Introduction to Linear Polynomials — Detailed Notes
An algebraic expression combines numbers, variables and operations — e.g. 4x + 5y + 3. Its parts are called terms (4x, 5y, 3); the letters x, y are variables; the numbers multiplying the variables (4, 5) are coefficients; and a number alone (3) is a constant.
2.1 Polynomials in one variable
An expression in one variable and its whole-number powers is a polynomial (univariate polynomial), e.g. 5y³ + y² + 2y − 1. The highest power of the variable is its degree.
| Degree | Name | Example |
|---|
| 0 | Constant polynomial | 8 |
| 1 | Linear polynomial | 3z + 7 |
| 2 | Quadratic polynomial | x² + 5x + 1 |
| 3 | Cubic polynomial | 5y³ + y² + 2y − 1 |
2.2 Linear polynomials & evaluating them
A linear polynomial has degree 1, e.g. the perimeter of a square of side x is 4x. A polynomial can be seen as an input–output (function) process: substitute a value of the variable to get the output. For 2x + 3: if x = 4, output = 2(4) + 3 = 11; if x = −6, output = 2(−6) + 3 = −9. Equating a linear polynomial to a constant gives a linear equation (e.g. 2x + 10 = 64 ⇒ x = 27).
2.3 Linear patterns
Linear pattern: a sequence of numbers in which the difference between consecutive terms is constant. Its n-th term is a linear expression in n. Example: tiles 1, 3, 5, 7, … → n-th term = 2n − 1 (constant difference 2).
2.4 Linear growth & linear decay
• Linear growth — a quantity increases by a constant amount over equal intervals (e.g. cost C(d) = 100 + 60d). • Linear decay — a quantity decreases by a constant amount over equal intervals (e.g. tank height h(t) = 3 − 0.5t).
2.5–2.6 Linear relationships & their graphs
A linear relationship between x and y is written y = ax + b, and its graph is a straight line. To draw it, find any two points (e.g. for y = 2x + 1, take (0, 1) and (3, 7)), plot and join them. A point lies on a line only if its coordinates satisfy the equation.
y = ax + b — a = slope, b = y-intercept (line cuts y-axis at (0, b))
Key observations: • Lines of the form y = ax pass through the origin. • The slope a measures steepness — larger |a| is steeper; positive slope = growth, negative slope = decay. • Changing b (keeping a fixed) shifts the line up/down, keeping it parallel — so lines with equal slopes but different y-intercepts are parallel. • The slope equals the constant difference of the corresponding linear sequence.
Key Terms
algebraic expression, term, variable, coefficient, constant, polynomial (univariate), degree, linear/quadratic/cubic/constant polynomial, linear equation, function (input–output), linear pattern, linear growth, linear decay, linear relationship, slope, y-intercept, parallel lines.
End-of-Chapter Exercises — Full Solutions
Complete worked solutions to all 14 end-of-chapter problems (starred * = HOTS), plus the key word problems from Exercise Sets 2.2 and 2.5.
1Write a polynomial of degree 3 in x in which the coefficient of the x² term is −7.
Any cubic polynomial whose x² coefficient is −7 works, for example p(x) = x³ − 7x² + 2x + 1. (It has degree 3 because the highest power is 3, and the coefficient of x² is −7.)
2Find the values: (i) 5x² − 3x + 7 at x = 1; (ii) 4t³ − t² + 6 at t = a.
(i) 5(1)² − 3(1) + 7 = 5 − 3 + 7 = 9. (ii) Substitute t = a: 4a³ − a² + 6.
3If we multiply a number by 5/2 and add 2/3 to the product, we get −7/12. Find the number.
Let the number be x. Then (5/2)x + 2/3 = −7/12 ⇒ (5/2)x = −7/12 − 8/12 = −15/12 = −5/4 ⇒ x = (−5/4) × (2/5) = −1/2. (Check: (5/2)(−1/2) + 2/3 = −5/4 + 2/3 = −7/12 ✓)
4A positive number is 5 times another. If 21 is added to both, one new number becomes twice the other. Find the numbers.
Let the smaller number be x and the larger be 5x. After adding 21: (5x + 21) and (x + 21). The larger becomes twice the smaller: 5x + 21 = 2(x + 21) ⇒ 5x + 21 = 2x + 42 ⇒ 3x = 21 ⇒ x = 7. So the numbers are 7 and 35. (Check: 35 + 21 = 56 = 2 × (7 + 21) = 2 × 28 ✓)
5You have ₹800 and save ₹250 every month. Find the amount after (i) 6 months, (ii) 2 years. Express as a linear pattern.
Amount after n months = 800 + 250n (a linear pattern, constant increase ₹250). (i) n = 6: 800 + 250×6 = 800 + 1500 = ₹2300. (ii) 2 years = 24 months: 800 + 250×24 = 800 + 6000 = ₹6800.
6(*) The digits of a two-digit number differ by 3. Reversing the digits and adding to the original gives 143. Find both numbers.
Let the tens digit be a and units digit b. Number + reversed = (10a + b) + (10b + a) = 11(a + b) = 143 ⇒ a + b = 13. Also |a − b| = 3. Solving a + b = 13 with a − b = ±3 gives the digit pair {8, 5}. So the numbers are 85 and 58 (both have digits differing by 3 and give 143 when added to their reverse).
7(*) Graph and find the slope, y-intercept and y-axis crossing of: (i) y = −3x + 4, (ii) 2y = 4x + 7, (iii) 5y = 6x − 10, (iv) 3y = 6x − 11. Which are parallel?
Write each as y = ax + b: (i) y = −3x + 4 → slope −3, y-intercept 4, cuts y-axis at (0, 4). (ii) y = 2x + 7/2 → slope 2, y-intercept 3.5, cuts at (0, 3.5). (iii) y = (6/5)x − 2 → slope 6/5, y-intercept −2, cuts at (0, −2). (iv) y = 2x − 11/3 → slope 2, y-intercept −11/3, cuts at (0, −11/3). Parallel: (ii) and (iv) both have slope 2, so they are parallel.
8(*) Given y = (9/5)(x − 273) + 32 (x in Kelvin, y in °F): (i) find y when x = 313 K; (ii) find x when y = 158 °F.
(i) y = (9/5)(313 − 273) + 32 = (9/5)(40) + 32 = 72 + 32 = 104 °F. (ii) 158 = (9/5)(x − 273) + 32 ⇒ 126 = (9/5)(x − 273) ⇒ x − 273 = 126 × 5/9 = 70 ⇒ x = 343 K.
9(*) Work = force × distance. Express as a linear equation in work w and distance d with force = 3 units; draw its graph; find the work when d = 2.
With constant force 3, w = 3d — a linear equation whose graph is a straight line through the origin with slope 3. When d = 2, w = 3 × 2 = 6 units (the point (2, 6) lies on the line).
10(*) The graph of a linear polynomial p(x) passes through (1, 5) and (3, 11). (i) Find p(x). (ii) Where does it cut the axes?
Let p(x) = ax + b. Then a + b = 5 and 3a + b = 11. Subtracting: 2a = 6 ⇒ a = 3, so b = 2. Thus p(x) = 3x + 2. (ii) y-axis: put x = 0 → (0, 2); x-axis: put p(x) = 0 → 3x + 2 = 0 → x = −2/3 → (−2/3, 0).
11(*) p(x) = ax + b, q(x) = cx + d with (i) p(0) = 5, (ii) p(x) − q(x) cuts the x-axis at (3, 0), (iii) p(x) + q(x) = 6x + 4. Find p and q.
From (i) b = 5. From (iii) (a + c)x + (b + d) = 6x + 4 ⇒ a + c = 6 and b + d = 4 ⇒ d = −1. From (ii) p(3) − q(3) = 0: (3a + 5) − (3c − 1) = 0 ⇒ 3(a − c) + 6 = 0 ⇒ a − c = −2. With a + c = 6 ⇒ a = 2, c = 4. So p(x) = 2x + 5 and q(x) = 4x − 1.
12(*) Hexagons made of matchsticks; each new hexagon shares one side with the previous. (i) matchsticks at stages 4, 5; (ii) table; (iii) rule for stage n; (iv) stage 15; (v) can 200 matchsticks form a stage?
Stage 1 uses 6 matchsticks; each extra hexagon adds 5 (one side is shared). (i) Stage 4 = 6 + 5×3 = 21, Stage 5 = 26. (ii) 6, 11, 16, 21, 26, … (iii) Rule: 5n + 1. (iv) Stage 15 = 5×15 + 1 = 76. (v) 5n + 1 = 200 ⇒ 5n = 199 ⇒ n = 39.8, not a whole number, so no — 200 matchsticks cannot form a stage.
13(*) p(x) = ax + b passes through (2, 3) and (6, 11); q(x) = cx + d passes through (4, −1) and is parallel to p. Find p and q and where each cuts the x-axis.
For p: 2a + b = 3, 6a + b = 11 ⇒ 4a = 8 ⇒ a = 2, b = −1 → p(x) = 2x − 1. Parallel ⇒ same slope, so c = 2. Through (4, −1): 2(4) + d = −1 ⇒ d = −9 → q(x) = 2x − 9. x-axis crossings: p: 2x − 1 = 0 → (1/2, 0); q: 2x − 9 = 0 → (9/2, 0). (Being parallel, the two lines never meet each other.)
14(*) What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Factor: f(x) = a(x + 1). Putting x = −1 gives f(−1) = a(0) = 0 for every a. So all these lines pass through the common point (−1, 0).
EKey word problems from the in-text exercises (Sets 2.2 & 2.5).
Set 2.2 Q3 — Salil = x, mother = 3x; (x+5)+(3x+5) = 70 ⇒ 4x = 60 ⇒ x = 15 → Salil 15, mother 45. Q4 — integers 2k, 5k; 5k − 2k = 63 ⇒ k = 21 → 42 and 105. Q5 — five-rupee coins x, two-rupee 3x; 5x + 2(3x) = 88 ⇒ 11x = 88 ⇒ x = 8 → 8 five-rupee and 24 two-rupee coins. Q6 — x + 4x = 300 ⇒ x = 60 → pieces 60 ft and 240 ft. Q7 — L = 2w + 3, perimeter 24 ⇒ L + w = 12 ⇒ 3w + 3 = 12 ⇒ w = 3, L = 9 → 9 cm × 3 cm. Set 2.5 Q1 — 400 = 10a + b, 500 = 14a + b ⇒ a = 25, b = 150 → y = 25x + 150. Q2 — 800 = 10a + b, 1100 = 15a + b ⇒ a = 60, b = 200 → y = 60x + 200. Q3 — °C = a·°F + b with (32, 0) and (212, 100): a = 5/9, b = −160/9 → °C = (5/9)(°F − 32).