Three guided lessons with explanations, visual steps, practice and a final check.
An introductory companion to the revised Grade 9 syllabus: terms and degree, substitution and linear models. This is a syllabus-topic map, not a verified whole-textbook chapter or full exercise solution bank.
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LESSON 1
Read the structure of an expression
Identify terms, coefficients and the degree of a polynomial.
A variable stands for a number; a coefficient multiplies a variable.
Separate signed terms
In 4x² − 3x + 8, the terms are 4x², −3x and 8. Keep the minus sign with −3x.
Read coefficients
The coefficients of x² and x are 4 and −3. The constant term is 8.
Find the degree
The largest exponent is 2, so this is a quadratic polynomial. The x term has exponent 1.
Test what is allowed
Polynomial exponents are nonnegative integers. Thus 2/x = 2x⁻¹ is not a polynomial in x. A nonzero constant has degree 0; the zero polynomial has no degree assigned here.
Worked example: For 7 − 2t + 5t³, the coefficients are 7, −2 and 5. Rearranging gives 5t³ − 2t + 7. Its degree is 3, not the number of written terms.
Watch for this: A negative coefficient is allowed; a negative power of the variable is not.
Discuss: Write two different degree-2 polynomials and explain what makes their degrees equal.
LESSON 2
Evaluate and check a value
Substitute a value consistently and respect the order of operations.
Square before multiplying a squared term by its coefficient.
Write the rule
Let p(x) = 2x² − 5x + 2. To evaluate p(−1), replace every x with −1.
Use parentheses
p(−1) = 2(−1)² − 5(−1) + 2. Parentheses show that the whole negative number is squared.
Calculate in order
(−1)² = 1, so the result is 2 + 5 + 2 = 9.
Check a proposed zero
For x = 2, p(2) = 2×4 − 10 + 2 = 0. Therefore 2 is a zero. This does not mean p is zero for every input.
Worked example: For q(t) = 3t − 12, q(5) = 15 − 12 = 3. Solving 3t − 12 = 0 gives t = 4, and q(4) = 0 confirms it.
Watch for this: −1² means −(1²), but (−1)² means (−1)×(−1). Parentheses matter.
Discuss: Invent a linear polynomial with zero 6, then test another input.
LESSON 3
Build a linear model
Explain a constant rate and a starting value using a linear expression.
A table can show how an output changes when the input increases by one.
Describe the situation
A craft kit costs ₹20 to prepare plus ₹6 for each card. For n cards, the total is C(n)=20+6n.
Build a table
For n=0,1,2,3 the totals are 20,26,32,38 rupees. Each extra card adds exactly ₹6.
Interpret a graph
Points (0,20),(1,26),(2,32) lie on the line C=6n+20. Here actual orders use nonnegative whole-number n.
Test the model
For five cards C(5)=20+30=₹50. The rule assumes the preparation charge and per-card cost stay fixed; bulk discounts would require a different rule.
Worked example: A tank begins with 90 L and loses 5 L each minute at a steady rate. V(t)=90−5t. After 8 minutes it contains 50 L. This model applies only while the tank is draining, 0≤t≤18.
Watch for this: The starting fee is added once. It is not multiplied by the number of cards.
Discuss: Compare a ₹10 setup charge plus ₹8 per item with a ₹30 setup charge plus ₹4 per item. When do the totals match?
An introductory companion to the revised Grade 9 syllabus: terms and degree, substitution and linear models. This is a syllabus-topic map, not a verified whole-textbook chapter or full exercise solution bank.
Original HelioLearn teaching and questions. Source-aligned companion; not an official NCERT/CBSE publication or full textbook solution bank. Independent teacher review and narrated video are pending.
