Introduction to Linear Polynomials

You have been writing algebraic expressions like $4x + 5y + 3$ for a couple of years now — combinations of numbers and letter-numbers glued together with plus and minus signs. This chapter narrows the lens onto a single, enormously useful family living inside that larger world: polynomials in one variable, and within them, the simplest and most important kind of all, linear polynomials. We start by sorting expressions properly into terms, coefficients, variables and constants, and then use the highest power of the variable — the degree — to name four kinds of one-variable polynomial: constant (degree $0$ ), linear (degree $1$ ), quadratic (degree $2$ ), cubic (degree $3$ ). A chess-club membership fee, pocket money that shrinks by a fixed amount every day, an auto-rickshaw fare that climbs steadily after the first two kilometres, a growing pattern of square tiles — every one of these everyday situations turns out to be described by exactly the same kind of expression, $ax + b$ , a linear polynomial in disguise. The chapter's real insight is that a linear polynomial is not just something to simplify — it is a rule, an input-output machine, that turns any input value of the variable into a single output value. Watching how the output changes as the input climbs by one unit at a time reveals the defining feature of every linear pattern: the difference between consecutive outputs stays exactly constant. That constant difference earns a name, the slope, and when a linear relationship $y = ax + b$ is drawn on the coordinate plane, it always comes out as a straight line, with the constant $b$ , the y-intercept, telling you exactly where the line crosses the y-axis. By the end of this chapter you will be able to name the degree and type of any one-variable polynomial, build and recognise a linear pattern from a table of values, tell linear growth from linear decay by the sign of the slope, recover the whole equation $y = ax + b$ from just two data points, and read the slope and y-intercept straight off a graph — skills the rest of your algebra and coordinate geometry will lean on for years.

Key takeaways

An algebraic expression combines numbers, variables and operation symbols; its parts are called terms, the number multiplying a variable is its coefficient, and a lone number is the constant term.
A univariate polynomial (polynomial in one variable) is an algebraic expression in a single variable whose powers are non-negative whole numbers; its degree is the highest power of the variable that appears.
By degree, one-variable polynomials are named: degree $0$ is a constant polynomial, degree $1$ a linear polynomial, degree $2$ a quadratic polynomial, degree $3$ a cubic polynomial.
A linear polynomial has the general form $ax + b$ with $a \neq 0$ ; equating it to a fixed number gives a linear equation.
Any polynomial can be read as an input-output machine, or function: substitute a value for the variable and get one output value back.
A linear pattern is a sequence in which the difference between consecutive terms stays constant; that constant equals the slope of the pattern's graph.
Linear growth is a linear pattern whose constant difference is positive; linear decay is one whose constant difference is negative.
A linear relationship between two variables is written $y = ax + b$ ; its graph is always a straight line, where $a$ is the slope and $b$ is the y-intercept, the point $(0, b)$ where the line meets the y-axis.
Any two points on the line are enough to recover the whole equation $y = ax + b$ , by solving for $a$ and then $b$ .
Key idea: any one-variable polynomial of degree 1, ax + b, models a constant rate of change - between successive integer inputs the output always changes by the same amount a, so its graph is a straight line with slope a and y-intercept b.

The core insight

A linear polynomial $ax + b$ ( $a \neq 0$ ) models any situation where a quantity changes by the same constant amount $a$ for every unit change in the variable, so its input-output graph is a straight line with slope $a$ and y-intercept $b$ , fully determined by any two of its points.

What this chapter covers

The chapter builds in six connected steps. It opens by revisiting algebraic expressions and sorting their pieces into terms, variables, coefficients and constants, before restricting attention to expressions in a single variable — univariate polynomials — and naming them by degree: constant, linear, quadratic, cubic. It then focuses on the star of the chapter, linear polynomials $ax + b$ , and the linear equations you get by setting one equal to a fixed number. The next two sections show linear polynomials at work as input-output functions and as the engine behind linear patterns, including the special cases of linear growth and linear decay. The chapter closes by learning to find a full linear relationship $y = ax + b$ from just two known points, and then to visualise any such relationship as a straight line, reading off its slope and y-intercept directly from the equation or the graph.

The core ideas

Algebraic expressions: terms, coefficients and constants

Before naming polynomials, it helps to name the pieces of any algebraic expression. Consider the expression built from Raju's pens and pencils: $x$ red boxes of $4$ pens each and $y$ blue boxes of $5$ pencils each, plus $3$ extra free pens.

4x + 5y + 3

where: $4x$ , $5y$ and $3$ are the terms of the expression; $x$ and $y$ are the variables; $4$ and $5$ , the numbers multiplying the variables, are their coefficients; and $3$ , the term with no variable attached, is the constant term. Every algebraic expression you meet in this chapter can be broken down the same way.

Polynomials in one variable, and their degree

This chapter restricts attention to expressions built from a single variable and its non-negative whole-number powers — a univariate polynomial, or simply a polynomial. Written in full generality, a polynomial in $x$ looks like

a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0, \qquad a_n \neq 0,

where: $n$ is the degree of the polynomial — the highest power of $x$ that appears with a nonzero coefficient — and $a_n, a_{n-1}, \ldots, a_0$ are its coefficients, with $a_0$ the constant term. Degree gives every one-variable polynomial a name: $5y^3 + y^2 + 2y - 1$ has degree $3$ and is a cubic polynomial; $x^2 + 5x + 1$ has degree $2$ and is a quadratic polynomial; $3z + 7$ has degree $1$ and is a linear polynomial; and a bare number like $8$ can be written $8x^0$ , so it has degree $0$ and is a constant polynomial.

Linear polynomials and linear equations

A polynomial of degree exactly $1$ is called a linear polynomial, and it always has the same shape once fully simplified:

p(x) = ax + b, \qquad a \neq 0,

where: $a$ is the coefficient of $x$ and $b$ is the constant term; both can be any real number — positive, negative, a fraction, or, for $b$ alone, zero — but $a$ can never be $0$ , since that would drop the degree to $0$ . When a linear polynomial is set equal to a fixed number, the result is a linear equation: for instance, the sentence "the sum of two numbers is $64$ , one being $10$ more than the other" translates to the linear polynomial $2x + 10$ equated to $64$ , giving the linear equation $2x + 10 = 64$ .

Polynomials as input-output functions

A polynomial can be thought of as a rule that turns any chosen input value of the variable into exactly one output value — an input-output machine. For the linear polynomial $2x + 3$ , feeding in $x = 4$ produces the output $2 \times 4 + 3 = 11$ , and feeding in $x = -6$ produces the output $2 \times (-6) + 3 = -9$ . This process is also called a function, and $2x + 3$ is described as a linear function of $x$ — in contrast to an expression such as $10x - x^2$ , whose highest power is $2$ , which is a quadratic function.

Figure

An input output machine turning the value x equals four into eleven and the value x equals negative six into negative nine using the linear expression two x plus three

Linear patterns, growth and decay

A linear pattern is a sequence of numbers where the difference between any two consecutive terms is constant. The growing pattern of square tiles — Stage $1, 2, 3, \ldots$ giving tile counts $1, 3, 5, 7, 9, \ldots$ — has the rule

t_n = 2n - 1,

where: $n$ is the stage (or term) number and $t_n$ is the tile count at that stage; the constant difference between consecutive tile counts is $2$ . The same idea, in general, is written $t_n = an + b$ , where the constant $a$ tells you exactly how much the pattern changes with every step. When that constant is positive, the quantity keeps increasing over equal intervals — this is linear growth, as in a journey costing $C(d) = 100 + 60d$ rupees for $d$ kilometres travelled, which rises by a fixed $\text{₹}60$ per kilometre. When the constant is negative, the quantity keeps decreasing over equal intervals — this is linear decay, as in the height of water in a tank following $h(t) = 3 - 0.5t$ metres, which falls by a fixed $0.5\ \text{m}$ every month.

Figure

The tile pattern two n minus one growing by exactly two tiles at every stage from one to nine

Linear relationships, slope and the y-intercept

A linear relationship between two variables $x$ and $y$ is written

y = ax + b,

where: $a$ is the slope of the line — how steep it is, and whether it rises ( $a > 0$ ) or falls ( $a < 0$ ) as $x$ increases — and $b$ is the y-intercept, the y-coordinate where the line crosses the y-axis, that is, the point $(0, b)$ . Any straight line always cuts the y-axis at exactly one point, and that point's height above (or below) the origin is $b$ ; when $b = 0$ , the line $y = ax$ passes through the origin itself. Two lines with the same slope $a$ but different intercepts $b$ never meet — they are parallel.

Figure

The line y equals two x plus one showing rise over run between the points zero one and three seven giving a slope of two

Derivation AFinding a and b for a linear relationship from two points

Section 2.5 of the chapter shows that when a monthly bill, or any other linear relationship, follows the rule $y = ax + b$ , knowing just two matching pairs of $x$ and $y$ values is always enough to pin down the exact values of the slope $a$ and the intercept $b$ , provided the two $x$ -values are different. This derivation carries out that recovery in full generality, rather than for one specific pair of numbers, so that the same steps can be reused for any two-point problem, including the end-of-chapter exercises that ask you to find a polynomial from its graph.

Step 1. Suppose the linear relationship $y = ax + b$ passes through two known points $(x_1, y_1)$ and $(x_2, y_2)$ , with $x_1 \neq x_2$ . Because both points lie on the same line, both pairs of coordinates must satisfy the equation, giving the two equations $y_1 = a x_1 + b$ and $y_2 = a x_2 + b$ .

Step 2. Subtract the first equation from the second to eliminate $b$ entirely, since $b$ appears identically in both: $y_2 - y_1 = (ax_2 + b) - (ax_1 + b) = a(x_2 - x_1)$ . Since $x_1 \neq x_2$ , divide both sides by $(x_2 - x_1)$ to isolate the slope: $a = \dfrac{y_2 - y_1}{x_2 - x_1}$ .

Step 3. With $a$ now known, substitute it back into either original equation to solve for $b$ . Using the first equation, $b = y_1 - a x_1$ . Using the second equation instead gives exactly the same value of $b$ , since both points lie on the same line — a useful check on your arithmetic.

Step 4. Apply the formula to NCERT's Example 11, where the two known points are $(10, 350)$ and $(20, 550)$ . Then $a = \dfrac{550 - 350}{20 - 10} = \dfrac{200}{10} = 20$ , and $b = 350 - 20(10) = 150$ , giving $y = 20x + 150$ — exactly the answer the textbook reaches by direct substitution. The general method and the substitution method always agree, because both are solving the very same pair of equations.

Go deeper · Derivation B · 1 minWhy the constant difference of a linear pattern equals the slope of its graph

Derivation BWhy the constant difference of a linear pattern equals the slope of its graph

Sections 2.3, 2.4 and 2.6 each describe the same number wearing three different costumes: the constant amount by which consecutive terms of a linear pattern differ, the fixed amount by which a growing or decaying quantity changes each interval, and the slope $a$ of the line $y = ax + b$ . This derivation proves once and for all that these three descriptions name exactly the same quantity, so that reading a pattern's table of values and reading its graph will always agree with each other.

Step 1. Let a linear pattern be generated by the rule $t_n = an + b$ , where $n = 1, 2, 3, \ldots$ names the term number — the tile-pattern stage, the day number, the month number — and $t_n$ is the pattern's value at that term, exactly the setup used for the tile pattern $2n - 1$ and for a linear-decay pattern like $100 - 5n$ .

Step 2. Write down the very next term by replacing $n$ with $n + 1$ : $t_{n+1} = a(n+1) + b = an + a + b$ , obtained simply by distributing $a$ across the bracket.

Step 3. Subtract the two consecutive terms to find how much the pattern changes in one step: $t_{n+1} - t_n = (an + a + b) - (an + b) = a$ . Every $an$ and every $b$ cancels exactly, leaving only $a$ — and crucially, this difference does not depend on $n$ at all, so it is the same number no matter which consecutive pair of terms you pick. This is precisely the constant-difference property that defines a linear pattern.

Step 4. Connect this result to the graph. The points $(n, t_n)$ and $(n+1, t_{n+1})$ both lie on the line $y = ax + b$ , and they differ by exactly $1$ unit in the $x$ -direction (run $= 1$ ) and by $a$ units in the $y$ -direction (rise $= t_{n+1} - t_n = a$ , from Step 3). The slope of the line, rise divided by run, is therefore $a / 1 = a$ — the identical number found simply by subtracting consecutive pattern values. This is also why $a > 0$ produces linear growth (a rising line, an increasing pattern) and $a < 0$ produces linear decay (a falling line, a decreasing pattern).

Worked examples

Board level - forming and solving a linear equation (Example 6 pattern)

The sum of two numbers is $64$ . One of the numbers is $10$ more than the other. Find the two numbers.

Step 1. Name the unknown. Let the smaller number be $x$ ; since the other number is $10$ more, the larger number is $x + 10$ .

Step 2. Translate "the sum of two numbers is $64$ " into a linear equation: $x + (x + 10) = 64$ .

Step 3. Simplify the linear polynomial on the left: $2x + 10 = 64$ . This is now a linear equation in the standard form $ax + b = c$ , with $a = 2$ , $b = 10$ , $c = 64$ .

Step 4. Solve for $x$ by transposing the constant: $2x = 64 - 10 = 54$ , so $x = 27$ . Answer: the two numbers are 27 and 37.

Board level - modelling and evaluating a linear function (Example 8 pattern)

An auto-rickshaw fare starts at ₹25 and stays the same for the first $2\ \text{km}$ ; after that it rises by ₹15 for every extra kilometre. Find the fare for a $10\ \text{km}$ ride, write the linear expression for the fare on an $n\ \text{km}$ ride ( $n \geq 2$ ), and find how many kilometres correspond to a fare of ₹130.

Step 1. For the first $2\ \text{km}$ the fare is a flat ₹25. Beyond that, every extra kilometre adds ₹15, so for $n$ kilometres travelled ( $n \geq 2$ ) the number of "extra" kilometres is $(n - 2)$ , and the total fare is $25 + 15(n - 2)$ .

Step 2. Simplify this into the standard linear-polynomial form: $25 + 15(n-2) = 25 + 15n - 30 = 15n - 5$ . So the fare function is $F(n) = 15n - 5$ rupees, a linear polynomial in $n$ with slope $15$ and constant term $-5$ .

Step 3. Evaluate at $n = 10$ : $F(10) = 15(10) - 5 = 150 - 5 = 145$ , matching the direct calculation $25 + 15 \times 8 = 145$ .

Step 4. To find $n$ for a fare of ₹130, set $F(n) = 130$ and solve the linear equation: $15n - 5 = 130$ , so $15n = 135$ , giving $n = 9$ . Answer: the fare for 10 km is ₹145, and a fare of ₹130 corresponds to a 9 km ride.

Higher-order thinking (HOTS / olympiad-style) - two linear polynomials tied together by three clues (End-of-Chapter pattern)

Let $p(x) = ax + b$ and $q(x) = cx + d$ be two linear polynomials such that $p(0) = 5$ ; the polynomial $p(x) - q(x)$ cuts the x-axis at $(3, 0)$ ; and the sum $p(x) + q(x)$ equals $6x + 4$ for every real $x$ . Find the polynomials $p(x)$ and $q(x)$ .

Step 1. Translate each clue into algebra one at a time. Since $p(x) = ax + b$ , the condition $p(0) = 5$ means $a(0) + b = 5$ , so $b = 5$ directly.

Step 2. The sum condition says $(ax + b) + (cx + d) = 6x + 4$ for every $x$ , so collecting like terms, $(a+c)x + (b+d) = 6x + 4$ . Since this holds for every $x$ , the coefficients of $x$ must match and the constant terms must match: $a + c = 6$ and $b + d = 4$ . Since $b = 5$ from Step 1, $d = 4 - 5 = -1$ .

Step 3. The polynomial $p(x) - q(x) = (ax+b) - (cx+d) = (a-c)x + (b-d)$ cuts the x-axis at $(3, 0)$ , which means it equals $0$ when $x = 3$ : $(a-c)(3) + (b-d) = 0$ . Using $b = 5$ and $d = -1$ , $b - d = 5 - (-1) = 6$ , so $3(a-c) + 6 = 0$ , giving $a - c = -2$ .

Step 4. Solve the pair of linear equations $a + c = 6$ and $a - c = -2$ simultaneously by adding them: $2a = 4$ , so $a = 2$ , and then $c = 6 - 2 = 4$ . Answer: p(x) = 2x + 5 and q(x) = 4x - 1. Timing note: budget about three to four minutes. The trap is trying to find $a, b, c, d$ all at once; instead peel off one unknown per clue in the order the information is easiest to use — $p(0) = 5$ gives $b$ immediately, matching coefficients in the sum gives two clean equations, and only then does the x-intercept condition need substituting. Always finish an olympiad-style polynomial problem like this with a quick check: verify $p(0) = 5$ , $p(3) - q(3) = (6+5) - (12-1) = 11 - 11 = 0$ , and $p(x) + q(x) = (2x+5) + (4x-1) = 6x + 4$ — all three clues confirmed.

Common traps

The most common slip is confusing an algebraic expression with a polynomial: an expression involving a variable in a denominator, like $\dfrac{1}{x}$ , or under a root, is not a polynomial at all, even though it looks similar. Students also frequently misread degree — the degree is the highest power present, not the number of terms, and a bare constant like $8$ has degree $0$ , not "no degree." Many assume the slope $a$ in $y = ax + b$ must be positive, forgetting that a negative slope is a perfectly valid straight line describing linear decay, not a mistake. Some treat $b$ as the point where the graph "starts," when a line actually extends infinitely in both directions and $b$ is only where it crosses the y-axis. Finally, a sequence that keeps growing is not automatically a linear pattern — the defining test is that the difference between every pair of consecutive terms is the same constant, not merely that the numbers keep increasing.

Common trapGrowing is not the same as linear - always check the difference is constant, not just increasing

The single biggest misconception in this chapter is assuming that any sequence or graph that looks roughly straight, or any quantity that keeps growing, must be linear. It is easy to be fooled: the area of a rectangle formed from a fixed-length wire, $10x - x^2$ , also increases for small $x$ , but it is a quadratic function, and its consecutive differences are not constant — they shrink as $x$ grows, and the graph is a curve, not a straight line. The only reliable test for a linear pattern is to compute the difference between several consecutive pairs of terms and check that every single one gives the same number. If even one pair gives a different difference, the pattern is not linear, no matter how straight it might look over a short stretch of values.

You should be able to…

Identify the terms, coefficients, variables and constant term of an algebraic expression.
State the degree of a one-variable polynomial and name it correctly as constant, linear, quadratic or cubic.
Recognise the general form of a linear polynomial, $ax + b$ with $a \neq 0$ , and distinguish it from a linear equation.
Evaluate any polynomial at a given value of the variable, treating it as an input-output function.
Build a table of values for a linear pattern and identify its constant difference.
Distinguish linear growth from linear decay by the sign of the constant difference, or slope.
Find the equation $y = ax + b$ of a linear relationship given any two of its points.
Plot a linear relationship on the coordinate plane using just two points.
Read the slope and y-intercept directly from the equation of a line, and interpret what each one means.
Recognise that lines with the same slope but different y-intercepts are parallel.

Ask the tutor: Want to practice finding a linear relationship y = ax + b from just two data points, step by step together?

What's the difference between an algebraic expression and a polynomial?

Every polynomial is an algebraic expression, but not every algebraic expression is a polynomial. An expression can involve a variable in a denominator, like $\dfrac{1}{x}$ , under a root, or with a negative power, and none of that is allowed in a polynomial. A polynomial in one variable only combines non-negative whole-number powers of that variable added or subtracted together with fixed coefficients. So $4x + 5y + 3$ is an algebraic expression, and $3z + 7$ is also a polynomial (specifically linear), but an expression like $\dfrac{5}{x} + 2$ is not a polynomial at all.

Why is a constant like 8 called a polynomial of degree 0?

A plain number such as $8$ can be written as $8x^0$ , since any nonzero number raised to the power $0$ equals $1$ , so $8x^0 = 8 \times 1 = 8$ . Written this way, $8$ is technically " $8$ times $x$ to the power zero" — a polynomial with only a constant term and no term where $x$ actually appears with a positive power. Because the highest, and only, power of $x$ present is $0$ , its degree is defined to be $0$ , not "no degree" or "undefined." This is why $8$ , $-5$ and $0.7$ are all constant polynomials.

Is a linear equation the same thing as a linear polynomial?

No, though the two are closely related. A linear polynomial, such as $2x + 3$ , is just an expression — it has no equality sign and no fixed value; you evaluate it for different inputs and get different outputs, like the input-output machine in this chapter. A linear equation is what you get when you set a linear polynomial equal to a specific number, such as $2x + 3 = 11$ . The equation has a definite solution (here, $x = 4$ ), while the polynomial itself does not — it takes a whole range of values depending on what you substitute for $x$ . Equations are solved; polynomials are evaluated or graphed.

How can slope be negative - doesn't growing by a constant amount always mean it goes up?

Slope simply measures the constant amount added to the output for each unit increase in the input, and that amount is allowed to be negative. When the slope $a$ is positive, each step adds to the value and the line rises left to right — linear growth, like a chess-club fee that climbs by ₹50 per match. When $a$ is negative, each step subtracts a fixed amount and the line falls left to right — linear decay, like a water tank whose height drops by $0.5\ \text{m}$ every month. Both are perfectly good linear relationships; only the direction of change differs, controlled entirely by the sign of $a$ .

Introduction to Linear Polynomials

What this chapter covers

The core ideas

Algebraic expressions: terms, coefficients and constants

Polynomials in one variable, and their degree

Linear polynomials and linear equations

Polynomials as input-output functions

Linear patterns, growth and decay

Linear relationships, slope and the y-intercept

Worked examples

Board level - forming and solving a linear equation (Example 6 pattern)

Board level - modelling and evaluating a linear function (Example 8 pattern)

Higher-order thinking (HOTS / olympiad-style) - two linear polynomials tied together by three clues (End-of-Chapter pattern)

Common traps

You should be able to…

Before this

Where this leads