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Polynomials

Introduction to polynomials

Polynomials

Consider this situation involving trains.

The speed of an express train is ten less than twice that of a passenger train. If each travels for as many hours as its speed, then what is the difference between the distances travelled by them?

Let the speed of the passenger train be x km/hr.

Then, travelling time of the train = x hours

Distance travelled by it = Speed × Time = x × x = x2 km

Now, speed of the express train = (2x − 10) km/hr

Its travelling time = (2x − 10) hours

Distance travelled by it = (2x − 10) (2x − 10) = (4x2 − 40x + 100) km

Thus, required difference = 4x2 − 40x + 100 − x2 = (3x2 − 40x + 100) km

The expression 3x2 − 40x + 100 is an example of a polynomial. Different real-life problems such as the one given above can be expressed in the form of polynomials. Go through this lesson to familiarize yourself with these useful expressions.

Topics to be covered in this lesson:

• Identifying polynomials

• Constant polynomials
• Classification of polynomials according to the number of terms

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Polynomials and Their Classification Based on the Number of Terms

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Ancient Babylonians developed a unique system to calculate things using formulae. These formulae consisted of letters, mathematical operators (+, −, ×, ÷ ) and numbers. It was this system that led to the development of algebra. The word ‘algebra’ is derived from the Arabic word ‘al-jabr’ meaning ‘the reunion of broken parts’. Another Arabian connection with algebra is the Arab mathematician Muhammad ibn Musa al-Khwarizmi, whose theories greatly influenced this branch of mathematics.

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Example 1:

i) 2x1/2 + 3x + 4

ii) 8x3 + 7 + x

iii)

iv)

v)

vi)

Solution:

i) 2x1/2 + 3x + 4

This expression is not a polynomial because the exponent of the first term is , which is not a whole number.

ii) 8x3 + 7 + x

This expression is a polynomial as all the coefficients are real numbers and the exponents of the variable x are whole numbers.

iii)

This expression is a polynomial as all the coefficients are real numbers and the exponents of the variables x and y are whole numbers.

iv)

This expression is a polynomial as all the coefficients are real numbers and the exponents of the variables x and y are whole numbers.

v)

This expression is a polynomial because it can be written as 5x + 9x2, in which all the coefficients are real numbers and the exponents of the variable x are whole numbers.

vi)

This expression is not a polynomial because it can be written as14x−2 − 9x2, in which the variable in the first term has a negative exponent.

Example 2:

For each of the given polynomials, state whether it is a monomial, binomial or trinomial.

i) 4x3

ii) 13y5y

iii) 29t3 + 14t − 9

iv) x − x2

Solution:

i) 4x3 is a monomial as it has only one term.

ii) 13y5y is a binomial as it has two terms.

iii) 29t3 + 14t − 9 is a trinomial as it has three terms.

iv) x − x2 is a binomial as it has two terms.

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Example 1:

For each of the given polynomials, state whether it is a monomial, binomial or trinomial.

i) (4x3 + 3y) − (3x3 + x2) + (yx3

ii) (t2 + 1) 2

iii) [(m − 1) (m + 1)] + 2m2 + 1

Solution:

i) (4x3 + 3y) − (3x3 + x2) + (yx3

= 4x3 + 3y − 3x3x2 + yx3

= (4x3 − 3x3x3) − x2 + (3y + y

= − x2 + 4y

The polynomial can be reduced to −x2 + 4y, which has two terms; so, it is a binomial.

ii) (t2 + 1)2

= (t2)2 + 2t2 + 12[∵ (a + b)2 = a2 + 2ab + b2]

= t4 + 2t2 + 1[∵ (am)n = am × n]

The polynomial can be reduced to t4 + 2 t2 + 1, which has three terms; so, it is a trinomial.

iii) [(m − 1) (m + 1)] + 2m2 + 1

= (m2 − 1) + 2m2 + 1 [∵ (ab) (a + b) = a2b2]

= m2 + 2m2 − 1 + 1

= 3m2

The polynomial can be reduced to 3m2, which has only one term; so, it is a monomial.

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Example 1:

State whether or not the following expressions are polynomials. Justify your answers.

i)

ii)

Solution:

i)

In the reduced form of the given expression, all the coefficients are real numbers and the exponents of the variables x and y are whole numbers. Hence, the given expression is a polynomial.

ii)

In the reduced form of the given expression, the exponent of the second term (i.e., −6) is not a whole number. Hence, the given expression is not a polynomial.

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Generalized Form of a Polynomial

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The word ‘polynomial’ is a combination of the Greek words ‘poly’ meaning ‘many’ and ‘nomos’ meaning ‘part or portion’. Thus, a polynomial is an algebraic expression having many parts.

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Example 1:

Find the coefficient of x in the following polynomials.

i)

ii)(2x2x) + 7 + 3x

Solution:

i)

This expression can also be written as

Thus, in the given polynomial, the coefficient of x is 0.

ii)(2x2x) + 7 + 3x

= 2x2x + 7 + 3x

= 2x2 + 2x + 7

Thus, in the given polynomial, the coefficient of x is 2.

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Example 1:

For each of the following polynomials, write the constant term and the coefficient of the variable having the highest exponent.

i)− 3+ (x + 2) − 4

ii)3 (y + 2)2 − 5y (y2 + ) + y3

Solution:

i) − 3+ (x + 2) − 4

= − 3x3 + (x + 2) − 4 [∵ (am)n = am × n and ]

= −x2 − 3x3 + x − 2

= − 3x3 x2 + x − 2

In this reduced form of the given polynomial, we have:

Constant term = −2

Term with the highest exponent = −3x3

So, coefficient of the variable having the highest exponent = −3

ii)3 (y + 2)2 − 5y (y2 + ) + y3

= 3 (y2+ 4y + 4) − 5y3 − 10 + y3

= 3y2+ 12y + 12 − 4y3 − 10

= −4y3 + 3y2+ 12y + 2

In this reduced form of the given polynomial, we have:

Constant term = 2

Term with the highest exponent = −4y3

So, coefficient of the variable having the highest exponent = 4

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Example 1

Find the coefficients of x3, x2, x, y3, y2 and y in the following polynomials. Also, find the real constants.

i)17x4 − 3yx (2x3 + x) + 3 (2y − 4x2 + 1) − 4

ii)

Solution:

i)17x4 − 3yx (2x3 + x) + 3 (2y − 4x2 + 1) − 4

= 17x4 − 3y − 2x4x2 + 6y − 12x2 + 3 − 4

= 15x4 − 13x2 + 3y − 1

This reduced form of the given polynomial can be further written as:

15x4 + 0.x3 − 13x2 + 0.x + 0.y3 + 0.y2 + 3y − 1

Therefore, we have:

Coefficient of x3 = 0 Coefficient of x2 = −13 Coefficient of x = 0

Coefficient of y3 = 0 Coefficient of y2 = 0 Coefficient of y = 3

Real constant = −1

ii)

This reduced form of the given polynomial can be further written as:

Therefore, we have:

Coefficient of x3 = 0 Coefficient of x2 = −7 Coefficient of x = 0

Coefficient of y3 =Coefficient of y2 = 0 Coefficient of y =

Real constant =

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Different Forms of a Polynomial

A polynomial can found and written in different forms. These forms are explained below.

Standard form: If the terms of a polynomial are written in descending order or ascending order of the powers of the variables then the polynomial is said to be in the standard form.

For example, the polynomial 3x + 15x4 − 1 − 13x2 is not in the standard form. It can be written in the standard form as 15x4 − 13x2 + 3x − 1 or −1 + 3x − 13x2+ 15x4.

Index form: Observe the polynomial x6 − 2x4 − 10x3 + 5. In this polynomial, terms having x5, x2 and x are missing. These terms can be added to the polynomial with coefficient 0. Thus, the obtained polynomial will be x6 + 0x5− 2x4 − 10x3 + 0x2 + 0x + 5.

The polynomial obtained on adding the missing terms is said to be in the index form.

Coefficient form: When the coefficients of all the terms of a polynomial are written in a bracket by separating with comma then the polynomial is said to be written in the coefficient form.

It should be noted that if a term is missing then its coefficient is taken as 0. So, it is better to write the given polynomial in the index form before wrting it in the coefficient form.

For example, to write the polynomial x6 − 2x4 − 10x3 + 5 in the coefficient form, we will first write it the index form as x6 + 0x5− 2x4 − 10x3 + 0x2 + 0x + 5.

Now, it can be written in the coefficient form as (1, 0, −2, −10, 0, 0, 5).

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Example 1:

Express the given polynomials in the standard form.

(i) −2y3 + 5y5 − 2 + y

(ii) 11a a6 − 2a3 + a2 − 1

Solution:

We know that if the terms of a polynomial are written in descending order or ascending order of the powers of the variables then the polynomial is said to be in the standard form.

Given polynomials can be written in the standard form as follows:

(i) Given form: −2y3 + 5y5 − 2 + y

Standard form: 5y5 − 2y3 + y − 2 or −2 + y − 2y3 + 5y5

(ii) Given form: 11a a6 − 2a3 + a2 − 1

Standard form: a6 − 2a3 + a2 + 11a − 1 or −1 + 11a + a2 − 2a3 a6

Example 2:

Express the given polynomials in the index form and coefficient form.

(i) 2m7 + 12m5− 7m2m

(ii) −4p6 + 3p3− 2p + 7

Soluti...

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