the options will do. Why so? Well, you

know that 3, 4 and 6 are factors of 12 from your knowledge of number

system, which you have learned long back. So, we can divide the 12

chocolates into these groups.

Factors

What are the factors? When it comes to integers, if a number has an

integral value and that particular value get divided completely by

another number(s) without leaving any remainder other than zero, then

the dividend is said to be a factor(s) of that integer.

OR

When an integer could be written as a product of two or more integers

each then such numbers will be its factor. Why not consider some

examples to understand more. What can be the factors for 12?

12 = 6 × 2

12 = 4 × 3

12 = 12 × 1

So, you can see that a number can be factorized in more than one way.

But it’s not applicable to all the numbers. For a number like 7, 3, 5, 11

(which are prime numbers), these numbers have only one factor other

than itself.

7 = 7 × 1

So, it can be factorized in only one possible way.

Factorization of Polynomials

The same case of numbers also exists with Polynomials. A polynomial

can be written as a product of two or more polynomials of degree less

than or equal to that of it. Each polynomial involved in the product

will be a factor of it.

Graph of a Polynomial (Source: Wikipedia)

The process involved in breaking a polynomial into the product of its

factors is known as the factorization of polynomials. As I’ve told in

the previous sections that Monomial, Binomials are just the other

names for Polynomials. Its always good to start with the smaller and

easiest one. So, let’s hit the Monomials.

Download NCERT Solutions for Class 10 Maths

Monomials

Monomials can be factorized in the same way as integers, just by

writing the monomial as the product of its constituent prime factors. In

the case of monomials, these prime factors can be integers as well as

other monomials which cannot be factorized further. Factorize:

● a3 = (a) × (a) × (a)

● 3abc = (3) × (a) × (b) × (c)

We will factorize monomials with binomials, trinomial and

polynomial. But before doing that, we will brush up concepts on GCD

or HCF.

GCD or HCF

For a given set of numbers, the Greatest Common Number that will

divide each of the numbers will be the GCD of that particular set of

numbers. It is also known as HCF of numbers .i.e. Highest Common

Factor.

Steps To Calculate the GCD of Integers

● Step 1: Break the number into the product of its prime factors.

● Step 2: Identify the common factors for the given set of

numbers.

● Step 3: The product of common factors will be gcd of the

number set.

● Step 4: If no common factor is found choose 1 as a common

factor.

Example: Find GCD of 15 and 24.

15 = 3 x 5

24 = 2 x 2 x 2 x 3

GCD of 15 and 24 is 3. This same procedure applies to polynomials.

Steps To Calculate the GCD of Polynomials

● Step 1: For a given set of polynomials, break the polynomial

into its factors such that each factor polynomial cannot be

factorized further.

● Step 2: Identify common terms or polynomials for a given set

of polynomials.

Example: Find GCD of 15ab and 3bc.

15ab = 3 x (5) x (a) x (b)

23bc = 3 x (b) x (c)

Here, GCD of 15ab and 3bc is 3b.

Factoring Binomials

Follow the 4 easy steps to factorize Binomials:

● Step 1: Break each term into its prime factors.

● Step 2: Find common factors or GCD for all individual terms.

● Step 3: Factor out the common factor or GCD.

● Step 4: The remaining terms in individual terms will form

another polynomial. Put them in separate bracket multiplied by

the GCD factored out initially.

● Step 5: The resulting product of terms will be the factors of the

initial polynomial (binomial).

Example: Factorize 15ab + 3bc

= 3 × (5) × (a) × (b) + 3 × (b) × (c) = 3b (5a + c)

Factoring Quadratic Polynomials

Case 1: When the coefficient of x2 is unity.

The general form of an equation is x2+bx+c. Every quadratic equation

can be expressed as: x2+bx+c= (x+d)(x+e). Here, b is the sum of d and

e & c is the product of d and e. Example: (x+2)(x+3) = x2+ 2x + 3x +

6 = x2+ 5x + 6.

Here, 5 = 2 + 3 = d + e = b in general form and 6 = 2 × 3 = d × e = c in

general form. To factorize quadratic polynomial, we shall be looking

for numbers which on multiplication will get equal to c and on

summation equal to b.

Example: Factorize x2+8x+12.

Solution: Below the steps are given for your understanding.

● Step 1: Factorize 12 as 12 = 2 × 6 or = 4 × 3. We have to find a

pair, such that its product is equal to 12 and summation is equal

to 8. Only one such pair is possible i.e. 2 and 6.

2 + 6 = 8

2 × 6 = 12

● Step 2: Break middle term in terms of the summation of a pair

of numbers such that its product is equal to c i.e. 12 in above

case. We will write 8=6+2

x2+ (6+2)x+ 12

= x2+ 6x +2x + 12

● Step 3: Form pairs of terms and factor out GCD of the two

pairs separately.

= x2+ 6x +2x + 12 = (x2+ 6x) +(2x + 12) = x(x+6)+2(x+6)

● Step 4: Again factor out GCD of remaining sum of products.

Follow factorization procedure of binomials as explained

earlier. Factor out (x+6) from sum of product,

=x(x+6)+2(x+6) = (x+6)(x+2)

Example: Do the factorization of polynomials: x2-5x-6

Soluiton: Factors of -6 are 2 × -3; – 3 × 2; 1 × -6; -1 × 6 and the pair

summing up to -5 is (-6,1) as required. Hence,

x2-5x-6 = x2-6x+x-6 = x(x-6)+(x-6) = (x-6)(x+1)

Case 2:When the coefficient of x2 a is an integer other than 1 or -1

General form is given by ax2+bx+c. Any quadratic of form ax2+bx+c

is expressible in the product of two linear polynomials:

ax2+bx+c= (a1x+b1)(a2x+b2)

where a is the product of a1 and a2, c is the product of b1 and b2 and b

is the sum of the product of a1b2 and a2b1. Consider one more

example,

(3x+2)(x+4)= 3x2+ 2x +12x +8 = 3x2+14x+ 8

Here, 3 = 3 × 1 = a1 × a2 = a in general form. 14 = 2 × 1 + 4 × 3 =

(a1b2) × (a2b1) = b in general form. 8 = 4 × 2 = b1 × b2= c in general

form. Example: Factorize 3x2 – 5x – 2

● Step 1: find factors of 3 × – 2 = -6.

● Step 2: Product of factors of -6 are = -3 × 2; -2 × 3; 6 × -1; -1 ×

6. Here, the possible pair of factor gives a summation of -1 is

(-6,1).

● Step 3: Break the middle term as the summation of two

numbers such that its product is equal to -6. Calculated above

such two numbers are -6 and 1.

● Step 4: Breaking the middle term: x2-6x+x-2

● Step 5: Making pairs of terms: (3x2-6x)+(x-2)

● Step 6: Factor out GCD for both the pairs: 3x(x-2)+1(x-2).

Note: 1 is factored out from the second bracket as there was no

other common factor.

● Step 7: Factor out GCD from the resulting summation of

products: =3x(x-2)+1(x-2) =(x-2)(3x+1)

Factorization of Polynomials by Grouping

We can factorize polynomial containing even number of terms by

forming pairs of terms.

● Step 1: Form pairs out of given even number of terms.

● Step 2: Try factoring out GCD from all the pairs separately.

● Step 3: Lastly, factor out the remaining common factor from

the products formed.

Note: In case you do not get common factors for the pairs formed, try

rearranging the terms and follow the same procedure again. Example:

Factorize x2+ 4xy+4y+x

= (x2 + 4xy) + (4y + x)

=x(x + 4y) + 1( 4y + x)

=(x + 4y)(x + 1)

Factorization of Polynomials by Perfect Squares

A trinomial which can be factored, such that both the factors are same.

Then, it will form a perfect square trinomial. For example, x2+2x+1 =

(x + 1)(x + 1) = (x + 1)2. There are certain identities which are

important for perfect square trinomials are as follows:

(a+b)2=(a2+2ab+b2)

Example:Do the factorization of polynomials, 4x2+12x+9

Solution: 4x2+12x+9 = (2x)2 + 2(2x)(3x) + 32 = (2x + 3)2

It can be seen from the identities and example above, that a trinomial

with first and last terms as perfect squares and the middle term can be

written as twice of the product of roots of first and last terms then the

trinomial can be expressed as a perfect square.

Factorization of Polynomials by Difference of Squares

This applies mainly to the pair of two polynomial terms which are a

perfect square and expressed as the difference between them.

a2-b2 = (a – b)(a + b)

Example: 9x2 – 4

Solution: 9x2 – 4 = (3x+2)(3x-2) { Using Identity }

Solved Example for You

Question: Factorise x2+11x+18

Solution: Consider the quadratic polynomial, x2+11x+18

=x2 + 9x + 2x + 18

=(x2 + 9x) + (2x + 18)

=x(x + 9) + 2(x + 9)

=(x + 9)(x + 2)

Remainder Theorem

When you divide one polynomial by another the process can be very

long. The Remainder and Factor Theorems help us avoid this long

division process by providing certain rules. We will learn about the

Remainder Theorem in this article. 

Introduction

Divide 15 by 6. What answer do you get? By using the simple division

process, we find that the quotient is 2 and the remainder is 3. Hence,

we write 15 = (2 x 6) + 3. Note: The remainder ‘3’ is less than the

divisor ‘6’. On the other hand, when we divide 12 by 6, we get a

quotient of 2 and remainder 0. In this case, we say that 6 is a factor of

12 OR 12 is a multiple of 6.

Dividing One Polynomial by Another

Let’s start by dividing a polynomial with a monomial as follows:

Dividend Polynomial: 2x3 + x2 + x

Divisor Monomial: x

We have, (2x3 + x2 + x) / x = (2x3)/x + x2/x + x/x = 2x2 + x + 1

Observe that ‘x’ is common to each term of the dividend polynomial.

We can also write the dividend as, x(2x2 + x + 1). Hence, ‘x’ and ‘2x2

+ x + 1’ are factors of ‘2x3 + x2 + x’.

==================================================

=================

Let’s look at another example.

Dividend polynomial: 3x2 + x + 1

Divisor Monomial: x

We have, (3x2 + x + 1) / x = (3x2)/x + x/x + 1/x = 3x + 1 + 1/x

Here, 1 is not divisible by x. So, we stop the division here and note

that 1 is the remainder. Hence, we have 3x2 + x + 1 = {x(3x + 1)} + 1.

So, the result of the division is,

Quotient: 3x + 1

Remainder: 1

Note that since the remainder is not zero, x is not a factor of 3x2 + x +

1.

Dividing a Polynomial by Any Non-zero Polynomial

Let’s do the following division: Dividend: x + 3x2 – 1 and Divisor: 1

+ x

Step 1.

We arrange the terms on the descending order of their degrees. Hence,

we have dividend: 3x2 + x – 1 and divisor: x + 1

Step 2.

Divide the first term of the dividend by the first term of the divisor:

3x2/x = 3x. This is the first term of the quotient.

Step 3.

Multiply the divisor by the first term of the quotient and subtract this

product from the dividend,

{3x2 + x – 1} – {3x(x + 1)}

= {3x2 + x – 1} – {3x2 + 3x}

= – 2x – 1

‘– 2x – 1’ is the remainder.

Step 4.

Now, the new dividend is ‘– 2x – 1’ and the divisor is still the same.

Repeat step 2 to get the next term of the quotient. Divide the first term

of the new dividend by the first term of the divisor: (– 2x)/x = – 2 =

the second term of the quotient

Step 5.

Multiply the divisor by the second term of the quotient and subtract

this product from the new dividend,

(– 2x – 1) – {– 2(x + 1)} = (– 2x – 1) – (– 2x – 2)

= – 2x – 1 + 2x + 2 = 1 = Remainder

Remember: This process continues until the degree of the new

dividend is less than the degree of the divisor.

Step 6.

Hence, we have quotient: 3x – 2 and remainder: 1. It is important to

note here that, 3x2 + x – 1 = (x + 1) (3x – 2) + 1 or Dividend =

(Divisor × Quotient) + Remainder

Therefore, we can conclude that, 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.

Next, let’s try to find a link between the remainder and the dividend.

Let’s find the value of the dividend polynomial at x = -1

p(x) = 3x2 + x – 1

p(-1) = 3(- 1)2 + (- 1) – 1

= 3 – 1 – 1

= 1 … which is the remainder!

So, the remainder of (3×2 + x – 1) / (x + 1) = Value of (3×2 + x – 1) at

x = – 1 (or the zero of the divisor [x + 1]). In other words, the

remainder obtained on dividing a polynomial by another is the same as

the value of the dividend polynomial at the zero of the divisor

polynomial. This brings us to the first theorem of this article.

Download NCERT Solutions for Class 10 Maths

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: p(x) is a polynomial with degree greater than or equal to one. It

is divided by a polynomial (x – a), where ‘a’ is a real number. Let’s

assume that the quotient is q(x) and the remainder is r(x). So, we can

write,

p(x) = (x – a)q(x) + r(x)

Now, the degree of (x – a) is 1. Also, since r(x) is the remainder, its

degree is less than the degree of the divisor: (x – a). Therefore, the

degree of r(x) = 0. In other words, r(x) is a constant. Let’s call the

constant ‘r’. Hence, for all values of ‘x’, r(x) = r. Therefore,

p(x) = (x – a) q(x) + r

Now, let’s find p(a) or the value of p(x) at x = a.

p(a) = (a – a) q(a) + r

= (0)q(a) + r

= r

We see that when a polynomial p(x) of a degree greater than or equal

to one is divided by a linear polynomial (x – a), where a is a real

number, then the remainder is r which is also equal to p(a). This

proves the Remainder Theorem.

For example, check whether the polynomial q(t) = 4t3 + 4t2 – t – 1 is a

multiple of 2t+1.

Solution: q(t) will be a multiple of 2t + 1 only, if 2t + 1 divides q(t)

with remainder zero. Let’s find the zero of the divisor polynomial:

2t + 1 = 0 Or t = – ½

Next, let’s find the value of q(t) at t= – ½

q(- ½) = 4(- ½)3 + 4(- ½)2 – (- ½) – 1

= 4(- 1/8) + 4(- ¼) + ½ – 1

= – ½ + 1 + ½ – 1 = 0.

Hence, we can conclude that the remainder obtained on dividing q(t)

by 2t + 1 is 0. And, (2t + 1) is a factor of ‘4t3 + 4t2 – t – 1’.

Solved Examples for You

Question: Find the remainder when x4 + x3 – 2x2 + x + 1 is divided by

(x – 1).

Solution: Dividend Polynomial = p(x) = x4 + x3 – 2x2 + x + 1 and

Divisor Polynomial = x – 1

Zero of the divisor polynomial is x – 1 = 0 or, x = 1.

Therefore, p(1) = (1)4 + (1)3 – 2(1)2 + 1 + 1 = 1 + 1 – 2 + 1 + 1 = 2.

So, by Remainder Theorem, the remainder is 2.

Question: Find the remainder when (x3 – ax2 + 6x – a) is divided by (x

– a).

Solution: Dividend Polynomial = p(x) = x3 – ax2 + 6x – a and Divisor

Polynomial = x – a

Zero of the divisor polynomial is x – a = 0 or, x = a.

Therefore, p(a) = (a)3 – a(a)2 + 6a – a = a3 – a3 + 6a – a = 5a

So, by Remainder Theorem, the remainder is 5a.

Factor Theorem

In this part, we will look at the Factor Theorem, which uses the

remainder theorem and learn how to factorise polynomials. Further,

we will be covering the splitting method and the factor theorem

method. 

Factor Theorem

If p(x) is a polynomial of degree n > 1 and a is any real number, then

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

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

Let’s look at an example to understand this theorem better.

Example:

Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6.

Solution: To begin with, we know that the zero of the polynomial (x +

2) is –2. Let p(x) = x3 + 3x2 + 5x + 6

Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6 = –8 + 12 – 10 + 6 = 0

According to the factor theorem, if p(a) = 0, then (x – a) is a factor of

p(x). In this example, p(a) = p(- 2) = 0

Therefore, (x – a) = {x – (-2)} = (x + 2) is a factor of ‘x3 + 3x2 + 5x +

6’ or p(x).

Download NCERT Solutions for Class 10 Maths

Factorisation of polynomials

You can factorise polynomials by splitting the middle term as follows:

to begin with, consider a polynomial ax2 + bx + c with factors (px + q)

and (rx + s). Therefore, we have ax2 + bx + c = (px + q) (rx + s). So,

ax2 + bx + c = prx2 + (ps + qr) x + qs

If we compare the coefficients of x2, we get a = pr. Also, on

comparing the coefficients of x, we get b = ps + qr. Finally, on

comparing the constants, we get c = qs. Hence, b is the sum of two

numbers ‘ps’ and ‘qr’, whose product is (ps)(qr) = (pr)(qs) = ac.

Therefore, to factorise ax2 + bx + c, we have to write b as the sum of

two numbers whose product is ‘ac’. Let’s look at an example to

understand this clearly.

Example

Factorise 6×2 + 17x + 5 by splitting the middle term.

Solution 1 (By splitting method): As explained above, if we can find

two numbers, ‘p’ and ‘q’ such that, p + q = 17 and pq = 6 x 5 = 30,

then we can get the factors.

After looking at the factors of 30, we find that numbers ‘2’ and ‘15’

satisfy both the conditions, i.e. p + q = 2 + 15 = 17 and pq = 2 x 15 =

30. So,

6x2 + 17x + 5 = 6x2 + (2 + 15)x + 5

= 6x2 + 2x + 15x + 5

= 2x(3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Therefore, the factors of (6x2 + 17x + 5) are (3x + 1) and (2x + 5) with

a remainder, zero.

Example

Factorise y2 – 5y + 6 by using the Factor Theorem.

Solution: Let p(y) = y2 – 5y + 6. Now, if p(y) = (y – a) (y – b), the

constant term will be ab as can be seen below,

p(y) = (y – a)(y – b)

= y2 – by – ay + ab

On comparing the constants, we get ab = 6. Next, the factors of 6 are

1, 2 and 3. Now, p(2) = 22– (5 × 2) + 6 = 4 – 10 + 6 = 0. So, (y – 2) is

a factor of p(y). Also, p(3) = 32 – (5 × 3) + 6 = 9 – 15 + 6 = 0. So, (y –

3) is also a factor of y2 – 5y + 6. Therefore, y2 – 5y + 6 = (y – 2)(y –

3)

More Solved Examples for You

Question: Factorise x3 – 23x2 + 142x – 120

Solution: Let p(x) = x3 – 23x2 + 142x – 120. To begin with, we will

start finding the factors of the constant ‘– 120’, which are:

±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60

and ±120

Further, by trial, we find that p(1) = 0. Hence, we conclude that (x – 1)

is a factor of p(x). Also, we see that

[x3 – 23x2 + 142x – 120] = x3 – x2 – 22x2 + 22x + 120x – 120

So, by removing the common factors, we have x3 – 23x2 + 142x – 120

= x2(x –1) – 22x(x – 1) + 120(x – 1)

Further, taking ‘x – 1’ common, we get x3 – 23x2 + 142x – 120 = (x –

1) (x2 – 22x + 120)

Therefore, x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)

Also, note that if we divide p(x) by ‘x – 1’, then the result will be (x2 –

22x + 120)

Going on, x2 – 22x + 120 can be factorised further. So, by splitting the

middle term, we get:

x2 – 22x + 120 = x2 – 12x – 10x + 120 … [ (– 12 – 10 = – 22) and

{(–12)( –10) = 120}]

= x(x – 12) – 10(x – 12)

= (x – 12) (x – 10)

Therefore, we have x3 – 23x2 – 142x – 120 = (x – 1)(x – 10)(x – 12)

Question: Factorise x3 – 2x2 – x + 2

Solution: Let p(x) = x3 – 2x2 – x + 2. To begin with, we will start

finding the factors of the constant ‘2’, which are: 1, 2

By trial, we find that p(1) = 0. Hence, we conclude that (x – 1) is a

factor of p(x).

So, by removing the common factors, we have x3 – 2x2 – x + 2 = x2(x

– 2) – (x – 2) = (x2 – 1)(x – 2)

= (x + 1)(x – 1)(x – 2) … [Using the identity (x2 – 1) = (x + 1)(x – 1)]

Therefore, the factors of x3 – 2x2 – x + 2 are (x + 1), (x – 1) and (x –

2)

Zeroes of Polynomial

We already know that a polynomial is an algebraic term with one or

many terms. Zeroes of Polynomial are the real values of the variable

for which the value of the polynomial becomes zero. So, real numbers,

‘m’ and ‘n’ are zeroes of polynomial p(x), if p(m) = 0 and p(n) = 0. 

Understanding Zeroes of Polynomial

Example 1

Let’s look at the polynomial, p(x) = 5x3 – 2x2 + 3x -2. Now, let’s find

the value of the polynomial(x) at x = 1, p(1) = 5(1)3 – 2(1)2 + 3(1) – 2

= 5 – 2 + 3 – 2 = 4. Therefore, we can say that the value of the

polynomial p(x) at x = 1 is 4.

Next, let’s find the value of the polynomial(x) at x = 0, p(0) = 5(0)3 –

2(0)2 + 3(0) – 2 = 0 – 0 + 0 – 2 = – 2. Therefore, we can say that the

value of the polynomial p(x) at x = 0 is – 2.

Example 2

Let’s look at another polynomial now, p(x) = x – 1. Let’s find the

value of the polynomial at x = 1, p(1) = 1 – 1 = 0. So, the value of the

polynomial p(x) at x = 1 is 0. Therefore, 1 is the zero of the

polynomial p(x).

Similarly, for the polynomial q(x) = x – 2, the zero of the polynomial

is 2. To summarize, zeroes of polynomial p(x) are numbers c and d

such that p(c) = 0 and p(d) = 0.

Download NCERT Solutions

for Class 10 Maths

Calculating Zeroes of polynomial

When we calculated zeroes of polynomial p(x) = x – 1, we equated it

to 0, x – 1 = 0 or, x = 1. Hence, we say that p(x) = 0 is the Polynomial

Equation. 1 is the Root of the Polynomial equation p(x) = 0. OR 1 is

the Zero of the Polynomial equation p(x) = x – 1 = 0.

Now, let’s look at a constant polynomial ‘5’. You can write this as

5x0. What is the Root of this constant polynomial? The answer is a

Non-zero constant polynomial has no zero. Also, every real number is

a zero of the Zero Polynomial.

Let’s look at the following linear polynomial to understand the

calculation of the roots or ‘zeroes of polynomial’: p(x) = ax + b …

where a ≠ 0. To find a zero, we must equate the polynomial to 0. [p(x)

= 0]. Hence, ax + b = 0 … where a ≠ 0. So, ax = – b or, x = – b/a.

Therefore, ‘- b/a’ is the only zero of p(x) = ax + b. We can also say

that a linear polynomial has only one zero.

Observations

● A zero of a polynomial need not be 0.

● 0 may be a zero of a polynomial.

● Every linear polynomial has one and only one zero.

● A polynomial can have more than one zero.

Learn Degree of Polynomials in detail.

More Solved Examples for You

Question: Find p(0), p(1) and p(2) for each of the following

polynomials:

● p(y) = y2 – y + 1

● p(t) = 2 + t + 2t2 – t3

● p(x) = x3

● p(x) = (x – 1) (x + 1)

Solution:

p(y) = y2 – y + 1

● p(0) = 02 – 0 + 1 = 1

● p(1) = 12 – 1 + 1 = 1

● p(2) = 22 – 2 + 1 = 3

p(t) = 2 + t + 2t2 – t3

● p(0) = 2 + 0 + 2(0)2 – (0)3 = 2 + 0 + 0 – 0 = 2

● p(1) = 2 + 1 + 2(1)2 – (1)3 = 2 + 1 + 2 – 1 = 4

● p(2) = 2 + 2 + 2(2)2 – (2)3 = 2 + 2 + 8 – 8 = 4

p(x) = x3

● p(0) = (0)3 = 0

● p(1) = (1)3 = 1

● p(2) = (2)3 = 8

p(x) = (x – 1) (x + 1)

● p(0) = (0 – 1)(0 + 1) = (-1)(1) = – 1

● p(1) = (1 – 1)(1 + 1) = (0)(2) = 0

● p(2) = (2 – 1)(2 + 1) = (1)(3) = 3

Verify the following

Question: Verify whether the following are zeroes of polynomials

indicated against them.

● p(x) = 3x + 1, x = – 1/3

● p(x) = 5x – π, x = 4/5

● p(x) = (x + 1) (x – 2), x = – 1, 2

● p(x) = 3x2 – 1, x = – 1/√3 , 2/√3

Solution:

● p(x) = 3x + 1, x = – 1/3

p(-1/3) = 3(-1/3) + 1 = – 1 + 1 = 0.

Hence, x = -1/3 is a zero of polynomial 3x + 1.

● p(x) = 5x – π, x = 4/5

p(4/5) = 5(4/5) – π = 4 – π ≠ 0.

Hence, x = 4/5 is not a zero of polynomial 5x – π.

● p(x) = (x + 1) (x – 2), x = – 1, 2

p(-1) = (– 1 + 1)(- 1 – 2) = (0)(- 3) = 0.

Hence, x = – 1 is a zero of polynomial (x + 1) (x – 2).

p(2) = (2 + 1)(2 – 2) = (3)(0) = 0.

Hence, x = 2 is a zero of polynomial (x + 1) (x – 2).

● p(x) = 3x2 – 1, x = – 1/√3 , 2/√3

p(- 1/√3) = 3(- 1/√3)2 – 1 = 3 (1/3) – 1 = 1 – 1 = 0.

Hence, x = – 1/√3 is a zero of polynomial 3x2 – 1.

p(2/√3) = 3(2/√3) – 1 = 3(4/3) – 1 = 4 – 1 = 3 ≠ 0.

Hence, x = 2/√3 is not a zero of polynomial 3x2 – 1.

Geometrical Representation of Zeroes of a Polynomial

In the study of polynomials, you are aware that a real number ‘k’ is a

zero of the polynomial p(x) if p(k) = 0. Remember, zero of a

polynomial is different from a zero polynomial. We have seen the

Remainder theorem use the concept of zeroes of a polynomial too. In

order to understand their importance, we will look at the geometrical

representations of linear and quadratic polynomials and the

geometrical meaning of their zeroes. 

Linear Polynomial

Let’s look at a linear polynomial ax + b, where a ≠ 0. You have

already studied that the graph of y = ax + b is a straight line. Let’s

look at the graph of y = 2x + 3.

x -2 2

y = 2x + 3 -1 7

The straight line y = 2x + 3 will pass through the points (- 2, – 1) and

(2, 7). Here is how the graph looks like:

Fig. 1

From the Fig.1 above, you can see that the graph of y = 2x + 3

intersects the x-axis at the point (- 3/2, 0). Now, the zero of (2x + 3) is

(- 3/2). Therefore, the zero of the linear polynomial (2x + 3) is the

x-coordinate of the point where the graph of y = 2x + 3 intersects the

x-axis. Hence, we can say,

For a linear polynomial ax + b, a ≠ 0, the graph of y = ax + b is a

straight line which intersects the x-axis at exactly one point, namely, (-

b/a, 0). Also, this linear polynomial has only one zero which is the

x-coordinate of the point where the graph of y = ax + b intersects the

x-axis.

Download NCERT Solutions for Class 10 Maths

Quadratic Polynomial

Let’s look at a quadratic polynomial, x2 – 3x – 4. To look at the graph

of y = x2 – 3x – 4, let’s list some values:

x – 2 – 1 0 1 2 3 4 5

y = x2 – 3x – 4 6 0 – 4 – 6 – 6 – 4 0 6

The graph of y = x2 – 3x – 4 will pass through (- 2, 6), (- 1, 0), (0, –

4), (1, – 6), (2, – 6), (3, – 4), (4, 0) and (5, 6). Here is how the graph

looks:

Fig. 2

For that matter, for any quadratic polynomial y = ax2 + bx + c, a ≠ 0,

the graph of y = ax2 + bx + c has either one of these two shapes:

● If a > 0, then it is open upwards like the one shown in the graph

above

● If a < 0, then it is open downwards.

These curves are parabolas. A quick look at the table above shows that

(-1) and (4) are zeroes of the quadratic polynomial. From the Fig. 2

above, you can see that (-1) and (4) are the x-coordinates of the points

where the graph of y = x2 – 3x – 4 intersects the x-axis. Therefore, we

can say,

The zeroes of a quadratic polynomial ax2 + bx + c, a ≠ 0, are

precisely the x-coordinates of the points where the parabola

representing y = ax2 + bx + c intersects the x-axis.

As far as the shape of the graph is concerned, the following three

cases are possible:

Case (i)

The graph cuts x-axis at two distinct points A and A′, where the

x-coordinates of A and A′ are the two zeroes of the quadratic

polynomial ax2 + bx + c, as shown below:

Fig. 3

Case (ii)

The graph intersects the x-axis at only one point, or at two coincident

points. Also, the x-coordinate of A is the only zero for the quadratic

polynomial ax2 + bx + c, as shown below:

Fig. 4

Case (iii)

The graph is either

● Completely above the x-axis or

● Completely below the x-axis.

So, it does not cut the x-axis at any point. Hence, the quadratic

polynomial ax2 + bx + c has no zero, as shown below:

Fig. 5

To summarize, a quadratic polynomial can have either:

● Two distinct zeroes (as shown in Case i)

● Two equal zeroes (or one zero as shown in Case ii)

● No zero (as shown in Case iii)

It can also be summarized by saying that a polynomial of degree 2 has

a maximum of 2 zeroes.

Cubic Polynomial

Let’s look at a cubic polynomial, x3 – 4x. Next, let’s list a few values

to plot the graph of y = x3 – 4x.

x – 2 – 1 0 1 2

y = x3 – 4x 0 3 0 – 3 0

The graph of y = x3 – 4x will pass through (- 2, 0), (- 1, 3), (0, 0), (1, –

3), and (2, 0). Here is how the graph looks like:

Fig. 6

From the table above, we can see that 2, 0 and – 2 are the zeroes of the

cubic polynomial x3 – 4x. You can also observe that the graph of y =

x3 – 4x intersects the x-axis at 2, 0 and – 2. Let’s take a quick look at

some examples:

Let’s plot the graph of the following two cubic polynomials:

1. x3

2. x3 – x2

The graphs of y = x3 and y = x3 – x2 look as follows:

Fig. 7

From the first graph, you can observe that 0 is the only zero of the

polynomial x3, since the graph of y = x3 intersects the x-axis only at 0.

Similarly, the polynomial x3 – x2 = x2(x – 1) has two zeroes, 0 and 1.

From the second diagram, you can see that the graph of y = x3 – x2

intersects the x-axis at 0 and 1.

Hence, we can conclude that there is a maximum of three zeroes for

any cubic polynomial. Or, any polynomial with degree 3 can have

maximum 3 zeroes. In general,

Given a polynomial p(x) of degree n, the graph of y = p(x) intersects

the x-axis at a maximum of n points. Therefore, a polynomial p(x) of

degree n has a maximum of n zeroes.

Solved Examples for You

Question: The graphs of y = p(x) are given in the figure below, for

some polynomials p(x). Find the number of zeroes of p(x), in each

case.

Fig. 8

Solution: The number of zeroes in each of the graphs above are:

● 0, since the line is not intersecting the x-axis.

● 1, since the line intersects the x-axis once.

● 2, since the line intersects the x-axis twice.

● 2, since the line intersects the x-axis twice.

● 4, since the line intersects the x-axis four times.

● 3, since the line intersects the x-axis thrice.