Introduction to Linear Polynomials
Why This Matters
Think about a small everyday problem. A chess club asks for ₹200 to join, then ₹50 for every match you play. How much will you pay after 5 matches? After 12 matches? After any number of matches?
You could work it out match by match. But there is a smarter way. You can write one short rule that works for every number of matches at once: 200 + 50m, where m is the number of matches. Put in m = 5, you get ₹450. Put in m = 12, you get ₹800. One little rule does all the work.
That short rule is a linear polynomial. This chapter is about these rules — how to write them, read them, find their values, and even draw them as a straight line. Once you can do this, a huge number of real situations become easy. A taxi fare. The water left in a tank. The pages left in a book. The growing edge of a tile pattern. All of them hide a linear polynomial inside.
A few words from earlier classes turn up again and again here — variable, term, coefficient, constant. Let’s refresh them so they feel familiar before we start.
The Big Idea
A linear polynomial is an algebraic expression in one variable whose highest power is 1 — it has the form ax + b (with a ≠ 0). Examples: 3z + 7, 2x − 5, 200 + 50m. You can find its value by putting a number in for the variable. You can find its zero by setting it equal to 0 and solving. And you can draw it as a straight line y = ax + b, where a is the slope (how steep) and b is the y-intercept (where it cuts the y-axis).
From expressions to polynomials
Let’s start with a real shop. Raju walks in. Red boxes hold 4 pens each. Blue boxes hold 5 pencils each. He buys x red boxes and y blue boxes, and gets 3 extra pens free. How many pens and pencils in total?
- x red boxes give 4x pens.
- y blue boxes give 5y pencils.
- Plus the 3 free pens.
So the total is the expression 4x + 5y + 3. Let’s name every part of it. Figure 2.1 below labels each piece so you can see exactly what a term, a coefficient, a variable and a constant are.
So in 4x + 5y + 3: the terms are 4x, 5y and 3; the variables are x and y; the coefficients are 4 and 5; and the constant is 3.
Here is a second example. A garden of length l metres and width w metres needs fencing and seeds. Working out each cost gives the expression 200l + 160w + 50lw for the total cost. Notice this one has a term, 50lw, where two variables are multiplied together.
Now look at a third example. A wire of length 20 cm is bent into a rectangle. If the length is x cm, the width must be (10 − x) cm, because length plus width is half the wire, which is 10. The area is x × (10 − x) = 10x − x². This expression has only one variable, x, but it has an x² term — a power of 2.
These three examples already show an important split. The first two expressions use two variables. The third uses just one variable. In this chapter we only study expressions with one variable.
What is a polynomial?
Expressions like 4x, x² + 1, 2y − 5, 5y³ + y² + 2y − 1, and 3z + 7 all use only one variable (x, y or z) and whole-number powers of it. Such expressions have a special name.
A one-variable polynomial (also called a univariate polynomial — “uni” means one) is an algebraic expression built from a single variable and its whole-number powers, joined by + and −.
The degree of a polynomial is the highest power of the variable in it. This one number gives the polynomial its family name. The classification tree in Figure 2.2 shows the four families you will meet.
So:
| Degree | Family name | Example |
|---|---|---|
| 0 | Constant | 8 |
| 1 | Linear | 3z + 7 |
| 2 | Quadratic | x² + 5x + 1 |
| 3 | Cubic | 5y³ + y² + 2y − 1 |
Why is a plain number like 8 a polynomial of degree 0? Because we can write it as 8x⁰, and anything to the power 0 equals 1, so 8x⁰ = 8 × 1 = 8. The power of x here is 0, so the degree is 0. NCERT states this quickly, so it is easy to miss why — but now you can see it: the hidden power is zero.
Let’s practise spotting the degree, the coefficient of a term, and the constant.
In the polynomial x⁴ − 3x³ + 6x² − 2x + 7, find (a) the degree, (b) the coefficient of x³, (c) the coefficient of x², and (d) the constant term.
- The degree is the highest power of x. The powers here are 4, 3, 2, 1 and 0 (the 7 is a constant, x⁰). The highest is 4, so the degree is 4.
- The coefficient of a term is the number multiplying that power. For x³ the term is −3x³, so its coefficient is −3. Keep the minus sign — it belongs to the coefficient.
- For x² the term is +6x², so the coefficient of x² is 6.
- The constant term is the plain number with no variable, which is 7. So: degree 4, coefficient of x³ is −3, coefficient of x² is 6, constant term 7.
What is the degree of the polynomial 4z − 3, and what kind of polynomial is it?
The highest power of z is 1 (in the term 4z). So the degree is 1, which makes it a linear polynomial.
Linear polynomials
A polynomial of degree 1 is a linear polynomial. Its general shape is ax + b, where a is not 0. From now on, this is our main character.
You meet linear polynomials all the time without noticing. The perimeter of a square of side x is 4x — a linear polynomial. The chess-club cost from the start of the chapter, 200 + 50m, is a linear polynomial in m. Let’s look at why it behaves so nicely.
The chess club charges ₹200 to join, then ₹50 per match. Here is the cost as the number of matches grows:
| Matches played | Amount paid (₹) |
|---|---|
| 1 | 250 |
| 2 | 300 |
| 3 | 350 |
| 4 | 400 |
| 5 | 450 |
| m | 200 + 50m |
Look at the right column: 250, 300, 350, 400, 450. Each value is ₹50 more than the one before. The jump is always the same. That steady, equal step is the signature of a linear polynomial. Every time the input goes up by 1, the output changes by a fixed amount.
A characteristic feature of linear polynomials: the difference between successive values (at equally-spaced inputs) is constant. For 200 + 50m, that constant difference is 50.
A linear equation comes from a linear polynomial
When we set a linear polynomial equal to a number, we get a linear equation — something we can solve. Let’s see this with a classic puzzle.
The sum of two numbers is 64. One number is 10 more than the other. Find the two numbers.
- Let the smaller number be x. The larger one is 10 more, so it is x + 10.
- Their sum is 64, so x + (x + 10) = 64.
- Simplify the left side: 2x + 10 = 64. Notice 2x + 10 is a linear polynomial; setting it equal to 64 makes a linear equation.
- Subtract 10 from both sides: 2x = 54. Divide by 2: x = 27.
- The smaller number is 27, and the larger is 27 + 10 = 37. The two numbers are 27 and 37. (Check: 27 + 37 = 64. Correct.)
The value of a polynomial — a number machine
A polynomial is like a little machine. You feed in a value of x, and it gives back one output. To find the output, you substitute the value of x and work it out.
Take the linear polynomial 2x + 3. If x = 4, the output is 2 × 4 + 3 = 11. If x = −6, the output is 2 × (−6) + 3 = −9. Figure 2.3 below shows this “input-output machine” idea, which is how mathematicians picture a polynomial.
Finding the value is just careful substitution. Watch the signs — that is where mistakes hide.
Find the value of the linear polynomial 5x − 3 when (i) x = 0, (ii) x = −1, (iii) x = 2.
- For x = 0: replace x with 0. We get 5 × 0 − 3 = 0 − 3 = −3.
- For x = −1: replace x with −1. We get 5 × (−1) − 3 = −5 − 3 = −8. (A negative times a positive is negative, so 5 × −1 = −5.)
- For x = 2: replace x with 2. We get 5 × 2 − 3 = 10 − 3 = 7. So the three values are −3, −8 and 7.
The same machine works for any polynomial, not just linear ones. For the quadratic 7s² − 4s + 6 at s = −3, you would get 7 × (−3)² − 4 × (−3) + 6 = 7 × 9 + 12 + 6 = 63 + 12 + 6 = 81. The only extra care is the power: (−3)² = 9, a positive number.
The zero of a linear polynomial
There is one input that is extra special: the input that makes the output 0. It is called the zero of the polynomial.
The zero of a polynomial p(x) is the value of x for which p(x) = 0.
For a linear polynomial ax + b, finding the zero is easy — set it to 0 and solve. Take 2x + 3:
2x + 3 = 0
2x = −3
x = −3/2 = −1.5
So the zero of 2x + 3 is x = −1.5. This number has a clear picture, which we will use later when we graph the line: it is exactly the spot where the line crosses the x-axis. Figure 2.4 shows it.
A linear polynomial ax + b (with a ≠ 0) always has exactly one zero, namely x = −b/a, because its graph is a straight line that crosses the x-axis at just one point.
What is the zero of the linear polynomial 5 − 4y?
Set it to 0: 5 − 4y = 0, so 4y = 5, giving y = 5/4 = 1.25. The zero is y = 1.25.
Linear patterns
Linear polynomials love to hide inside growing patterns. Look at the pattern of square tiles in Figure 2.5. Each new stage just adds two more tiles to the stage before it.
How many tiles at stage n? Look for the rule. At stage 2 there are 3 tiles, and 2 × 2 − 1 = 3. At stage 5 there are 9 tiles, and 2 × 5 − 1 = 9. The pattern is: tiles = 2n − 1.
This is a linear polynomial (degree 1). And the sequence it makes, 1, 3, 5, 7, 9, …, has a constant difference of 2 between terms. That constant difference is the fingerprint of a linear pattern.
A linear pattern is a sequence of numbers where the difference between two consecutive terms is constant.
Let’s use such a rule to model real life and answer a real question.
Bela has ₹100 pocket money. She spends ₹5 every day. After how many days will she be left with ₹40?
- Each day she loses ₹5. After n days she has spent 5n rupees. So the amount left is 100 − 5n. This is a linear polynomial in n.
- We want the amount left to be 40. So set up the equation: 100 − 5n = 40.
- Subtract 100 from both sides: −5n = 40 − 100 = −60.
- Divide both sides by −5: n = (−60) / (−5) = 12. After 12 days she is left with ₹40. (Check: 100 − 5 × 12 = 100 − 60 = 40. Correct.)
Sometimes the rule needs a little care because the pattern only “kicks in” after a few steps. An auto fare is a good example: it stays flat for the first 2 km, then climbs.
An auto fare is ₹25 for the first 2 km. After that it rises by ₹15 per km. What is the fare for a 10 km trip?
- The first 2 km cost ₹25 (flat). Only the kilometres after the 2nd one add ₹15 each.
- For a 10 km trip, the number of extra kilometres beyond 2 is 10 − 2 = 8.
- Those 8 extra km add 15 × 8 = ₹120. So the total fare is 25 + 120 = ₹145.
- In general, for n km (with n ≥ 2) the fare is 25 + 15 × (n − 2) = 15n − 5. Putting n = 10 gives 15 × 10 − 5 = 145. The 10 km fare is ₹145.
Linear growth and linear decay
Linear polynomials are perfect for modelling things that grow or shrink by a fixed step. There are two cases, and a single picture makes them clear. Figure 2.6 puts them side by side.
Linear growth is a pattern where a quantity increases by a constant amount over equal intervals.
Linear decay is a pattern where a quantity decreases by a constant amount over equal intervals.
Here is growth. A journey costs ₹100 to start plus ₹60 per km, so the cost function is C(d) = 100 + 60d, where d is the distance in km.
| Distance d (km) | Cost C (₹) |
|---|---|
| 0 | 100 |
| 1 | 160 |
| 2 | 220 |
| 3 | 280 |
| 4 | 340 |
| 5 | 400 |
Every extra km adds exactly ₹60. The cost climbs by the same step each time — this is linear growth.
Now decay. A water tank starts at 3 m and drops by 0.5 m each month, so the height is h(t) = 3 − 0.5t, where t is months.
| Month t | Height h (m) |
|---|---|
| 0 | 3 |
| 1 | 2.5 |
| 2 | 2 |
| 3 | 1.5 |
| 4 | 1 |
Every month the height falls by exactly 0.5 m. The same step down each time — this is linear decay. Notice the minus sign in 3 − 0.5t. That minus sign is what makes it decay instead of growth.
The expression 100 − 5n is not a linear polynomial because it has a minus sign and the number goes down.
A minus sign feels like the opposite of the neat plus-shaped form ax + b you first saw, so it is tempting to think a decreasing rule must be a different kind of object.
100 − 5n is exactly a linear polynomial. It is the same as −5n + 100, which fits ax + b with a = −5 and b = 100. The sign of a just decides whether the value grows (a > 0) or decays (a < 0); both are linear.
Linear relationships — finding y = ax + b
So far we knew the rule and used it. Sometimes we have to find the rule from a couple of facts. A linear relationship between two quantities x and y is written as:
y = ax + b
Here a and b are fixed numbers we may need to discover. The trick: if you know two pairs of (x, y) values, you can find a and b by building two equations and solving them.
A telecom plan has a fixed monthly fee plus a cost per GB. Using 10 GB gave a bill of ₹350. Using 20 GB gave a bill of ₹550. The bill follows y = ax + b, where x is GB used and y is the bill in rupees. Find a and b.
- Put the two facts into y = ax + b. When x = 10, y = 350, so 350 = 10a + b. When x = 20, y = 550, so 550 = 20a + b.
- From the first equation, write b on its own: b = 350 − 10a.
- Substitute this b into the second equation: 550 = 20a + (350 − 10a).
- Simplify the right side: 20a − 10a = 10a, so 550 = 10a + 350. Then 10a = 550 − 350 = 200, giving a = 20.
- Put a = 20 back into b = 350 − 10a: b = 350 − 200 = 150. So the relationship is y = 20x + 150. (Here 20 is the cost per GB and 150 is the fixed monthly fee.)
Visualising a linear relationship — drawing the line
Every linear relationship y = ax + b can be drawn as a straight line. Before we plot, let’s refresh how the coordinate plane works, since we lean on it heavily here.
To draw the line y = 2x + 1, pick any two x-values and find their y-values:
- When x = 0: y = 2 × 0 + 1 = 1. So A (0, 1) is on the line.
- When x = 3: y = 2 × 3 + 1 = 7. So B (3, 7) is on the line.
Plot A and B, join them with a ruler, and extend in both directions. Figure 2.7 shows the finished line.
A key fact to remember: a point lies on a line exactly when its coordinates satisfy the equation. To test whether (7, 15) is on y = 2x + 1, put x = 7: y = 2 × 7 + 1 = 15. It matches, so yes, (7, 15) is on the line.
The slope a — how steep the line is
Let’s see what the number a does. Below are three lines that all pass through the origin: y = ½x, y = x and y = 2x. Figure 2.8 draws them together.
So we learn two things from Figure 2.8:
- Every line of the form y = ax passes through the origin (0, 0), because putting x = 0 gives y = 0.
- The number a is the slope — it controls how steep the line is. Bigger a means a steeper climb.
The slope also connects back to patterns. For the tile pattern, the line is y = 2x − 1 and its slope is 2 — exactly the constant difference (2 tiles each stage) we found earlier. So the slope is the steady step of the pattern.
And the sign of the slope tells growth from decay: a positive slope means a rising line (linear growth), a negative slope means a falling line (linear decay). That matches Figure 2.6 exactly.
The y-intercept b — where the line cuts the y-axis
Now let’s see what b does. Keep the slope fixed at 2 and change b: draw y = 2x − 1, y = 2x + 1 and y = 2x + 5. Figure 2.9 shows all three.
Look closely: all three lines are parallel. They never cross. What makes the difference is where each cuts the y-axis. Putting x = 0 into y = 2x + b always gives y = b. So:
Any line y = ax + b cuts the y-axis at the point (0, b). This value b is called the y-intercept.
For y = 2x + 5 the y-intercept is 5 (it cuts the y-axis 5 units above the origin). For y = 2x − 1 the y-intercept is −1 (1 unit below the origin).
Putting it all together, here is what the two numbers in y = ax + b do:
| Number | Name | What it controls |
|---|---|---|
| a | Slope | How steep the line is; positive a rises (growth), negative a falls (decay) |
| b | y-intercept | Where the line cuts the y-axis, at the point (0, b) |
And two neat rules fall out of all this:
- Change a (keep b fixed): the line tilts to a new steepness, but still cuts the y-axis at the same point (0, b).
- Change b (keep a fixed): the line slides up or down but keeps the same direction. Lines with equal slopes but different y-intercepts are parallel.
Let’s use these ideas to recover a polynomial from its graph.
The graph of a linear polynomial p(x) = ax + b passes through the points (1, 5) and (3, 11). Find p(x).
- The graph is y = ax + b. Each point must satisfy it. From (1, 5): 5 = a × 1 + b, so 5 = a + b. From (3, 11): 11 = a × 3 + b, so 11 = 3a + b.
- Subtract the first equation from the second to remove b: (11 − 5) = (3a − a) + (b − b), which gives 6 = 2a.
- So a = 3. The slope is 3.
- Put a = 3 into 5 = a + b: 5 = 3 + b, so b = 2. Therefore p(x) = 3x + 2. (Check (3, 11): 3 × 3 + 2 = 11. Correct.)
Common Mistakes
Linear polynomials are simple, but a few slips happen again and again. Spotting them now will save you marks later.
In 5y³ + y² − 8, the coefficient of y³ is y³, or the degree is the number of terms (which is 3).
The words 'degree' and 'coefficient' sound technical and similar, and a cubic happens to have 3 terms here, so it is easy to mix counting terms with reading the highest power.
The degree is the highest power of the variable, which is 3 (from y³) — not the number of terms. The coefficient of y³ is the number in front of it, which is 5.
When finding the value of 5x − 3 at x = −1, you write 5 × −1 − 3 = 5 − 3 = 2.
It is tempting to ignore the minus sign on the input and just multiply 5 by 1, because handling a negative number feels like an extra, skippable step.
You must keep the sign: 5 × (−1) = −5, so 5 × (−1) − 3 = −5 − 3 = −8. Substitute the value with its sign, then do the arithmetic carefully.
The lines y = 2x + 1 and y = 3x + 1 are parallel because they both have + 1 at the end.
The matching '+ 1' jumps out at the eye, and it is natural to think the part that looks the same is what makes lines parallel.
Parallel lines must have the same slope (the number a in front of x), not the same y-intercept. Here the slopes are 2 and 3, which are different, so the lines are not parallel — they cross. The '+ 1' only means both cut the y-axis at the same point (0, 1).
The graph of y = 2x + 5 passes through the origin (0, 0).
Many of the first lines you draw, like y = 2x or y = x, do pass through the origin, so it is easy to assume every line does.
Only lines of the form y = ax (with no constant term, b = 0) pass through the origin. y = 2x + 5 has b = 5, so it cuts the y-axis at (0, 5), not at the origin.
Quick Check
Try these to test yourself. Click an option to see the answer and why.
What is the degree of the polynomial 4z − 3?
What is the value of the linear polynomial 5x − 3 when x = −1?
Which line is parallel to y = 3x − 4?
Practice Problems
Work each out yourself first, then reveal the full solution.
Easy
Find the degree of each polynomial: (i) 2x² − 5x + 3, (ii) y³ + 2y − 1, (iii) −9, (iv) 4z − 3.
The degree is the highest power of the variable. (i) Highest power is 2, so degree 2 (quadratic). (ii) Highest power is 3, so degree 3 (cubic). (iii) −9 is a constant, which can be written as −9x⁰, so degree 0 (constant). (iv) Highest power of z is 1, so degree 1 (linear).
Find the value of the quadratic polynomial 7s² − 4s + 6 when (i) s = 0, (ii) s = −3, (iii) s = 4.
Substitute carefully, and remember a power of a negative number is positive. (i) s = 0: 7 × 0 − 4 × 0 + 6 = 6. (ii) s = −3: 7 × (−3)² − 4 × (−3) + 6 = 7 × 9 + 12 + 6 = 63 + 12 + 6 = 81. (iii) s = 4: 7 × 4² − 4 × 4 + 6 = 7 × 16 − 16 + 6 = 112 − 16 + 6 = 102.
A student has ₹500 in a savings account and gets ₹150 every month. Write a linear expression for the amount she has at the end of the nth month, and find the amount after 4 months.
She starts with ₹500 and adds ₹150 each month. After n months she has added 150n rupees. So the amount is 500 + 150n. After 4 months: 500 + 150 × 4 = 500 + 600 = ₹1100. This is linear growth — the amount rises by a fixed ₹150 each month, so the slope is 150.
Medium
Ruby has 3 times as many two-rupee coins as five-rupee coins. The total value of her coins is ₹88. How many of each coin does she have?
Let the number of five-rupee coins be x. Then the number of two-rupee coins is 3x. Value of five-rupee coins: 5 × x = 5x rupees. Value of two-rupee coins: 2 × 3x = 6x rupees. Total value: 5x + 6x = 11x. Set this equal to 88: 11x = 88, so x = 8. So she has 8 five-rupee coins and 3 × 8 = 24 two-rupee coins. (Check: 8 × 5 + 24 × 2 = 40 + 48 = 88. Correct.)
The present age of Salil's mother is three times Salil's present age. After 5 years their ages add up to 70. Find their present ages.
Let Salil’s present age be x. His mother’s present age is 3x. After 5 years: Salil is x + 5, his mother is 3x + 5. Their sum then is 70: (x + 5) + (3x + 5) = 70. Simplify: 4x + 10 = 70, so 4x = 60, giving x = 15. So Salil is 15 years old and his mother is 3 × 15 = 45 years old. (Check after 5 years: 20 + 50 = 70. Correct.)
A telecom plan has a fixed monthly fee plus a charge per GB. Using 10 modules a bill was ₹400; using 14 modules it was ₹500. The bill is y = ax + b where x is modules used. Find a and b.
Put both facts into y = ax + b. When x = 10, y = 400: 400 = 10a + b. When x = 14, y = 500: 500 = 14a + b. Subtract the first from the second to remove b: 100 = 4a, so a = 25. Put a = 25 into 400 = 10a + b: 400 = 250 + b, so b = 150. So a = 25 and b = 150, giving y = 25x + 150. (Here ₹25 is the cost per module and ₹150 is the fixed fee.)
Challenge
The temperature relationship is °C = a × °F + b. Ice melts at 0 °C and 32 °F; water boils at 100 °C and 212 °F. Find a and b, and write the relationship.
Use the two facts in °C = a × °F + b. Ice: when °F = 32, °C = 0, so 0 = 32a + b. Boiling: when °F = 212, °C = 100, so 100 = 212a + b. Subtract the first from the second to remove b: 100 = (212 − 32)a = 180a. So a = 100/180 = 5/9. Put a = 5/9 into 0 = 32a + b: b = −32 × 5/9 = −160/9. So °C = (5/9) × °F − 160/9, which is the same as °C = (5/9)(°F − 32). (Check 212 °F: (5/9)(212 − 32) = (5/9)(180) = 100 °C. Correct.)
Look at a growing pattern of hexagons made with matchsticks. Stage 1 uses 6 matchsticks. Each new hexagon shares one side with the previous one, so every extra hexagon adds 5 matchsticks. Find a rule for the number of matchsticks at stage n, the number at stage 15, and check whether 200 matchsticks can form a stage.
Stage 1: 6 matchsticks (one full hexagon). Each later stage adds one hexagon that shares a side, so it adds only 5 new matchsticks (6 sides minus the 1 shared side). After stage 1, the number of extra hexagons added is (n − 1), each adding 5. So matchsticks at stage n = 6 + 5 × (n − 1) = 6 + 5n − 5 = 5n + 1. This is a linear polynomial. At stage 15: 5 × 15 + 1 = 75 + 1 = 76 matchsticks. Can 200 form a stage? Set 5n + 1 = 200, so 5n = 199, giving n = 39.8. Since n must be a whole number, 200 matchsticks cannot form a stage — the nearest stages are stage 39 (196) and stage 40 (201).
Two linear polynomials are p(x) = ax + b and q(x) = cx + d. Given p(0) = 5, the polynomial p(x) − q(x) cuts the x-axis at (3, 0), and p(x) + q(x) = 6x + 4 for all x. Find p(x) and q(x).
From p(0) = 5: putting x = 0 into ax + b gives b = 5. From p(x) + q(x) = 6x + 4: (a + c)x + (b + d) = 6x + 4. Matching parts, a + c = 6 and b + d = 4. Since b = 5, we get d = 4 − 5 = −1. p(x) − q(x) = (a − c)x + (b − d) = (a − c)x + (5 − (−1)) = (a − c)x + 6. This cuts the x-axis at (3, 0), so its value is 0 when x = 3: (a − c)(3) + 6 = 0, giving 3(a − c) = −6, so a − c = −2. Now solve a + c = 6 and a − c = −2. Adding: 2a = 4, so a = 2, then c = 4. So p(x) = 2x + 5 and q(x) = 4x − 1. (Check sum: (2x + 5) + (4x − 1) = 6x + 4. Correct.)
Summary
After this chapter, you can now explain:
- What an algebraic expression is, and how to name its terms, coefficients, variables and the constant (in 4x + 5y + 3, the terms are 4x, 5y, 3; the coefficients are 4 and 5; the constant is 3).
- What a one-variable polynomial is, and how its degree (the highest power of the variable) sorts it into constant (0), linear (1), quadratic (2) and cubic (3) families.
- Why a linear polynomial has the form ax + b, and that its signature is a constant difference between successive values.
- How to find the value of a polynomial by substituting a number for the variable (treat it like an input-output machine, and watch the signs).
- How to find the zero of a linear polynomial by setting ax + b = 0 and solving to get x = −b/a — the point where its line crosses the x-axis.
- How linear patterns, linear growth and linear decay all come from a fixed step, and how to model real situations (pocket money, fares, costs, tank height) with a linear polynomial.
- How to draw y = ax + b as a straight line using two points, and read off the slope a (steepness, and growth vs decay) and the y-intercept b (the point (0, b)), and why lines with equal slopes are parallel.
What’s Next
Next we widen our view from polynomials to the numbers themselves. In Chapter 3 — The World of Numbers you will explore the different kinds of numbers — natural numbers, integers, rationals and irrationals — and how they all fit together on the number line. The careful substitution and equation-solving you practised here will keep paying off as the numbers get richer.
Frequently Asked Questions
What makes a polynomial a linear polynomial?
A linear polynomial is a polynomial of degree 1. That means the highest power of the variable is 1, so it looks like ax + b where a is not 0. For example, 3z + 7 and 2x − 5 are linear polynomials. There is no x squared or any higher power in them.
How do you find the value of a polynomial like 5x minus 3 when x equals 2?
You substitute the value of x into the expression and work out the answer. For 5x − 3 with x = 2, you get 5 times 2 minus 3, which is 10 − 3 = 7. So the value of the polynomial at x = 2 is 7. Think of the polynomial as a machine: put a number in, get a number out.
What is the zero of a linear polynomial?
The zero of a linear polynomial is the value of x that makes the polynomial equal to 0. To find it, set the polynomial equal to zero and solve. For 2x + 3, solving 2x + 3 = 0 gives x = −3/2. On a graph, this is the x-value where the line crosses the x-axis.
What does the slope and y-intercept mean in y equals ax plus b?
In y = ax + b, the number a is the slope and tells you how steep the line is and whether it rises or falls. The number b is the y-intercept — the point (0, b) where the line cuts the y-axis. A bigger a makes a steeper line, and changing b just slides the line up or down.
Why are two lines like y equals 2x plus 1 and y equals 2x plus 5 parallel?
Both lines have the same slope, which is 2. The slope decides the direction and steepness of a line. When two lines have equal slopes but different y-intercepts, they go in exactly the same direction but cut the y-axis at different points, so they never meet. Lines that never meet are parallel.
What is the difference between linear growth and linear decay?
Linear growth is when a quantity increases by the same fixed amount over equal intervals, like a cost that goes up by ₹60 for every extra km. Linear decay is when a quantity decreases by the same fixed amount, like a phone's value dropping by ₹800 each year. Growth gives a line with positive slope; decay gives a line with negative slope.