The zeroes of \(x^2 - 2x - 15\) are
- (a)\(3, -5\)
- (b)\(-3, -5\)
- (c)\(3, 5\)
- (d)\(-3, 5\)
Polynomials Class 10: notes and important questions
Revision notes, 39 board-style questions with the step marks shown, and a four-step route from the basics to 95+, each step ending in a short checkpoint.
- 39 questions
- 12 multiple choice, 3 assertion–reason, 8 very short answer, 8 short answer, 5 long answer, 3 case study
- About 8 hours to master
Algebra unit: 20 of 80 theory marks (Polynomials, Pair of Linear Equations, Quadratic Equations, Arithmetic Progressions).
Revision notes
Polynomials — revision notes
1. Basics
- Degree = highest power of the variable. Linear (degree 1), quadratic (degree 2), cubic (degree 3).
- A zero of \(p(x)\) is a number \(k\) with \(p(k) = 0\). A polynomial of degree \(n\) has at most \(n\) zeroes.
2. Geometrical meaning of zeroes
- The zeroes of \(p(x)\) are the \(x\)-coordinates of the points where the graph of \(y = p(x)\) meets the \(x\)-axis.
- \(y = ax + b\) is a straight line: exactly one zero, \(-\dfrac{b}{a}\).
- \(y = ax^2 + bx + c\) is a parabola, opening upward if \(a \gt 0\) and downward if \(a \lt 0\). It can meet the \(x\)-axis at 2 points, touch it at 1 point, or miss it (0 zeroes).
3. Zeroes and coefficients (quadratic \(ax^2 + bx + c\))
\[\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}\]
Useful identities: \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\); \((\alpha - \beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta\); \(\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)\); \(\dfrac1\alpha + \dfrac1\beta = \dfrac{\alpha+\beta}{\alpha\beta}\).
4. Forming a quadratic polynomial
With sum \(S\) and product \(P\) of zeroes: \(k\left[x^2 - Sx + P\right]\), \(k \neq 0\). Multiply by a suitable \(k\) to clear fractions.
Worked example 1
Zeroes of \(3x^2 + 5x - 2\): split \(-6 = 6 \times (-1)\): \(3x^2 + 6x - x - 2 = (3x - 1)(x + 2)\). Zeroes \(\dfrac13, -2\). Check: sum \(-\dfrac53 = -\dfrac{b}{a}\), product \(-\dfrac23 = \dfrac{c}{a}\).
Worked example 2
Polynomial with zeroes \(3 \pm \sqrt2\): \(S = 6\), \(P = 9 - 2 = 7\), so \(x^2 - 6x + 7\).
Worked example 3
If \(\alpha, \beta\) are zeroes of \(x^2 - 4x + 1\), find \(\alpha^2 + \beta^2\): \(16 - 2 = 14\).
Common errors
- Sign slip: the sum of zeroes is \(-b/a\), not \(b/a\).
- Writing \(x^2 + Sx + P\) instead of \(x^2 - Sx + P\).
- Stating that a quadratic “has exactly 2 zeroes” — it may have 0, 1 or 2 real zeroes.
- When verifying the relationship, forgetting to compare with \(-b/a\) and \(c/a\) explicitly.
Board-exam tips
- “Find zeroes and verify” questions are 2–3 marks: factorise by splitting the middle term, then show both checks.
- For expressions in \(\alpha, \beta\), never solve for the zeroes unless asked — use the identities.
- Graph-reading MCQs: count intersections with the \(x\)-axis only (not the \(y\)-axis).
Topics in this chapter: Geometrical meaning of the zeroes · Zeroes of a polynomial · Relationship between zeroes and coefficients · Forming a quadratic polynomial.
Route to 95: four steps
Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.
Step 1
Secure the basics
You can find the zeroes of a quadratic by splitting the middle term and read the number of zeroes from a graph.
Step 2
Board standard
You can use sum = −b/a and product = c/a confidently, verify the relationship, and build a quadratic from given zeroes.
Step 3
Full marks on long answers
You can write full-marks long answers and case studies, including completing a table, plotting the graph and reading off the zeroes.
Step 4
95+ stretch (HOTS)
You can handle expressions like α² + β² or α/β + β/α without solving, and build new polynomials whose zeroes are changed versions of old ones.
Practice questions
Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 3 of the 39 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.
If \(2\) is a zero of \(3x^2 - 8x + k\), then the value of \(k\) is
- (a)\(4\)
- (b)\(-4\)
- (c)\(28\)
- (d)\(-28\)
Which of the following is a quadratic polynomial?
- (a)\(x^2 + \dfrac1x\)
- (b)\(\sqrt{x} + 4\)
- (c)\(3 - 2x + \sqrt2\,x^2\)
- (d)\(x^3 - x\)
Where marks are lost in Polynomials
- Sign slip in the sum of zeroes: writing b/a instead of −b/a. Fix: write 'α + β = −b/a = …' with the formula before substituting.
- Forming the polynomial as x² + Sx + P. Fix: memorise k[x² − (sum)x + (product)] and include 'k' when asked for 'a' quadratic polynomial.
- In 'find the zeroes and verify', stopping after the zeroes. Fix: show both checks, sum vs −b/a and product vs c/a, each on its own line; they carry separate marks.
- Solving for α and β when the question only needs α² + β² or 1/α + 1/β. Fix: use identities (α² + β² = (α + β)² − 2αβ), which is shorter and is what the scheme rewards.
- Graph MCQs: counting where the curve meets the y-axis, or counting a touching point twice. Fix: count only distinct points on the x-axis.
- Surd coefficients (e.g. 2√3x² − 5x + √3): splitting the middle term with the wrong product. Fix: multiply a × c first (here 6) and look for factors of that number.
Polynomials in our sample papers
Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).
- Sample paper 1 (Board standard): questions 2, 19, 22
- Sample paper 2 (Board standard): questions 1, 2
- Sample paper 3 (Above board): questions 2, 3, 22
- Sample paper 4 (Above board): questions 2, 3
- Sample paper 5 (95+ challenge): questions 2, 22