Introduction to Polynomials (titled Introduction to Linear Polynomials in the 2026-27 NCERT book) builds algebraic expressions from real situations, defines a polynomial in one variable and its degree, then studies linear polynomials in depth. You learn to evaluate them, turn word problems into linear equations, find the nth term of a linear pattern, model linear growth and decay, find y = ax + b from two readings, and draw it as a straight line with slope a and y-intercept b. These skills feed directly into sequences, linear equations in two variables and coordinate geometry.
Key Concepts
1. Algebraic Expressions: Terms, Variables, Coefficients and Constants
Raju at the shop. Red boxes hold 4 pens each and blue boxes hold 5 pencils each. Raju buys x red boxes and y blue boxes, and gets 3 extra pens free. The x red boxes hold 4x pens and the y blue boxes hold 5y pencils, so the total number of pens and pencils is
4x + 5y + 3
- Terms: 4x, 5y and 3 (the parts joined by + or - signs).
- Variables: x and y. Earlier books called these letter-numbers; from now on the word used is variable.
- Coefficients: 4 is the coefficient of x and 5 is the coefficient of y.
- Constant: 3, a term with no variable.
Fencing a garden. A rectangular garden is l metres long and w metres wide. A wire fence along the length costs ₹100 per metre, a wooden fence along the width costs ₹80 per metre, and seeds cost ₹50 per square metre of the garden.
Wire fence: 2l × 100 = 200l. Wooden fence: 2w × 80 = 160w. Seeds: 50 × l × w = 50lw. Total cost = ₹(200l + 160w + 50lw).
Here the terms are 200l, 160w and 50lw, the variables are l and w, and the coefficients are 200, 160 and 50. The difference from Raju’s expression: this one has a term, 50lw, in which two variables are multiplied together, and it has no constant term.
Bending a wire. A wire 20 cm long is bent into rectangles: 7 cm by 3 cm, or 5.5 cm by 4.5 cm, and many more. Since the perimeter is always 20 cm, length + width = 10 cm. If the length is x cm, the width is (10 - x) cm, and the area is
x(10 - x) = 10x - x2
This expression has only one variable, x, but it contains a power of x (x2). The coefficient of x is 10 and the coefficient of x2 is -1. The whole chapter now restricts itself to expressions in one variable.
2. Polynomials in One Variable and Their Degree
Expressions such as 4x, x2 + 1, 2y - 5, 5y3 + y2 + 2y - 1 and 3z + 7 each use only one variable (x, y or z) together with its powers. Such expressions are called one-variable polynomials, univariate polynomials (“univariate” means having one variable), or simply polynomials when the context is clear.
Degree. The highest power of the variable in a polynomial is its degree. In x2 + 5x + 1 the highest power of x is 2. In 5y3 + y2 - 8 the highest power of y is 3.
Reading coefficients. In 5y3 + y2 + 2y - 1 the coefficient of y3 is 5, that of y2 is 1 (no number written means 1), that of y is 2, and the constant term is -1.
Names by degree.
| Degree | Name | Example from the chapter |
|---|---|---|
| 0 | Constant polynomial | 8, because 8 = 8x0 and the power of x is 0 |
| 1 | Linear polynomial | 3z + 7 |
| 2 | Quadratic polynomial | x2 + 5x + 1 |
| 3 | Cubic polynomial | 5y3 + y2 + 2y - 1 |
Missing terms. A power that does not appear has coefficient 0. In 4z3 + 5z2 - 11 there is no z term, so the coefficient of z is 0. In x4 - 3x3 + 6x2 - 2x + 7 the coefficient of x3 is -3 and that of x2 is 6.
A polynomial of degree 1 is a linear polynomial. The rest of the chapter studies these.
3. Evaluating a Polynomial and Forming Linear Equations
A chess club charges a joining fee of ₹200 plus ₹50 for every match played.
| Matches played | 1 | 2 | 3 | 4 | 5 | m |
|---|---|---|---|---|---|---|
| Amount paid (₹) | 250 | 300 | 350 | 400 | 450 | 200 + 50m |
The amount 200 + 50m is a linear polynomial in m, and it rises by the constant ₹50 for every extra match. If a player paid ₹750, then 200 + 50m = 750, so 50m = 550 and m = 11 matches.
This is the characteristic feature of linear polynomials: the difference between successive values at integers is constant. Patterns with this property are called linear patterns.
From polynomial to equation. When a linear polynomial in one variable is set equal to a constant, the result is a linear equation. Example: two numbers add up to 64 and one is 10 more than the other. Let the smaller be x, so the larger is x + 10. Then x + (x + 10) = 64, which gives 2x + 10 = 64. The left side, 2x + 10, is a linear polynomial; equating it to 64 gives the equation 2x = 54, so x = 27. The numbers are 27 and 37.
Polynomials as input-output processes. For every value of x, the polynomial 2x + 3 gives one output: x = 4 gives 2 × 4 + 3 = 11 and x = -6 gives 2 × (-6) + 3 = -9. Think of a machine: a value of x goes in, the value of 2x + 3 comes out. This process is called a function, and 2x + 3 is a function of the variable x. Functions are studied in detail in later classes.
The rectangle made from the 20 cm wire also works as a machine: its area 10x - x2 is a function of x. At x = 6 cm, the area is 10 × 6 - 62 = 60 - 36 = 24 cm2. Note that 2x + 3 is a linear function, whereas 10x - x2 is a quadratic function.
4. Linear Patterns and the nth Term
A growing pattern of square tiles. The chapter shows a pattern of square tiles in which each stage is made by adding two more tiles to the previous stage, starting from a single tile at Stage 1.
| Stage | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Number of square tiles | 1 | 3 | 5 | 7 | 9 | 11 | 13 |
The number of tiles at each stage is one less than twice the stage number: Stage 2 has 2 × 2 - 1 = 3 tiles, Stage 5 has 2 × 5 - 1 = 9 tiles. So Stage n has 2n - 1 tiles.
2n - 1 has degree 1, so it is a linear polynomial, and consecutive terms differ by the constant 2. The relation between stage number and number of tiles is a linear relationship.
Using the rule: Stage 15 has 2 × 15 - 1 = 29 tiles and Stage 26 has 2 × 26 - 1 = 51 tiles. To find which stage has 21 tiles, solve 2n - 1 = 21, so n = 11. For 47 tiles, 2n - 1 = 47 gives n = 24.
Spending pocket money. Bela has ₹100 and spends ₹5 every day, so she has 95, 90, 85, 80, … rupees left after days 1, 2, 3, 4, and 100 - 5n after n days. She has ₹40 left when 100 - 5n = 40, that is on day 12. On day 15 she has 100 - 75 = ₹25. The money runs out when 100 - 5n = 0, that is after 20 days.
A fare with a starting charge. An auto-rickshaw fare is ₹25 for the first 2 km, then rises by ₹15 for every further km.
The fares for 1 to 6 km are ₹25, 25, 40, 55, 70 and 85.
For n km with n ≥ 2 the fare is 25 + 15 × (n - 2) = 15n - 5. A 10 km ride costs 25 + 15 × 8 = ₹145. A fare of ₹130 means 15n - 5 = 130, so 15n = 135 and n = 9 km.
Definition. A linear pattern is a sequence of numbers where the difference between two consecutive terms is constant. In every example above, the nth term is a linear expression in n. Such sequences are studied again in the chapter on Sequences and Progressions.
How to find the nth-term rule quickly. If the terms change by the same amount d each time (d is negative when they go down), the rule has the form dn + c. Find c by checking one term. For 1, 3, 5, … the difference is 2, so the rule is 2n + c; at n = 1 the term is 1, so 2 + c = 1 and c = -1, giving 2n - 1.
5. Linear Growth and Linear Decay
Linear growth: cost of a journey. The cost of a journey is C(d) = 100 + 60d, where C is the total cost in rupees and d the distance in km.
| Distance d (km) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Cost C (₹) | 100 | 160 | 220 | 280 | 340 | 400 |
Each extra km adds a fixed ₹60. For 15 km the cost is 100 + 60 × 15 = ₹1000. The cost is ₹700 when 100 + 60d = 700, so 60d = 600 and d = 10 km.
Linear decay: water in a tank. The water in a cylindrical tank is 3 m high at the start of summer. After t months its height is h(t) = 3 - 0.5t metres.
| Month t | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Height h (m) | 3 | 2.5 | 2 | 1.5 | 1 |
Each month the height falls by the same 0.5 m. After 5 months, h = 3 - 2.5 = 0.5 m.
Telling them apart. Growth has the form start + (fixed amount) × t, decay has the form start - (fixed amount) × t. Check the sign of the coefficient of the time variable: positive for growth, negative for decay.
Building a model from words. Pick the starting value (the value when time is 0) and the fixed change per interval, then write start ± change × t.
6. Finding a Linear Relationship y = ax + b
A linear relationship between two variables x and y can be written as y = ax + b. For the tile pattern, with x the stage number and y the number of tiles, the relationship is y = 2x - 1.
The data bill. A telecom company charges a fixed monthly fee plus a cost per GB of data. With 10 GB the bill was ₹350; with 20 GB it was ₹550. The bill y depends on data x (in GB) by y = ax + b.
- Substitute x = 10, y = 350: 350 = 10a + b.
- Substitute x = 20, y = 550: 550 = 20a + b.
- From the first equation, b = 350 - 10a.
- Put this into the second: 550 = 20a + (350 - 10a) = 10a + 350, so 10a = 200 and a = 20.
- Then b = 350 - 10 × 20 = 150.
So y = 20x + 150. The number 20 is the charge per GB (₹20) and 150 is the fixed monthly fee (₹150), the bill when no data is used.
A shortcut check. Going from 10 GB to 20 GB (10 more GB) raised the bill by ₹200, so each GB costs 200 ÷ 10 = ₹20. This matches a. The value of a is always the change in y divided by the change in x.
Temperature scales. Celsius and Fahrenheit are linked by °C = a °F + b. Ice melts at 0 °C and 32 °F, and water boils at 100 °C and 212 °F. This gives 0 = 32a + b and 100 = 212a + b. Subtracting, 180a = 100, so a = 5/9. Then b = -32 × 5/9 = -160/9. So °C = (5/9) °F - 160/9, which is the same as °C = (5/9)(°F - 32).
7. Plotting a Linear Relationship as a Straight Line
Every linear relationship y = ax + b can be drawn as a straight line on the coordinate plane. Two points are enough to fix a line.
Plotting y = 2x + 1.
- Choose x = 0: y = 1. Call this point A (0, 1). Its x-coordinate is 0 and its y-coordinate is 1.
- Choose x = 3: y = 7. Call this point B (3, 7).
- Plot A and B on graph paper, join them with a ruler and extend the line in both directions.
More points on the same line: (1, 3), (2, 5), (5, 11), (7, 15), (9, 19), (12, 25) and (20, 41).
Point on a line test. A point lies on a line exactly when its coordinates satisfy the equation. (7, 15) lies on y = 2x + 1 because 2 × 7 + 1 = 15. The point (4, 10) does not, because 2 × 4 + 1 = 9, not 10.
Guessing the equation from points. The points (-1, -3), (0, 0), (1, 3), (3, 9) and (4, 12) all lie on one straight line. In each, the y-coordinate is three times the x-coordinate, so the line is y = 3x.
Tip: plot a third point as a check. If the three points do not line up, one of your calculations is wrong.
8. Slope, y-intercept and Parallel Lines
Lines of the form y = ax. The chapter draws y = (1/2)x, y = x and y = 2x on the same axes. For y = (1/2)x, the points (0, 0) and (4, 2) lie on the line, since half of 4 is 2. Observations:
- Every line y = ax passes through the origin (0, 0), since x = 0 gives y = 0.
- The line y = x is equally inclined to both axes.
- For a > 1 the line is steeper than y = x (for example y = 2x or y = 3x).
- For a < 1 (with a positive) the line is less steep than y = x (for example y = (1/2)x or y = (1/3)x).
The number a is called the slope of the line y = ax. When a is negative, as in y = -(1/3)x, y = -x and y = -3x, the lines still pass through the origin but fall from left to right: as x increases, y decreases.
Growth, decay and the sign of the slope. Linear growth is represented by a straight line with positive slope; linear decay by a line with negative slope. For the tile pattern y = 2x - 1, the slope 2 equals the constant difference between consecutive terms.
Comparing y = 3x + 1 and y = -3x + 1. Both cut the y-axis at (0, 1). The first rises (slope 3) and the second falls (slope -3), and they are equally steep.
Changing b with a fixed. The lines y = 2x - 1, y = 2x + 1 and y = 2x + 5 all have slope 2. Drawn on the same axes, they are parallel; changing b only shifts the line up or down.
The y-intercept. Any line y = ax + b cuts the y-axis at the point (0, b), because x = 0 gives y = b. The number b is the y-intercept.
So y = 2x + 5 cuts the y-axis at A (0, 5), y = x + 3 at B (0, 3), and y = 3x - 2 at C (0, -2).
A y-intercept of 3 means the line cuts the y-axis 3 units from the origin in the positive direction. A y-intercept of -2 means it cuts 2 units from the origin in the negative direction.
Rearrange first. Read a and b only after y is alone on one side. For 2y = 4x + 7, divide by 2: y = 2x + 7/2, so the slope is 2 and the y-intercept is 7/2.
Formula and Theorem Sheet
| Topic | Result | Example |
|---|---|---|
| Univariate polynomial | Expression in one variable and its powers | 5y3 + y2 + 2y - 1 |
| Degree | Highest power of the variable | x2 + 5x + 1 has degree 2 |
| Constant, linear, quadratic, cubic | Degree 0, 1, 2, 3 | 8, 3z + 7, x2 + 5x + 1, 5y3 + 1 |
| Linear equation | Linear polynomial equated to a constant | 2x + 10 = 64 |
| Linear pattern | Constant difference d; nth term = dn + c | 1, 3, 5, …: 2n - 1 |
| Linear growth | Fixed increase over equal intervals | C(d) = 100 + 60d |
| Linear decay | Fixed decrease over equal intervals | h(t) = 3 - 0.5t |
| Linear relationship | y = ax + b | y = 20x + 150 |
| Finding a from two readings | a = (change in y) ÷ (change in x) | (550 - 350) ÷ (20 - 10) = 20 |
| Slope and y-intercept | y = ax + b: slope a, meets y-axis at (0, b) | y = x + 3: slope 1, (0, 3) |
| Sign of slope | Positive: growth (line rises); negative: decay (line falls) | y = -x falls |
| Parallel lines | Same a, different b | y = 3x - 1, y = 3x, y = 3x + 1 |
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Important Definitions
- Coefficient: the number multiplying the variable part of a term. In 2x2 + 5xy - 3y2 the coefficients are 2, 5 and -3.
- Constant term: the term with no variable, such as -1 in 5y3 + y2 + 2y - 1.
- Univariate polynomial (polynomial in one variable): an algebraic expression in one variable and its powers, such as x2 + 5x + 3.
- Degree of a polynomial: the highest power of the variable in it.
- Constant, quadratic and cubic polynomials: polynomials of degree 0, 2 and 3, such as 8, x2 + 5x + 1 and 5y3 + y2 + 2y - 1.
- Linear polynomial: a polynomial of degree 1, such as 2x + 3 or 5 - 4y.
- Linear pattern: a sequence of numbers in which the difference between consecutive terms is constant.
- Linear growth: a pattern in which a quantity increases by a fixed amount over equal intervals.
- Linear decay: a pattern in which a quantity decreases by a fixed amount over equal intervals.
- Linear relationship: a relationship between two variables x and y of the form y = ax + b, represented by a straight line.
- Slope: the number a in y = ax + b; it measures how steep the line is and whether it rises or falls.
- y-intercept: the number b in y = ax + b; the line cuts the y-axis at (0, b).
Solved Examples (NCERT-Based)
Example 1: Degrees of polynomials (Exercise Set 2.1)
Question: Find the degrees of the following polynomials: (i) 2x2 - 5x + 3 (ii) y3 + 2y - 1 (iii) -9 (iv) 4z - 3
Solution:
- (i) Powers of x present: 2, 1, 0. Highest is 2, so the degree is 2 (quadratic).
- (ii) Powers of y present: 3, 1, 0. Degree 3 (cubic).
- (iii) -9 = -9x0. Degree 0 (constant polynomial).
- (iv) 4z - 3 has highest power 1. Degree 1 (linear).
Example 2: Coefficients and constant terms (Exercise Set 2.1)
Question: (a) What are the coefficients of x2 and x3 in the polynomial x4 - 3x3 + 6x2 - 2x + 7? (b) What is the coefficient of z in the polynomial 4z3 + 5z2 - 11? (c) What is the constant term of the polynomial 9x3 + 5x2 - 8x - 10?
Solution:
- (a) The x2 term is +6x2, so its coefficient is 6. The x3 term is -3x3, so its coefficient is -3.
- (b) There is no term in z (power 1). The coefficient of z is 0.
- (c) The term with no variable is -10. The constant term is -10.
Example 3: Values of polynomials (Exercise Set 2.2)
Question: (a) Find the value of the linear polynomial 5x - 3 if: (i) x = 0 (ii) x = -1 (iii) x = 2. (b) Find the value of the quadratic polynomial 7s2 - 4s + 6 if: (i) s = 0 (ii) s = -3 (iii) s = 4.
Solution:
- (a) x = 0: 0 - 3 = -3. x = -1: 5 × (-1) - 3 = -8. x = 2: 10 - 3 = 7.
- (b) s = 0: 6. s = -3: 7 × (-3)2 - 4 × (-3) + 6 = 63 + 12 + 6 = 81. s = 4: 7 × 16 - 16 + 6 = 102.
Example 4: Ages of Salil and his mother (Exercise Set 2.2)
Question: 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.
Solution: Let Salil’s present age be x years. His mother’s age is 3x years. After 5 years their ages are x + 5 and 3x + 5.
(x + 5) + (3x + 5) = 70, so 4x + 10 = 70, 4x = 60 and x = 15.
Salil is 15 years old and his mother is 45 years old. Check: after 5 years, 20 + 50 = 70.
Example 5: Ruby’s coins (Exercise Set 2.2)
Question: 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?
Solution: Let the number of five-rupee coins be n. Then she has 3n two-rupee coins.
Value: 5 × n + 2 × 3n = 5n + 6n = 11n. So 11n = 88 and n = 8.
She has 8 five-rupee coins and 24 two-rupee coins. Check: 40 + 48 = 88.
Example 6: Dimensions of a rectangle (Exercise Set 2.2)
Question: 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?
Solution: Let the width be w cm. The length is (2w + 3) cm.
Perimeter = 2(length + width) = 2(2w + 3 + w) = 2(3w + 3) = 6w + 6.
6w + 6 = 24, so 6w = 18 and w = 3. Length = 2 × 3 + 3 = 9.
The rectangle is 9 cm by 3 cm. Check: 2(9 + 3) = 24.
Example 7: A growing plant (Exercise Set 2.4)
Question: 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.
Solution:
Start with the expression from part (iii): the height begins at 1.75 feet and adds 0.5 feet each month, so h = 1.75 + 0.5t.
(i) After 7 months: h = 1.75 + 3.5 = 5.25 feet.
(ii) Table: at t = 0, 1, 2, 3, …, 10 the height is 1.75, 2.25, 2.75, 3.25, …, 6.75 feet, rising by 0.5 feet every month.
The height increases by the same 0.5 feet over each equal interval of one month, and h is a linear polynomial in t with positive coefficient 0.5. So it is linear growth.
Example 8: Finding a and b from two bills (Exercise Set 2.5)
Question: A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.
Solution:
- x = 10, y = 800: 800 = 10a + b, so b = 800 - 10a.
- x = 15, y = 1100: 1100 = 15a + b = 15a + 800 - 10a = 5a + 800.
- 5a = 300, so a = 60.
- b = 800 - 600 = 200.
The relationship is y = 60x + 200: ₹60 per hour of court use and a fixed fee of ₹200.
Example 9: Slopes, intercepts and parallel lines (End-of-Chapter Exercises)
Question: Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis. (i) y = -3x + 4 (ii) 2y = 4x + 7 (iii) 5y = 6x - 10 (iv) 3y = 6x - 11. Are any of the lines parallel?
Solution: First make y the subject of each equation. To draw each graph, plot the point (0, b) and one more point from the equation (for (i), x = 1 gives y = 1, so (1, 1)), join them and extend the line.
| Line | In the form y = ax + b | Slope | y-intercept | Cuts the y-axis at |
|---|---|---|---|---|
| (i) | y = -3x + 4 | -3 | 4 | (0, 4) |
| (ii) | y = 2x + 7/2 | 2 | 7/2 | (0, 7/2) |
| (iii) | y = (6/5)x - 2 | 6/5 | -2 | (0, -2) |
| (iv) | y = 2x - 11/3 | 2 | -11/3 | (0, -11/3) |
Lines (ii) and (iv) have the same slope 2 and different y-intercepts, so (ii) and (iv) are parallel.
Example 10: Two linear polynomials from conditions (End-of-Chapter Exercises)
Question: Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: (i) p(0) = 5. (ii) The polynomial p(x) - q(x) cuts the x-axis at (3, 0). (iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x. Find the polynomials p(x) and q(x).
Solution:
- p(0) = a × 0 + b = b, so b = 5.
- p(x) + q(x) = (a + c)x + (b + d) = 6x + 4 for all x. So a + c = 6 and b + d = 4, giving d = -1.
- p(x) - q(x) = (a - c)x + (b - d) = (a - c)x + 6. It cuts the x-axis at (3, 0), so its value at x = 3 is 0: 3(a - c) + 6 = 0, giving a - c = -2.
- Adding a + c = 6 and a - c = -2: 2a = 4, so a = 2 and c = 4.
p(x) = 2x + 5 and q(x) = 4x - 1. Check: sum = 6x + 4; difference = -2x + 6, which is 0 at x = 3.
Competency-Based Questions (with answers)
1. Case-based: The cab company
A cab company charges a fixed booking fee plus a fixed rate per km. A 6 km ride costs ₹170 and a 10 km ride costs ₹250. The fare y (in ₹) for x km follows y = ax + b.
(a) Find a and b. 170 = 6a + b and 250 = 10a + b. Subtracting, 4a = 80, so a = 20. Then b = 170 - 120 = 50. The fare is y = 20x + 50.
(b) What do a and b mean here? a = ₹20 is the charge per km; b = ₹50 is the booking fee (the fare for 0 km).
(c) How far can a rider travel for ₹390? 20x + 50 = 390, so 20x = 340 and x = 17 km.
2. Case-based: Draining a pond
A school pond holds 900 litres of water. For cleaning, water is pumped out at a steady 45 litres per minute.
(a) Write the amount of water W left after t minutes. W = 900 - 45t.
(b) How much water is left after 12 minutes? 900 - 540 = 360 litres.
(c) When will the pond be empty? 900 - 45t = 0 gives t = 20 minutes.
3. Case-based: A row of squares made of matchsticks
Stage 1 is one square made of 4 matchsticks. At every new stage one more square is added to the end of the row, sharing one side with the last square, so each new square needs 3 more matchsticks.
(a) Write the number of matchsticks for Stages 1 to 5. 4, 7, 10, 13, 16.
(b) Find the rule for Stage n. Constant difference 3, so the rule is 3n + c. At n = 1: 3 + c = 4, so c = 1. Rule: 3n + 1.
(c) Can a stage use exactly 100 matchsticks? 3n + 1 = 100 gives 3n = 99, n = 33. Yes, Stage 33.
4. Assertion-Reason
Assertion (A): The number 7 is a polynomial of degree 0.
Reason (R): 7 can be written as 7x0, in which the power of the variable is 0.
Options: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a). The chapter shows that a constant such as 8 equals 8x0, so it is a constant polynomial of degree 0. R gives exactly this reason.
5. Assertion-Reason
Assertion (A): The lines y = 4x + 3 and y = 4x - 5 are parallel.
Reason (R): Two lines of the form y = ax + b with the same y-intercept are parallel.
Answer: (c). A is true: both lines have slope 4 and different y-intercepts. R is false: lines with the same y-intercept meet at that point on the y-axis. Parallel lines need the same slope.
6. Assertion-Reason
Assertion (A): The sequence 2, 5, 9, 14, … is a linear pattern.
Reason (R): In a linear pattern, the difference between consecutive terms is constant.
Answer: (d). R is the definition of a linear pattern, so it is true. The differences in A are 3, 4, 5, which are not constant, so A is false.
7. Data reasoning: Is it linear?
Two tables are recorded.
| t | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Table P | 18 | 15 | 12 | 9 | 6 |
| Table Q | 1 | 2 | 4 | 8 | 16 |
(a) Which table shows a linear relationship? Justify. Table P: the differences are all -3, a constant, so it is linear. Table Q: the differences are 1, 2, 4, 8, which are not constant, so it is not linear.
(b) Write the rule for the linear table and say whether it is growth or decay. P = 18 - 3t. It falls by a fixed 3 each step, so it is linear decay.
8. Source-based: The cost of a garden
Read: “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.”
(a) Write the total cost as an algebraic expression. 200l + 160w + 50lw.
(b) Is this a univariate polynomial? Explain. No. It has two variables, l and w, and a univariate polynomial has only one variable.
(c) The owner fixes the width at 10 m. Write the cost as a polynomial in l and state its degree. 200l + 1600 + 500l = 700l + 1600. It is a linear polynomial in l (degree 1).
Important Questions for Exams
1-Mark Questions
Q1. Write the degree of the polynomial 7 - 2x + 5x3.
Answer: The highest power of x is 3, so the degree is 3.
Q2. Write the coefficient of x in 3x2 - 9.
Answer: There is no x term, so the coefficient of x is 0.
Q3. Classify 5 - 4y by degree.
Answer: Degree 1, so it is a linear polynomial.
Q4. Write the y-intercept of the line y = 3x - 2.
Answer: -2. The line cuts the y-axis at (0, -2).
2-Mark Questions
Q5. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x2 term is -7.
Answer: x3 - 7x2 + 2x + 1 (any cubic with -7x2 works, for example 2x3 - 7x2). Its highest power is 3 and its x2 coefficient is -7.
Q6. Find the values of the following polynomials at the indicated values of the variables: (i) 5x2 - 3x + 7 if x = 1 (ii) 4t3 - t2 + 6 if t = a.
Answer: (i) 5 - 3 + 7 = 9. (ii) 4a3 - a2 + 6.
Q7. 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?
Answer: Shorter = x, longer = 4x. x + 4x = 300, so 5x = 300 and x = 60. The pieces are 60 feet and 240 feet.
3-Mark Questions
Q8. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.
Answer: 400 = 10a + b and 500 = 14a + b. Subtracting, 4a = 100, so a = 25. Then b = 400 - 250 = 150. So y = 25x + 150: ₹25 per module and a fixed fee of ₹150.
Q9. 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.
Answer: (iii) v = 10000 - 800t. (i) After 3 years, v = 10000 - 2400 = ₹7,600. (ii) For t = 0 to 8: 10000, 9200, 8400, 7600, 6800, 6000, 5200, 4400, 3600. The value falls by the same ₹800 over each equal interval of one year, and the coefficient of t is negative, so this is linear decay.
Q10. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.
Answer: Let the tens digit be a and the units digit b. The number is 10a + b and the reversed number is 10b + a. Their sum is 11a + 11b = 143, so a + b = 13. The digits differ by 3, so taking a > b, a - b = 3. Adding, 2a = 16, so a = 8 and b = 5. The numbers are 85 and 58. Check: 85 + 58 = 143.
5-Mark Questions
Q11. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11). (i) Find the polynomial p(x). (ii) Find the coordinates where the graph of p(x) cuts the axes.
Answer: Let p(x) = ax + b. From (1, 5): a + b = 5. From (3, 11): 3a + b = 11. Subtracting, 2a = 6, so a = 3 and b = 2. So p(x) = 3x + 2. It cuts the y-axis where x = 0: (0, 2). It cuts the x-axis where p(x) = 0: 3x + 2 = 0, x = -2/3, so (-2/3, 0).
Q12. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: (i) The graph of p(x) passes through the points (2, 3) and (6, 11). (ii) The graph of q(x) passes through the point (4, -1). (iii) The graph of q(x) is parallel to the graph of p(x). Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.
Answer: From (2, 3) and (6, 11): 2a + b = 3 and 6a + b = 11. Subtracting, 4a = 8, so a = 2, and b = 3 - 4 = -1. So p(x) = 2x - 1. Parallel lines have equal slopes, so c = 2. From (4, -1): -1 = 8 + d, so d = -9 and q(x) = 2x - 9. p(x) = 0 at x = 1/2, so p meets the x-axis at (1/2, 0). q(x) = 0 at x = 9/2, so q meets it at (9/2, 0).
Q13. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = (9/5)(x - 273) + 32. (i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K. (ii) If the temperature is 158 °F, then find the temperature in Kelvin.
Answer: (i) x = 313: x - 273 = 40, (9/5) × 40 = 72, y = 72 + 32 = 104 °F. (ii) y = 158: 158 - 32 = 126 = (9/5)(x - 273). Multiply by 5/9: x - 273 = 70, so x = 343 K.
Common Mistakes and Examiner Tips
- Taking the number of terms as the degree. 2x2 - 5x + 3 has three terms but degree 2. Fix: look only at the highest power of the variable.
- Dropping the sign of a coefficient. In x4 - 3x3 + … the coefficient of x3 is -3, not 3. Fix: the sign in front of a term belongs to its coefficient.
- Saying a missing term has “no coefficient”. The coefficient of z in 4z3 + 5z2 - 11 is 0. Fix: write the answer as 0.
- Calling a constant “not a polynomial” or giving it degree 1. -9 is a constant polynomial of degree 0. Fix: remember -9 = -9x0.
- Substituting a negative value without brackets. 7s2 at s = -3 is 7 × 9 = 63, not -63. Fix: write 7 × (-3)2 with brackets every time.
- Wrong nth-term rule from a table. For 1, 3, 5, 7, … writing 2n + 1 gives 3 at n = 1. Fix: after writing a rule, test it at n = 1 and n = 2.
- Forgetting the part already paid for. In the auto fare, ₹25 covers the first 2 km, so the extra charge is 15(n - 2), not 15n. Fix: count only the units after the fixed part.
- Mixing up growth and decay. h(t) = 3 - 0.5t is decay even though the numbers are positive. Fix: check the sign of the coefficient of the time variable.
- Reading slope and intercept before making y the subject. For 2y = 4x + 7 the slope is 2 (not 4) and the y-intercept is 7/2 (not 7). Fix: divide through so the equation reads y = ax + b.
- Thinking equal y-intercepts make lines parallel. y = 3x + 1 and y = -3x + 1 meet at (0, 1). Fix: parallel lines have equal slopes and different y-intercepts.
Quick Revision Points
- A univariate polynomial has one variable and its powers.
- Degree = highest power of the variable.
- Degree 0 constant, 1 linear, 2 quadratic, 3 cubic.
- To evaluate a polynomial, substitute the value and simplify; bracket negative values.
- A linear polynomial equated to a constant gives a linear equation.
- A linear pattern has a constant difference between consecutive terms.
- The tile pattern 1, 3, 5, 7, … has nth term 2n - 1.
- Linear growth: fixed increase over equal intervals, such as C(d) = 100 + 60d.
- Linear decay: fixed decrease over equal intervals, such as h(t) = 3 - 0.5t.
- A linear relationship is y = ax + b; two readings give a and b.
- a = change in y ÷ change in x.
- y = ax passes through the origin; a > 1 is steeper than y = x, 0 < a < 1 is less steep.
- Positive slope: line rises, growth. Negative slope: line falls, decay.
- y = ax + b cuts the y-axis at (0, b); b is the y-intercept.
- Same slope, different y-intercepts: parallel lines.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, Unit II, Algebra, carries 20 marks out of the 80-mark theory paper and is allotted 66 teaching periods. That 20 marks is shared by four chapters: Introduction to Polynomials, Sequences and Progressions, Exploring Algebraic Identities, and Linear Equations in Two Variables. The course structure does not give a separate mark for this chapter alone. The remaining 20 marks of the subject are internal assessment (pen-paper tests and multiple assessment 10, portfolio 5, lab practical 5).
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
The question paper design in the course structure weights questions by typology: about 54% remembering and understanding, 24% applying, and 22% analysing, evaluating and creating.
The course structure lists the key concepts as algebraic expressions, the definition and degree of a polynomial, linear polynomials and applications, linear patterns, linear growth and decay, linear relationships and their graphs, and the slope and y-intercept of y = ax + b. The chapter prepares you for sequences and progressions, linear equations in two variables and Class 10 algebra.
Class 9 Maths · Chapter 2 – swipe through all 10 cards to understand the whole chapter.
Parts of an Expression
Algebraic expressions are built from terms made of numbers and variables.
Variables were called letter-numbers in earlier classes.
- Terms: 4x, 5y and 3.
- Coefficients: 4 of x and 5 of y.
- Constant term: 3, the term with no variable.
Degree of a Polynomial
A univariate polynomial uses one variable and its powers; its degree is the highest power.
The number of terms never decides the degree.
- Degree 0 constant, 1 linear, 2 quadratic, 3 cubic.
- 8 = 8x0, so a constant has degree 0.
- A missing power has coefficient 0.
Evaluating a Polynomial
Substitute a value for the variable and simplify to get the output.
Always bracket a negative input before squaring.
- At x = 4, 2x + 3 = 11.
- 7s2 − 4s + 6 at s = −3 gives 63 + 12 + 6 = 81.
- The process is a function of the variable.
Linear Polynomial to Linear Equation
Equate a linear polynomial to a constant to get a linear equation.
Write a let-statement, the equation, the answer and a check.
- Two numbers with sum 64, one 10 more: 27 and 37.
- Salil 15 years, mother 45 years.
- Rectangle with perimeter 24 cm: 9 cm by 3 cm.
Linear Patterns and the nth Term
A linear pattern has a constant difference between consecutive terms.
Test your rule at n = 1 and n = 2.
- Stage 15 of the tile pattern has 29 tiles.
- Bela’s money after n days: 100 − 5n.
- Auto fare for n km (n ≥ 2): 15n − 5.
Linear Growth
A quantity increases by a fixed amount over equal intervals.
The coefficient of the time or distance variable is positive.
- 15 km costs 100 + 900 = ₹1000.
- Plant: h = 1.75 + 0.5t, so 5.25 feet after 7 months.
- Graph: a line with positive slope.
Linear Decay
A quantity decreases by a fixed amount over equal intervals.
Set the rule equal to 0 to find when the quantity runs out.
- Water level after 5 months: 0.5 m.
- Prepaid balance b(x) = 600 − 15x runs out in 40 days.
- Graph: a line with negative slope.
Finding y = ax + b
Two pairs of values give two equations that fix a and b.
a = change in y ÷ change in x.
- Subtracting gives a = 20, then b = 150.
- y = 20x + 150: ₹20 per GB, ₹150 fixed fee.
- Celsius and Fahrenheit: °C = (5/9)(°F − 32).
Slope and y-intercept
The line y = ax + b has slope a and cuts the y-axis at (0, b).
Make y the subject before reading a and b.
- y = ax passes through the origin (0, 0).
- a > 1 is steeper than y = x; a between 0 and 1 is less steep.
- y = 3x − 2 cuts the y-axis at (0, −2).
Same Slope, Parallel Lines
Changing b with a fixed shifts the line without turning it.
Equal y-intercepts make lines meet on the y-axis.
- Parallel lines: same a, different b.
- 2y = 4x + 7 and 3y = 6x − 11 both have slope 2.
- y = 3x + 1 and y = −3x + 1 meet at (0, 1).
📝 Practice Introduction to Polynomials - 10 questions
Concept-check questions · with answers & explanations
Start →Close ✕
Why A: The Assertion is true; NCERT states that the constant 8 is a polynomial of degree 0 and calls such polynomials constant polynomials. The Reason is also true, and it is exactly the justification NCERT gives: 8 = 8x⁰, and the highest (only) power of x is 0, so the degree is 0. R correctly explains A.
Why not B: R is not merely a true side fact; the power 0 in 8x⁰ is precisely the reason the degree is 0.
Why not C: R is true; x⁰ = 1, so 8x⁰ = 8.
Why not D: A is true; a constant is a polynomial of degree 0, not a non-polynomial.
Remember: A constant c = c·x⁰ is a polynomial of degree 0.
Why D: Step 1 (assign the variable): Salil’s age x, mother’s 3x. Step 2 (ages after 5 years): x + 5 and 3x + 5. Step 3 (form the equation): (x + 5) + (3x + 5) = 70, so 4x + 10 = 70. Step 4 (solve): 4x = 60, x = 15, mother 45. Check: in 5 years, 20 + 50 = 70.
Why not A: 17.5 and 52.5 come from 4x = 70, ignoring that 5 years are added to each age.
Why not B: 20 and 60 come from 4x – 10 = 70, subtracting the years instead of adding them.
Why not C: 15 and 50 mixes Salil’s present age with his mother’s age 5 years from now.
Remember: After n years, add n to EVERY person’s age.
Why B: Setting 2n – 1 = 200 gives 2n = 201 and n = 100.5, which is not a whole number, so no stage has 200 tiles. Equivalently, 2n – 1 is odd for every whole number n and 200 is even.
Why not A: Stage 100 has 2 × 100 – 1 = 199 tiles, not 200.
Why not C: Stage numbers must be positive whole numbers; there is no stage 100.5.
Why not D: The pattern never stops; it simply skips every even number.
Remember: A value occurs only if solving for n gives a positive whole number.
Why A: The Assertion is true: h(0) = 3, h(1) = 2.5, h(2) = 2, h(3) = 1.5, each month lowers h by exactly 0.5. The Reason is true and explains it: in any linear function b + at, increasing t by 1 changes the value by exactly a; here a = -0.5, so every month the height changes by -0.5 m. R is the correct explanation of A.
Why not B: R is not a separate fact; the constant coefficient is precisely why the monthly drop is constant.
Why not C: R is true; the coefficient of t is -0.5, which is negative.
Why not D: A is true; substituting t = 0, 1, 2, 3 shows the drop is always 0.5.
Remember: In b + at, each unit step in t changes the value by exactly a.
Why C: Step 1 (substitute the two pairs, with °F as x and °C as y): 0 = 32a + b and 100 = 212a + b. Step 2 (subtract to eliminate b): 100 = 180a, so a = 100/180 = 5/9. Step 3 (find b): b = -32a = -32 × 5/9 = -160/9. So °C = (5/9)°F – 160/9, the same as °C = (5/9)(°F – 32). Check: 212 × 5/9 – 160/9 = (1060 – 160)/9 = 100.
Why not A: 9/5 and 32 belong to the reverse relation °F = (9/5)°C + 32, where °C is the input.
Why not B: a is right, but b = 32 ignores the sign and the multiplication by a; at 32 °F this would give 32 × 5/9 + 32, not 0.
Why not D: 9/5 is the change in x over the change in y, the wrong way round.
Remember: a = (change in output) ÷ (change in input); keep fractions exact.
Why A: Step 1 (slope of p from two points): a = (11 – 3) ÷ (6 – 2) = 8 ÷ 4 = 2, and 3 = 2 × 2 + b gives b = -1, so p(x) = 2x – 1. Step 2 (parallel means equal slope): c = 2, and -1 = 2 × 4 + d gives d = -9, so q(x) = 2x – 9. Step 3 (x-axis means y = 0): 2x – 9 = 0, x = 9/2. The point is (9/2, 0).
Why not B: 1/2 is where p(x) = 2x – 1 meets the x-axis, not q(x).
Why not C: (0, -9) is where q(x) meets the y-axis, not the x-axis.
Why not D: -9/2 comes from solving 2x + 9 = 0, with the sign of d wrong.
Remember: On the x-axis y = 0; on the y-axis x = 0; parallel lines share the slope.
Why A: The perimeter is 20 cm, so length + width = 10 cm and the width is (10 – x) cm. Area = length × width = x(10 – x) = 10x – x². The highest power of x is 2, so it is a quadratic polynomial.
Why not B: 20x – x² uses the whole perimeter 20 as the sum of length and width instead of half of it.
Why not C: 10 – x is the width, not the area.
Why not D: 4x is the perimeter of a square of side x, a different figure.
Remember: In a rectangle, length + width = half the perimeter; area = length × width.
Why D: Step 1 (square the input): (-3)² = 9, so 7s² = 63. Step 2 (product of two negatives): -4 × (-3) = +12. Step 3 (add all terms): 63 + 12 + 6 = 81.
Why not A: 57 comes from 63 – 12 + 6, treating -4 × (-3) as -12.
Why not B: -45 comes from treating (-3)² as -9, giving -63 + 12 + 6.
Why not C: 51 is 63 – 12, an arithmetic slip that both mis-signs the middle term and forgets the constant 6.
Remember: (-3)² = +9 and (-4)(-3) = +12; two negatives multiply to a positive.
Why C: Step 1 (set up the equation): 2n – 1 = 47. Step 2 (undo the -1): 2n = 48. Step 3 (undo the ×2): n = 24. Check: 2 × 24 – 1 = 47.
Why not A: 23 is (47 – 1) ÷ 2, subtracting 1 instead of adding it when undoing.
Why not B: 93 is 2 × 47 – 1, applying the rule forwards to 47 instead of reversing it.
Why not D: 48 is the value of 2n, one step short of the answer.
Remember: To find the stage, reverse the rule: add 1, then halve.
Why A: Step 1 (build the model): starting value 10,000 with a fixed decrease of 800 per year gives v = 10000 – 800t. Step 2 (evaluate at t = 3): 10000 – 800 × 3 = 10000 – 2400 = Rs 7,600. The negative coefficient shows linear decay.
Why not B: The value is right but v = 10000 + 800t would make the phone gain value each year.
Why not C: 2,400 is only the amount lost, 800 × 3, and 800t is the loss, not the value.
Why not D: 9,200 is the value after 1 year, not 3.
Remember: Value = start – loss per interval × number of intervals.
Chapter Navigation
Previous: Coordinate Geometry Class 9 Notes
Next: Number System Class 9 Notes
Related Chapters in Class 9 Maths
- Coordinate Geometry Class 9 Notes
- Number System Class 9 Notes
- Exploring Algebraic Identities Class 9 Notes
- Circles Class 9 Notes
- Area and Perimeter Class 9 Notes
- Introduction to Probability Class 9 Notes
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Frequently Asked Questions
The degree of a polynomial in one variable is the highest power of the variable in it. For example, 5y³ + y² + 2y − 1 has degree 3 and x² + 5x + 1 has degree 2. The number of terms and the size of the coefficients do not affect the degree. A constant such as 8 has degree 0 because it can be written as 8x⁰.
A linear polynomial is a polynomial of degree 1, such as 3z + 7, 2x + 3 or 5 − 4y. Its value changes by the same amount every time the variable increases by 1. For 2x + 3 the value rises by 2 for each step of 1 in x. Equating a linear polynomial to a constant, as in 2x + 10 = 64, gives a linear equation.
A linear pattern is a sequence of numbers in which the difference between consecutive terms is constant. The tile pattern 1, 3, 5, 7, 9 has a constant difference of 2 and its nth term is 2n − 1, a linear expression in n. The sequence 2, 5, 9, 14 is not a linear pattern because its differences 3, 4, 5 keep changing.
In linear growth a quantity increases by a fixed amount over equal intervals, as in C(d) = 100 + 60d, where the cost rises by ₹60 for each km. In linear decay a quantity decreases by a fixed amount over equal intervals, as in h(t) = 3 − 0.5t, where the water level falls by 0.5 m each month. On a graph, growth has positive slope and decay has negative slope.
Substitute each pair of values into y = ax + b to get two equations. For bills of ₹350 at 10 GB and ₹550 at 20 GB: 350 = 10a + b and 550 = 20a + b. Subtracting gives 10a = 200, so a = 20, and then b = 350 − 200 = 150. The relation is y = 20x + 150: ₹20 per GB plus a fixed fee of ₹150.
For a line written as y = ax + b, the slope is a and the y-intercept is b. The line cuts the y-axis at the point (0, b). For y = 3x − 2 the slope is 3 and the line cuts the y-axis at (0, −2). If the equation is 2y = 4x + 7, first divide by 2 to get y = 2x + 7/2, so the slope is 2 and the y-intercept is 7/2.
Lines of the form y = ax + b are parallel when they have the same slope a and different y-intercepts b. For example, y = 2x − 1, y = 2x + 1 and y = 2x + 5 are parallel: changing b only shifts the line up or down. Lines with the same y-intercept but different slopes, such as y = 3x + 1 and y = −3x + 1, meet on the y-axis.
Class 9 has no board exam, so the chapter is tested in school exams. In the CBSE Class IX course structure for 2026-27 it belongs to Unit II, Algebra, which carries 20 of the 80 theory marks together with three other algebra chapters. Expect questions on degree and coefficients, word problems, nth-term rules, growth and decay models, and graphs with slope and y-intercept.
