CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Exemplar Solutions Class 10 Maths Chapter 2 Polynomials
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Boards begin February 2027
Pre-boards are in December
Polynomials is in Algebra, which carries 20 of the 80 marks in the 2026-27 CBSE Maths paper
9 NCERT Exemplar questions from Chapter 2
Exercise 2.1
Exemplar Ex 2.1 Q11 mark
If one of the zeroes of the quadratic polynomial (k – 1)x² + kx + 1 is −3, then the value of k is
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Assertion-Reason1 mark
Assertion (A): If both zeroes of a quadratic polynomial ax² + bx + c are negative, then a, b and c all have the same sign.
Reason (R): For ax² + bx + c with zeroes α, β: α + β = −b/a and αβ = c/a.
Step 1α, β < 0 ⇒ α + β < 0 and αβ > 0. So −b/a < 0 (i.e., b/a > 0) and c/a > 0.
Step-by-step solution
α, β < 0 ⇒ α + β < 0 and αβ > 0. So −b/a < 0 (i.e., b/a > 0) and c/a > 0.
b/a > 0 ⇒ b and a have the same sign; c/a > 0 ⇒ c and a have the same sign. Hence all three same sign. A true.
Assertion (A): If the zeroes of ax² + bx + c (with c ≠ 0) are equal, then c and a have the same sign.
Reason (R): For real, equal zeroes the discriminant b² − 4ac = 0 and the common root is −b/(2a), whose square equals c/a.
Step 1Equal zeroes α = β ⇒ αβ = α² ≥ 0, and > 0 if α ≠ 0. αβ = c/a > 0 ⇒ c and a have same sign. A true.
Step-by-step solution
Equal zeroes α = β ⇒ αβ = α² ≥ 0, and > 0 if α ≠ 0. αβ = c/a > 0 ⇒ c and a have same sign. A true.
R is the standard discriminant relation — true.
R justifies why c/a is positive (= α²), which directly gives A.
A bridge’s arch is modelled by p(x) = x² + (k + 1)x + b. Surveyors find that the arch lands on the x-axis at x = 2 and x = −3.
(i) Find b using the product of zeroes.
(ii) Find k using the sum of zeroes.
(iii) Write the polynomial explicitly.
(iv) Verify that x = 2 is indeed a zero of your polynomial.
Step 1(i) Product of zeroes 2 × (−3) = −6 = b ⇒ b = −6.
Step-by-step solution
(i) Product of zeroes 2 × (−3) = −6 = b ⇒ b = −6.
(ii) Sum of zeroes 2 + (−3) = −1 = −(k + 1) ⇒ k + 1 = 1 ⇒ k = 0.
(iii) p(x) = x² + (0 + 1)x − 6 = x² + x − 6.
(iv) p(2) = 4 + 2 − 6 = 0 ✓.
Answer:(i) b = −6; (ii) k = 0; (iii) p(x) = x² + x − 6; (iv) verified.
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Questions from the NCERT Class 10 Mathematics Exemplar Problems (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.