Chapter 2: Introduction to Linear Polynomials - Ultimate 100/100 Concept & Practice Guide

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Topic 1: Algebraic Expressions & Real-Life Mathematical Modeling

An algebraic expression combines variables and constants using arithmetic operations ($+, -, \times, \div$).

Anatomy of Algebraic Expressions

Terms: Parts separated by $+$ or $-$ signs (e.g. In $4x + 5y + 3$, terms are $4x, 5y, 3$).
Variables: Symbols (letters like $x, y, z, m, t$) representing varying quantities.
Coefficients: The numbers multiplied by variables (e.g. In $4x$, $4$ is the coefficient of $x$).
Constant: A term containing no variable (e.g. $3$).

Real-Life Modeling Scenarios (NCERT Examples)

Scenario 1: Pens & Pencils Box Pricing (NCERT Example 1)
Raju buys $x$ red boxes (4 pens each) and $y$ blue boxes (5 pencils each), plus 3 free pens $\implies \mathbf{4x + 5y + 3}$.

Scenario 2: Garden Fencing & Seed Cost (NCERT Example 2)
Wire fence length $2l$ at ₹100/m, wooden fence width $2w$ at ₹80/m, seeds area $lw$ at ₹50/m$^2$ $\implies \mathbf{200l + 160w + 50lw}$.

Scenario 3: Bent Wire Area (NCERT Example 3)
Wire of length $20\text{ cm}$ bent into rectangle of length $x$ and width $(10 - x) \implies \text{Area } \mathbf{10x - x^2}$.

Immediate Solved Application

Example 1.1: Constructing Expressions & Identifying Components NCERT

Question: Form an algebraic expression for: "The perimeter of a rectangular garden whose length is $5\text{ metres}$ more than twice its width $w$". Identify terms and coefficients.

Solution:

Width $= w$. Length $= 2w + 5$.
Perimeter $= 2(\text{Length} + \text{Width}) = 2(2w + 5 + w) = 2(3w + 5) = \mathbf{6w + 10}$.

Terms: $6w$ and $10$. Variable: $w$. Coefficient of $w$: $6$. Constant: $10$.

Topic 1: Test Your Understanding

Write an expression for the total cost of $x$ notebooks at ₹40 each and $y$ pens at ₹15 each with a ₹20 delivery fee. [Ans: $40x + 15y + 20$]
Identify terms and coefficients in $7x^2 - 4xy + 9$. [Ans: Terms: $7x^2, -4xy, 9$; Coeff of $x^2$ is $7$, coeff of $xy$ is $-4$]

Topic 2: Univariate Polynomials, Exponent Rules, & Degree

Definition & Degree Rule

An expression in one variable where all exponents are non-negative integers ($\mathbb{W} = \{0, 1, 2, 3, \dots\}$) is a Univariate Polynomial.

$$\mathbf{p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \quad (a_n \neq 0)}$$

The Degree is the highest power of the variable occurring in the polynomial.

Is it a Polynomial? Exam Checklist

$\sqrt{x} + 5 = x^{1/2} + 5 \implies$ NOT a Polynomial (Exponent $1/2$ is not an integer).
$\frac{1}{x} + 3 = x^{-1} + 3 \implies$ NOT a Polynomial (Negative exponent $-1$).
$3y^2 + \sqrt{2}y - 7 \implies$ IS a Polynomial (Square root is on coefficient $\sqrt{2}$, variable exponents are $2, 1$).
$0 \implies$ Zero Polynomial (Degree is undefined!).

Degree ($n$)	Polynomial Type	Standard Form	Example	Zeroes Count
Undefined	Zero Polynomial	$p(x) = 0$	$0$	Infinitely many
Degree 0	Constant Polynomial	$p(x) = c \quad (c \neq 0)$	$8$, $-9$, $\sqrt{5}$	No zeroes
Degree 1	Linear Polynomial	$p(x) = ax + b \quad (a \neq 0)$	$3z + 7$, $5x - 3$	Exactly 1 zero
Degree 2	Quadratic Polynomial	$p(x) = ax^2 + bx + c$	$x^2 + 5x + 1$	At most 2 zeroes
Degree 3	Cubic Polynomial	$p(x) = ax^3 + bx^2 + cx + d$	$5y^3 + y^2 + 2y - 1$	At most 3 zeroes

Immediate Solved Application

Example 2.1: Classifying Polynomials & Finding Coefficients NCERT

Question: State degree and coefficient of $x^3$ for: (i) $x^4 - 3x^3 + 6x^2 - 2x + 7$ (ii) $4z - 3$.

Solution:

(i) Highest power is $4 \implies$ Degree = 4 (Biquadratic). Coefficient of $x^3 = \mathbf{-3}$.

(ii) Highest power is $1 \implies$ Degree = 1 (Linear). Coefficient of $x^3 = \mathbf{0}$ (term is absent).

Topic 2: Test Your Understanding

Which of these are polynomials? (a) $x^2 + \frac{3}{x}$, (b) $y^3 - 4y + \sqrt{7}$, (c) $z^{2/3} + 5$. [Ans: Only (b)]
What is the degree of the polynomial $p(x) = (x^2 + 1)(x^3 - 5)$? [Ans: Degree $5$]

Topic 3: Zeroes of a Linear Polynomial & Geometrical Meaning

Zero of linear polynomial $p(x) = ax + b \quad (a \neq 0)$: $$ax + b = 0 \implies \mathbf{x = -\frac{b}{a} = -\frac{\text{Constant Term}}{\text{Coefficient of } x}}$$

Geometrical Meaning

The zero of $ax + b$ is the $x$-coordinate of the point $\left(-\frac{b}{a}, 0\right)$ where the line $y = ax + b$ cuts the $X$-axis.

Immediate Solved Application

Example 3.1: Finding Zeroes & Polynomial Evaluation NCERT

Question: (i) Find the zero of $p(x) = 5x - 15$. (ii) Evaluate $p(x) = 5x - 3$ at $x = -1$.

Solution:

(i) Set $5x - 15 = 0 \implies 5x = 15 \implies \mathbf{x = 3}$. (Zero is $3$, cuts $X$-axis at $(3, 0)$).

(ii) Substitute $x = -1 \implies p(-1) = 5(-1) - 3 = -5 - 3 = \mathbf{-8}$.

Topic 3: Test Your Understanding

Find the zero of $p(x) = 7x + 21$. [Ans: $x = -3$]
If $x = 2$ is a zero of $p(x) = kx - 8$, find the value of $k$. [Ans: $2k - 8 = 0 \implies k = 4$]

Topic 4: Linear Functions & Input-Output Machine Model

Fig 2.1: Input-Output Process of a Linear Function $y = 2x + 3$

Immediate Solved Application

Example 4.1: Evaluating Function Machine NCERT

Question: For the function machine $y = 3x - 5$, find the output $y$ when input $x = -2$, and find input $x$ when output $y = 16$.

Solution:

1. Input $x = -2 \implies y = 3(-2) - 5 = -6 - 5 = \mathbf{-11}$.

2. Output $y = 16 \implies 3x - 5 = 16 \implies 3x = 21 \implies \mathbf{x = 7}$.

Topic 4: Test Your Understanding

A function machine is defined by $y = 40 + 15x$. Find $y$ when $x = 8$. [Ans: $y = 160$]

Topic 5: Linear Growth, Linear Decay, & Sequence Patterns

Growth vs Decay Patterns

Linear Growth: Quantity increases by a constant amount over equal intervals (Positive Slope $a > 0$).
Examples: Journey cost $C(d) = 100 + 60d$, Tile pattern $2n - 1$, Hexagon matchsticks $5n + 1$.
Linear Decay: Quantity decreases by a constant amount over equal intervals (Negative Slope $a < 0$).
Examples: Water tank height $h(t) = 3 - 0.5t$, Mobile value $v(t) = 10000 - 800t$, Pocket money $100 - 5n$.

Immediate Solved Application

Example 5.1: Modeling Linear Decay NCERT

Question: Bela starts with ₹100 pocket money and spends ₹5 every day. Find the linear expression for money left after $n$ days and find money left on Day 12.

Solution:

Expression: $\mathbf{A(n) = 100 - 5n}$ (linear decay with rate $-5$).

On Day 12: $A(12) = 100 - 5(12) = 100 - 60 = \text{₹}\mathbf{40}$.

Topic 5: Test Your Understanding

A bank account starts with ₹500 and gets ₹150 added every month. Write the linear growth expression for month $m$. [Ans: $500 + 150m$]
A phone bought for ₹10,000 depreciates by ₹800/year. Find its value after 3 years. [Ans: $10000 - 800(3) = \text{₹}7600$]

Topic 6: Graph Geometry of Linear Equations ($y = ax + b$)

Slope & Intercept Rules

Slope ($a$): Steepness/rate of change. $a > 1 \implies$ steeper than $y = x$; $0 < a < 1 \implies$ less steep.
Y-intercept ($b$): Point $(0, b)$ where line intersects $Y$-axis.
Parallel Lines Condition: Two lines are parallel if and only if they have equal slopes ($a_1 = a_2$) and different $y$-intercepts.
Lines Through Origin: Line $y = ax$ ($b = 0$) passes through $(0, 0)$.

Immediate Solved Application

Example 6.1: Telecom Data Line Relationship NCERT

Question: Find $y = ax + b$ if for $x = 10\text{ GB}$, bill $y = 350$ and for $x = 20\text{ GB}$, bill $y = 550$.

Solution:

Equations: $10a + b = 350$ and $20a + b = 550$.

Subtracting: $10a = 200 \implies \mathbf{a = 20}$ (Slope = ₹20/GB). $b = 350 - 10(20) = \mathbf{150}$ (Y-intercept = ₹150 fixed fee).

Formula: $\mathbf{y = 20x + 150}$.

Topic 6: Test Your Understanding

Are the lines $y = 4x + 7$ and $y = 4x - 3$ parallel? Explain. [Ans: Yes, both have slope $a = 4$]
Find slope and $y$-intercept of $2y = 6x - 10$. [Ans: Divide by 2 $\implies y = 3x - 5$. Slope $= 3$, Y-intercept $= -5$]

Topic 7: Case Studies & Exam HOTS Problems

Case Study 1: Telecom Internet Billing CASE STUDY

A telecom company charges $y = ax + b$, where $x$ is data in GB and $y$ is bill in Rupees. For $10\text{ GB}$, bill is ₹350; for $20\text{ GB}$, bill is ₹550.

Questions:
1. Derive $a$ and $b$. 2. Find bill for $35\text{ GB}$. 3. Find GB usable for ₹1150 budget.

Solution:

1. $y = 20x + 150$.

2. At $x = 35$: $y = 20(35) + 150 = 700 + 150 = \text{₹}\mathbf{850}$.

3. $20x + 150 = 1150 \implies 20x = 1000 \implies x = \mathbf{50\text{ GB}}$.

HOTS 1: Simultaneous Polynomial Systems HOTS / RD Sharma

Question: Let $p(x) = ax + b$ and $q(x) = cx + d$. If $p(0) = 5$, $p(x) - q(x)$ cuts $X$-axis at $(3, 0)$, and $p(x) + q(x) = 6x + 4$, find $p(x)$ and $q(x)$.

Solution:

1. $a + c = 6, b + d = 4$. $p(0) = 5 \implies b = \mathbf{5} \implies d = \mathbf{-1}$.

2. $p(x) - q(x) = (a - c)x + (5 - (-1)) = (a - c)x + 6$. Intersects at $(3, 0) \implies 3(a - c) + 6 = 0 \implies a - c = -2$.

3. Solving $a + c = 6$ and $a - c = -2 \implies \mathbf{a = 2}, \mathbf{c = 4}$.

Thus, $\mathbf{p(x) = 2x + 5}$ and $\mathbf{q(x) = 4x - 1}$.

8. 100/100 Quick Revision Checklist

Master Checklist

Polynomial Condition: Exponents MUST be non-negative integers ($\{0, 1, 2, \dots\}$).
Degree Rules: Constant = 0, Linear = 1, Quadratic = 2, Cubic = 3. Zero polynomial degree is undefined.
Zero of Linear Polynomial: $x = -\frac{b}{a}$, which is the $X$-intercept $\left(-\frac{b}{a}, 0\right)$.
Linear Growth vs Decay: Growth has slope $a > 0$; Decay has slope $a < 0$.
Parallel Lines: Equal slope $a$, different $y$-intercept $b$.