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Introduction to Linear Polynomials 2.1 Introduction We have learnt about algebraic expressions in earlier grades. In this chapter, we will learn about the special types of algebraic expressions called linear polynomials. Let us first consider a few examples of algebraic expressions. Example 1: Raju went to a shop where there were sealed boxes of different colours on sale. The shop owner told him that the red boxes have 4 pens each and the blue boxes have 5 pencils each. Now, if Raju bought x red boxes and y blue boxes, how can he quickly figure out the total quantity of pens and pencils? Also, if he got 3 extra pens free, how many pens and pencils did he get altogether? Observe that x red boxes will have 4x pens and y blue boxes will have 5y pencils. Also, he got 3 extra pens free. Thus, the total number of pens and pencils is given by the algebraic expression 4x + 5y + 3. In this example, 4x, 5y and 3 are terms of the expression, x and y are letter-numbers, the numbers 4 and 5 are the coefficients of x and y, respectively, and 3 is a constant. From now onwards, we will use a widely used alternate word for letter-numbers: variables. Thus in the expression 4x + 5y + 3, we say that the variables used are x and y. Example 2: A rectangular garden of length l metres and width w metres has to be fenced and decorated. A wire fence is to be laid along the length costing `100 per metre and a wooden fence is to be built along the width costing `80 per metre. Special seeds have to be sown throughout the garden which will cost `50 per square metre. 2 Fig. 2.1 Chapter 2.indd 16 4/15/2026 2:52:57 PMIntroduction to Linear Polynomials 17 Fig. 2.2 What will be the total cost incurred? Cost of wire fencing along the garden length = 2l × 100 = `200l Cost of wooden fencing along the garden width = 2w × 80 = `160w Cost of sowing seeds throughout the entire garden (depends on the area) = 50 × l × w = `50lw Total cost = ` (200l + 160w + 50lw). Thus, 200l + 160w + 50lw is the algebraic expression for the total cost. Think and Reflect 1. Can you identify the terms, variables and coefficients of this algebraic expression? 2. How is it different from the algebraic expression in Example 1? Example 3: A wire of length 20 cm is bent in different ways to form rectangles. For example, we can have a rectangle with length 7 cm and width 3 cm. We can also have one of length 5.5 cm and width 4.5 cm. (Think of a few more ways of forming such rectangles.) Can you write an expression for the area of such rectangles? If the length of the rectangle is x cm, then the width is (10 – x) cm. The expression for the area of these rectangles is x(10 – x) or 10x – x2 . Think and Reflect 1. Can you identify the terms, variables and coefficients of this algebraic expression? 2. Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3? Chapter 2.indd 17 4/15/2026 2:52:58 PM18 Ganita Manjari | Grade 9 | Part I Note that the algebraic expressions in Example 1 and Example 2 involve two variables, whereas the algebraic expression in Example 3 involves only one variable. Expressions such as 4x, x2 + 1, 2y – 5, 5y3 + y2 + 2y – 1, 3z + 7 are algebraic expressions that involve only one variable: x, y or z. In this chapter, we will restrict our discussion to algebraic expressions involving only one variable. You may have noticed that in an algebraic expression, the powers of a variable also appear. For example, in the expression x2 + 5x + 1, the highest power of x is 2, whereas in the expression 5y3 + y2 – 8, the highest power of the variable y is 3. Further, in the expression 5y3 + y2 + 2y – 1, the coefficient of y3 is 5, that of y2 is 1, that of y is 2 and the constant term is –1. Such algebraic expressions involving one variable and its powers are called one-variable polynomials, univariate polynomials, or when the context is clear, simply polynomials. (‘univariate’ means ‘having one variable’). The highest power of the variable in a polynomial is called its degree. For example: (i) 5y3 + y2 + 2y – 1 is a polynomial of degree 3. Such polynomials are called cubic polynomials. (ii) x2 + 5x + 1 is a polynomial of degree 2. Such polynomials are called quadratic polynomials. (iii) 3z + 7 is a polynomial of degree 1. Such polynomials are called linear polynomials. (iv) The constant 8 is a polynomial of degree 0 as it can be written as 8x0 in which the power of the variable x is 0. Such polynomials are called constant polynomials. Exercise Set 2.1 1. Find the degrees of the following polynomials: (i) 2x2 – 5x + 3 (ii) y3 + 2y – 1 (iii) – 9 (iv) 4z – 3 2. Write polynomials of degrees 1, 2 and 3. 3. What are the coefficients of x2 and x3 in the polynomial x4 – 3x3 + 6x2 – 2x + 7? Chapter 2.indd 18 4/15/2026 2:52:58 PMIntroduction to Linear Polynomials 19 4. What is the coefficient of z in the polynomial 4z3 + 5z2 – 11? 5. What is the constant term of the polynomial 9x3 + 5x2 – 8x –10? Recall that polynomials of degree 1 are called linear polynomials. In this chapter, we shall study linear polynomials. 2.2 Linear Polynomials We begin with some examples involving linear polynomials. Example 4: The perimeter of a square of side x is 4x, which is a linear polynomial in the variable x. Think and Reflect Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm? Example 5: A chess club charges a joining fee of `200 plus `50 for every match played. The following table shows the amount a player will have to pay as the number of matches varies. Number of matches played 1 2 3 4 5 … m Amount paid (`) 250 300 350 400 450 … 200 + 50m Hence, if m is the number of matches played, the total cost will be `(200 + 50m). Observe that 200 + 50m is a linear polynomial in the variable m. The amount paid increases by the constant value of `50 for every additional match played. Think and Reflect If a player paid `750, how many matches did he play? The examples given above highlight a characteristic feature of linear polynomials — that the difference between the successive values at integers is constant. In Example 4, the perimeters increase by 2 cm each time the side of the square increases by 0.5 cm. Similarly in Example 5, Chapter 2.indd 19 4/15/2026 2:52:58 PM20 Ganita Manjari | Grade 9 | Part I the amount paid by a player increases by `50 for every additional match played. Such patterns are called linear patterns. When we equate a linear polynomial in one variable to a constant, we get a linear equation. Let us consider the following example. Example 6: The sum of two numbers is 64. One of the numbers is 10 more than the other. What are the two numbers? Let the smaller number be x. Then the larger number must be x + 10. Since their sum is 64, we have the linear equation x + (x + 10) = 64. This implies that 2x + 10 = 64. Note that 2x + 10 is a linear polynomial. By equating it to 64 we get a linear equation 2x = 54 or x = 27. The numbers are therefore, 27 and 37. Polynomials can also be thought of as input-output processes. For instance, consider the linear polynomial 2x + 3. For every x, there is a corresponding value of the polynomial 2x + 3. For instance, if x = 4, we substitute 4 in the expression to get 2 × 4 + 3 = 11. If x = –6, then we substitute –6 in the expression to get 2 × –6 + 3 = –9. Fig. 2.3 shows this process as an input-output machine where the input is the value of x and the output is the value of 2x + 3. This process can be referred to as a function where the expression 2x + 3 is a function of the variable x. You will learn more about functions in later grades. Fig. 2.3: A linear expression as an input-output process Think and Reflect We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x – x2 , is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm? Chapter 2.indd 20 4/15/2026 2:52:58 PMIntroduction to Linear Polynomials 21 Note that 2x + 3 is a linear function, whereas 10x – x2 is a quadratic function. Exercise Set 2.2 1. Find the value of the linear polynomial 5x – 3 if: (i) x = 0 (ii) x = –1 (iii) x = 2 2. Find the value of the quadratic polynomial 7s2 – 4s + 6 if: (i) s = 0 (ii) s = –3 (iii) s = 4 3. The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present ages. 4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers. 5. Ruby has 3 times as many two-rupee coins as she has five rupee-coins. If she has a total `88, how many coins does she have of each type? 6. A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces? 7. If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle? 2.3 Exploring linear patterns Observe the following growing pattern of square tiles. Stage 1 Stage 2 Stage 3 Stage 4 Fig. 2.4: A growing pattern of square tiles Chapter 2.indd 21 4/15/2026 2:52:58 PM22 Ganita Manjari | Grade 9 | Part I Think and Reflect Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern. Each stage is obtained by adding two more tiles to the previous stage. The table mentions the number of tiles for the first seven stages. Stage 1 2 3 4 5 6 7 Number of the square tiles 1 3 5 7 9 11 13 To generalise this pattern, we observe that the number of squares at each stage is one less than twice the number of the term. For example, in Term 2, the number of squares is 2 × 2 – 1 = 3, in Term 5, the number of squares is 2 × 5 – 1 = 9 and so on. This leads us to conclude that the number of squares at Stage n is given by 2n – 1. The polynomial 2n – 1 has degree 1. Hence, it is an example of a linear polynomial. Also, the difference between consecutive terms in the sequence of the number of squares, that is, 1, 3, 5, 7, 9 … is the constant value 2. Thus, with each stage, the number of squares increases by 2. The relationship between the number of the stage and the number of square tiles is a linear relationship. Think and Reflect Using the expression 2n – 1, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles? Example 7: Bela has `100 for pocket money. She spends `5 every day. After how many days will she be left with `40? Day Number 0 1 2 3 4 Amount left (`) 100 100 – 1 × 5 = 95 100 – 2 × 5 = 90 100 – 3 × 5 = 85 100 – 4 × 5 = 80 Chapter 2.indd 22 4/15/2026 2:52:58 PMIntroduction to Linear Polynomials 23 Observe that the amount left on the nth day will be `(100 – 5n). Therefore, on the 12th day the amount left will be `(100 – 12 × 5) = `40. Think and Reflect What amount will be left on the 15th day? How many days will it take for the entire amount to be spent? Example 8: An auto-rikshaw fare starts at `25 and remains the same for the initial 2 km. Then it increases by `15 per km. What will be the fare for a travel of 10 km? For the initial 2 km, the fare is `25. For every kilometre (km) thereafter, the fare will increase by `15. Therefore, the total fare for a travel of 10 km will be `25 + 15 × 8 = `145. Km travelled 1 2 3 4 5 6 Fare (`) 25 25 25 + 1 × 15 = 40 25 + 2 × 15 = 55 25 + 3 × 15 = 70 25 + 4 × 15 = 85 Observe that the total fare for a travel of n km will be `25 + 15 × (n – 2) = 15n – 5, when n ≥ 2. Here, the fare for a specific distance covered is a function of the distance n km. Think and Reflect For how many km will the fare be `130? Note that in all the above examples, the nth term is a linear expression in n. A linear pattern is a sequence of numbers where the difference between two consecutive terms is constant. We will learn more about linear patterns in the chapter on Sequences and Progressions. Exercise Set 2.3 Solve the following: 1. A student has `500 in her savings bank account. She gets `150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the nthmonth. Chapter 2.indd 23 4/15/2026 2:52:58 PM24 Ganita Manjari | Grade 9 | Part I 2. A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3, … hours? Find a linear expression to represent the number of members at the end of the nth hour. 3. Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area of the rectangle. 4. Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern representing the volume of the rectangular box. 5. Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern. 2.4. Linear growth and linear decay Linear expressions help to model situations or phenomena where there is growth or decline. Consider the following examples: Example 9: The cost of a journey is given by the linear function C(d) = 100 + 60d, where C indicates total cost in rupees and d the distance travelled in km. Let us make a table of values for d varying from 0 to 10 km and show how the cost increases for every km. Distance travelled, d (km) 0 1 2 3 4 5 Cost, C (`) 100 160 220 280 340 400 In this example, as the value of d increases by one km, the value of the cost function C, increases by a fixed amount of `60. This is an example of linear growth. Think and Reflect What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be `700? Example 10: The height of water in a cylindrical tank is 3 m at the start of summer. The height h m at the end of t months is given by the linear function h(t) = 3 – 0.5t. Chapter 2.indd 24 4/15/2026 2:52:58 PMIntroduction to Linear Polynomials 25 Month, t 0 1 2 3 4 Height, h (m) 3 2.5 2 1.5 1 In the example above, as the value of t increases by a fixed number (one month), the value of the height h decreases by a fixed number (0.5). Therefore, this example represents linear decay. Think and Reflect What will be the height of the water at the end of 5 months? Linear growth describes a linear pattern where a quantity increases by a constant amount over equal intervals. Similarly, linear decay describes a linear pattern where a quantity decreases by a constant amount over equal intervals. Exercise Set 2.4 1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month. (i) Find the height after 7 months. (ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month. (iii) Find an expression that relates h and t, and explain why it represents linear growth. 2. A mobile phone is bought for `10,000. Its value decreases by `800 every year. (i) Find the value of the phone after 3 years. (ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time. (iii) Find an expression that relates v and t, and explain why it represents linear decay. 3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village. (i) Find the population of the village after 6 years. Chapter 2.indd 25 4/15/2026 2:52:58 PM