1. Introduction to Polynomials
A polynomial is an algebraic expression built from a variable, non-negative whole-number powers of that variable, and numbers. The general form of a polynomial in one variable \(x\) is:
\[p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0, \qquad a_n \neq 0\]
Every part of this expression has a name, and using the names precisely is the first step to accuracy in the exam.
Variable
The letter (usually \(x\)) on which the expression depends.
Term
Each separate part of the expression, e.g. in \(2x^2 + 5x - 3\) the terms are \(2x^2\), \(5x\), \(-3\).
Coefficient
The number multiplying a power of the variable, e.g. the coefficient of \(x^2\) in \(2x^2 + 5x - 3\) is \(2\), and the coefficient of \(x\) is \(5\).
Constant term
The term with no variable. In \(2x^2 + 5x - 3\), the constant term is \(-3\).
Degree
The highest power of the variable that appears. It is always a whole number. In \(2x^2 + 5x - 3\) the degree is \(2\).
Standard form
Terms written from the highest power down to the constant, e.g. \(3x^2 + 2x - 5\), not \(2x - 5 + 3x^2\).
\[p(x) = 2x^2 + 5x - 3 \qquad \text{(degree 2)}\]
\[p(x) = 3x - 7 \qquad \text{(degree 1)}\]
\[p(x) = 8 \qquad \text{(degree 0)}\]
The exponents of the variable are whole numbers (2, 1, and 0) in every case, so all three expressions are polynomials.
Expressions such as \(2x + \dfrac{3}{x}\), \(\sqrt{x}\), \(x^2 + \dfrac{1}{x+1}\) are not polynomials, because the exponent of the variable is negative (\(x^{-1}\)), a fraction \(\left(x^{\frac{1}{2}}\right)\), or the variable sits in a denominator. Powers of a variable in a polynomial must be non-negative whole numbers.
2. Types of Polynomials
Polynomials are classified by their degree. For Class 10 you need three types, all of which already appeared above.
| Name | Degree | General form | Example |
|---|---|---|---|
| Constant polynomial | 0 | \(a_0\) | \(8,\; -5,\; 0\) |
| Linear polynomial | 1 | \(ax + b,\; a \neq 0\) | \(3x - 7\) |
| Quadratic polynomial | 2 | \(ax^2 + bx + c,\; a \neq 0\) | \(2x^2 + 5x - 3\) |
In the constant polynomial \(p(x) = 8\), the power of \(x\) hidden inside the constant is \(x^0 = 1\), so the degree is \(0\). A non-zero constant has degree \(0\); the zero polynomial \(p(x) = 0\) is given no degree.
In a quadratic polynomial \(ax^2 + bx + c\), remember \(a \neq 0\). If \(a = 0\), the expression would become linear, not quadratic.
3. Zeroes of a Polynomial
A real number \(k\) is called a zero of a polynomial \(p(x)\) if
\[p(k) = 0\]
In words: substituting \(k\) in place of \(x\) makes the whole expression equal to zero.
What the definition really means
- The zero is a number, not a point on the graph and not a factor. A zero \(k\) satisfies \(p(k) = 0\).
- \(\dfrac{(x-2)(x-3)}{x-2} = x - 3\) is a good example of why context matters: after cancelling the factor \((x-2)\) the expression is only a polynomial when we remember it is not defined at \(x = 2\). In this chapter we work with expressions that are genuinely polynomials, so no such exceptions appear.
- Verifying a proposed zero is always possible: just substitute and check whether the value becomes \(0\).
How to verify a zero
For \(p(x) = x^2 - 5x + 6\), check \(x = 2\) and \(x = 4\).
\(p(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0\), so \(2\) is a zero.
\(p(4) = (4)^2 - 5(4) + 6 = 16 - 20 + 6 = 2 \neq 0\), so \(4\) is not a zero.
Zero of a polynomial ≠the value of the polynomial. Writing "the zero of \(p(x)\) is \(p(0)\)" is wrong. The zero is the input \(k\) that makes the output zero. So the zero of \(p(x) = 2x + 3\) is \(x = -\dfrac{3}{2}\), not the number \(p(0) = 3\).
4. Geometrical Interpretation of Zeroes
Every polynomial \(p(x)\) has a graph: the set of points \((x, y)\) where \(y = p(x)\). A real zero \(k\) means the point \((k, 0)\) lies on the graph — that is, the graph intersects or touches the x-axis at \(x = k\). Counting the intersections of the graph with the x-axis is the same as counting the real zeroes.
The linear polynomial
The graph of a linear polynomial \(ax + b\) is a straight line. A non-horizontal straight line crosses the x-axis at exactly one point, so a linear polynomial has exactly one real zero.
The quadratic polynomial
The graph of a quadratic polynomial \(ax^2 + bx + c\) is a smooth curve called a parabola. Because a parabola bends, three cases are possible.
Two distinct real zeroes
The graph crosses the x-axis at two different points.
Two equal real zeroes
The graph just touches the x-axis at one point; the two zeroes coincide.
No real zeroes
The graph never touches the x-axis at all.
Number of real zeroes = number of times the graph meets the x-axis. A quadratic polynomial has at most two real zeroes. It can have two distinct, two equal, or no real zeroes — never more than two.
Do not assume every quadratic has two different real zeroes. Some parabolas merely touch the axis (equal zeroes) and some never reach it (no real zeroes). Always check the actual zeroes by factorising or substituting.
5. Relationship Between Zeroes and Coefficients
For a quadratic polynomial
\[p(x) = ax^2 + bx + c, \qquad a \neq 0\]
whose zeroes are \(\alpha\) and \(\beta\), the two most important results of this chapter are:
\[\alpha + \beta = -\frac{b}{a}\qquad \text{and}\qquad \alpha\beta = \frac{c}{a}\]
Write the polynomial in standard form \(ax^2 + bx + c\) first. Take \(a\) as the coefficient of \(x^2\) (never forget it in the denominator of \(-\dfrac{b}{a}\)), \(b\) as the coefficient of \(x\) including its sign, and \(c\) as the constant term including its sign. For \(2x^2 - 5x + 3\): \(a = 2\), \(b = -5\), \(c = 3\).
Verifying the relationship when zeroes are known
Verify the relationship for \(p(x) = x^2 - 5x + 6\).
Find the zeroes. \(x^2 - 5x + 6 = (x - 2)(x - 3)\), so \(\alpha = 2\), \(\beta = 3\).
Check the sum. \(\alpha + \beta = 2 + 3 = 5\). Formula: \(-\dfrac{b}{a} = -\dfrac{(-5)}{1} = 5\). ✔
Check the product. \(\alpha\beta = 2 \times 3 = 6\). Formula: \(\dfrac{c}{a} = \dfrac{6}{1} = 6\). ✔
Finding the sum and product without finding the zeroes
For \(p(x) = 2x^2 - 4x + 1\), \(a = 2\), \(b = -4\), \(c = 1\).
Sum of zeroes \(= -\dfrac{b}{a} = -\dfrac{(-4)}{2} = 2\).
Product of zeroes \(= \dfrac{c}{a} = \dfrac{1}{2}\).
Determining an unknown coefficient
If one zero of \(x^2 - 5x + k\) is \(2\), find \(k\).
Since \(2\) is a zero, \(p(2) = 0\): \(4 - 10 + k = 0\), so \(k = 6\). Check: \(x^2 - 5x + 6 = (x - 2)(x - 3)\), and the other zero is \(3\).
Checking whether given numbers can be zeroes
Can \(-2\) and \(-3\) be the zeroes of \(x^2 - 5x + 6\)?
If they were, the sum would be \(-5\), but \(-\dfrac{b}{a} = 5\). The product would be \(6\), which agrees with \(\dfrac{c}{a} = 6\) — but the product alone is not enough. The sum fails, so \(-2\) and \(-3\) cannot be the zeroes. The actual zeroes are \(2\) and \(3\).
6. Finding the Zeroes of a Quadratic Polynomial
The standard Class 10 method is factorisation: express the quadratic as a product of two linear factors using the split-the-middle-term technique, then set each factor to zero.
For \(ax^2 + bx + c\), find two numbers whose product is \(ac\) and whose sum is \(b\). Split \(b\) into the sum of those two numbers, group the four terms, and factor each group.
Step 1. \(a = 1\), \(b = -5\), \(c = 6\). Product \(ac = 6\); we need two numbers with product \(6\) and sum \(-5\). Those numbers are \(-2\) and \(-3\).
Step 2. Split the middle term: \(x^2 - 2x - 3x + 6\).
Step 3. Factor by grouping: \(x(x - 2) - 3(x - 2) = (x - 2)(x - 3)\).
Step 4. Set each factor to zero: \(x - 2 = 0\) gives \(x = 2\); \(x - 3 = 0\) gives \(x = 3\).
\[x = 2, \ 3\]
Verify. Sum \(= 5 = -\dfrac{b}{a}\), product \(= 6 = \dfrac{c}{a}\). ✔
More patterns to practise
- Negative zeroes: \(x^2 + 5x + 6 = (x + 2)(x + 3)\) gives zeroes \(-2\) and \(-3\).
- A leading constant: \(2x^2 - 6x = 2x(x - 3)\) gives zeroes \(0\) and \(3\).
- Different coefficients: \(3x^2 - x - 4 = (3x - 4)(x + 1)\) gives zeroes \(\dfrac{4}{3}\) and \(-1\).
- A square: \(4x^2 - 4x + 1 = (2x - 1)^2\) gives the two equal real zeroes \(\dfrac{1}{2}, \dfrac{1}{2}\).
When splitting the middle term, the two numbers must multiply to give \(a \times c\), not just \(c\). For \(6x^2 - x - 2\), \(a = 6\) so the product needed is \(6 \times (-2) = -12\); the numbers whose product is \(-12\) and sum is \(-1\) are \(-4\) and \(3\): \(6x^2 - 4x + 3x - 2 = (3x - 2)(2x + 1)\).
Always verify your zeroes using the relationships \(\alpha + \beta = -\dfrac{b}{a}\) and \(\alpha\beta = \dfrac{c}{a}\). It catches most sign and arithmetic slips.
7. Forming a Quadratic Polynomial from its Zeroes
If \(\alpha\) and \(\beta\) are the zeroes, then the factors are \((x - \alpha)\) and \((x - \beta)\), so the polynomial is:
\[p(x) = x^2 - (\alpha + \beta)x + \alpha\beta\]
This is a quadratic with leading coefficient \(1\). Multiply the whole expression by any non-zero constant to get another quadratic with the same zeroes. When the question fixes a certain leading coefficient, include that multiplier.
\(\alpha + \beta = 5\), \(\alpha\beta = 6\). So \(p(x) = x^2 - 5x + 6\).
\(\alpha + \beta = -2 + 5 = 3\), \(\alpha\beta = (-2)(5) = -10\). So \(p(x) = x^2 - 3x - 10\). Check: \((x + 2)(x - 5) = x^2 - 3x - 10\). ✔
Zeroes \(1\) and \(\dfrac{2}{3}\): sum \(= \dfrac{5}{3}\), product \(= \dfrac{2}{3}\).
\(p(x) = x^2 - \dfrac{5}{3}x + \dfrac{2}{3}\). Multiply by 3: \(3x^2 - 5x + 2\). Check: \((x - 1)(3x - 2) = 3x^2 - 5x + 2\). ✔
When only the sum and product are given
Direct substitution gives \(p(x) = x^2 - 5x + 6\). No extra calculation is needed — the formula is exactly built for this.
Double-check the sign of the \(x\)-coefficient: the formula uses \(x^2 - (\text{sum})x + \text{product}\). If the sum is negative, then \(-(\text{sum})\) becomes positive.
8. Application-Based Learning
In exam questions, a quadratic polynomial is often hiding inside a real situation. The skill is to translate the situation into a polynomial first, and only then use factorisation or the zeroes–coefficients relationships.
Problem. A rectangular hall has area \(x^2 + 8x + 15\) square metres. If its length is \((x + 5)\) metres, find its breadth.
Method. Area = length × breadth, so breadth = area ÷ length.
Working. \((x + 5)(x + 3) = x^2 + 5x + 3x + 15 = x^2 + 8x + 15\). So breadth \(= x + 3\) metres.
Answer. The breadth is \((x + 3)\) metres.
Interpretation. For the measurements to be positive we need \(x > 0\); the area itself only makes sense then.
Problem. The product of two consecutive positive integers is 306. Form a quadratic polynomial to represent the situation.
Working. Let the integers be \(x\) and \(x + 1\). Then \(x(x + 1) = 306\), i.e. \(x^2 + x - 306 = 0\). Factorising: \((x - 17)(x + 18) = 0\).
Answer. The integers are 17 and 18. (Check: \(17 \times 18 = 306\). ✔)
Problem. A rectangular plot has length \((x + 6)\) m and breadth \((x + 1)\) m. For what value of \(x\) is the area 36 m²?
Working. Area \(= (x + 6)(x + 1) = x^2 + 7x + 6\). Set it equal to 36: \(x^2 + 7x + 6 = 36\), so \(x^2 + 7x - 30 = 0\). Factorising: \((x + 10)(x - 3) = 0\).
Answer. \(x = -10\) is not possible for a length; \(x = 3\). Then length \(= 9\) m and breadth \(= 4\) m, area \(= 36\) m². ✔
Problem. The weekly profit (in hundreds of rupees) of a stall is modelled by \(P(x) = 30x - x^2\), where \(x\) is the number of items sold in dozens.
Working. Factorise \(P(x) = x(30 - x)\). The profit is zero when \(x = 0\) or \(x = 30\).
Interpretation. No items sold gives zero profit; selling exactly 30 dozens also gives zero profit (the market is saturated). The model gives a positive profit only for \(0 < x < 30\), which is the sensible operating range.
Problem. The side of a square is \((2x + 1)\) m. Express its area and find it when \(x = 2\).
Working. Area \(= (2x + 1)^2 = 4x^2 + 4x + 1\). At \(x = 2\): side \(= 5\) m, area \(= 25\) m².
Verify. \(4(4) + 4(2) + 1 = 16 + 8 + 1 = 25\). ✔
9. Case-Based Questions — Polynomials
Case-based questions describe a small real-world situation and then ask 2–3 questions about it. Read the situation fully, write down the polynomial it suggests, and use factorisation or the zeroes–coefficients rules to answer.
Case Study 1 — School Garden
A school plans a rectangular garden whose length is \((x + 6)\) metres and whose breadth is \((x + 2)\) metres. The area of the garden is modelled by the quadratic polynomial \(A(x) = x^2 + 8x + 12\).
Question 1
Confirm that the given area polynomial matches \((x + 6)(x + 2)\).
Question 2
Find the zeroes of \(A(x)\) and verify the relationship between the zeroes and the coefficients.
Question 3
Interpret the zeroes in the context of the garden. If the actual area of the garden is 45 m², find \(x\) and the two dimensions.
Solution. (1) \((x + 6)(x + 2) = x^2 + 2x + 6x + 12 = x^2 + 8x + 12\). ✔
(2) \(A(x) = (x + 6)(x + 2)\), so the zeroes are \(x = -6\) and \(x = -2\). Here \(a = 1\), \(b = 8\), \(c = 12\). Sum \(= -8 = -\dfrac{b}{a}\) ✔; product \(= 12 = \dfrac{c}{a}\) ✔.
(3) Both zeroes are negative, so no positive value of \(x\) makes the area zero — the model is meaningful only for \(x > 0\). Geometrically, \(x = -2\) corresponds to breadth \(x + 2 = 0\), and \(x = -6\) to length \(x + 6 = 0\). For area 45 m²: \(x^2 + 8x + 12 = 45 \Rightarrow x^2 + 8x - 33 = 0 \Rightarrow (x + 11)(x - 3) = 0\). So \(x = 3\): length 9 m, breadth 5 m, area \(9 \times 5 = 45\) m². ✔
Case Study 2 — Classroom Arrangement
Benches in a hall are arranged in \(x\) rows with \((x + 5)\) benches in each row. The total number of benches is given by the quadratic polynomial \(B(x) = x^2 + 5x\).
Question 1
Explain why \(B(x) = x(x + 5)\) represents the total number of benches.
Question 2
Find the zeroes of \(B(x)\). Which zero is meaningful for the classroom? Give a reason.
Question 3
The hall has 36 benches. Find the number of rows and the number of benches in each row.
Solution. (1) Total benches = rows × benches per row \(= x \times (x + 5) = x^2 + 5x\). ✔
(2) \(x(x + 5) = 0\) gives zeroes \(0\) and \(-5\). \(x = 0\) (no rows, no benches) is trivially meaningful; \(x = -5\) is not, because the number of rows cannot be negative.
(3) \(x^2 + 5x = 36 \Rightarrow x^2 + 5x - 36 = 0 \Rightarrow (x + 9)(x - 4) = 0\). So \(x = 4\) rows and \(x + 5 = 9\) benches per row; total \(= 4 \times 9 = 36\). ✔
Case Study 3 — Sports Ground
A rectangular sports ground has length \((2x + 1)\) metres and breadth \((x + 4)\) metres. Its area is \(2x^2 + 9x + 4\) square metres.
Question 1
Show that the product of the two dimensions gives the given area polynomial.
Question 2
Find the zeroes of \(2x^2 + 9x + 4\) and verify the sum and product relationships.
Question 3
When \(x = 5\), find the length, breadth and area of the ground.
Solution. (1) \((2x + 1)(x + 4) = 2x^2 + 8x + x + 4 = 2x^2 + 9x + 4\). ✔
(2) \((2x + 1)(x + 4) = 0\) gives zeroes \(-\dfrac{1}{2}\) and \(-4\). Here \(a = 2\), \(b = 9\), \(c = 4\). Sum \(= -\dfrac{1}{2} - 4 = -\dfrac{9}{2} = -\dfrac{b}{a}\) ✔; product \(= \left(-\dfrac{1}{2}\right)(-4) = 2 = \dfrac{4}{2} = \dfrac{c}{a}\) ✔.
(3) At \(x = 5\): length \(= 2(5) + 1 = 11\) m, breadth \(= 5 + 4 = 9\) m, area \(= 11 \times 9 = 99\) m². Check: \(2(25) + 9(5) + 4 = 99\). ✔
Case Study 4 — Product / Revenue Model
A fruit seller finds that his daily revenue (in rupees) is modelled by \(R(x) = x(200 - 2x)\), where \(x\) is the price (in rupees) of a basket.
Question 1
Write \(R(x)\) in the standard form \(ax^2 + bx + c\), stating \(a\), \(b\) and \(c\).
Question 2
Find the values of \(x\) for which the revenue becomes zero, and verify the sum and product of these values.
Question 3
Interpret the zeroes in the context of the seller. Over what range of prices is the revenue positive?
Solution. (1) \(R(x) = -2x^2 + 200x\), so \(a = -2\), \(b = 200\), \(c = 0\).
(2) \(x(200 - 2x) = 0\) gives \(x = 0\) and \(x = 100\). Sum \(= 100 = -\dfrac{200}{-2}\) ✔; product \(= 0 = \dfrac{0}{-2}\) ✔.
(3) Revenue is zero when nothing is sold (\(x = 0\)) and again at a price of ₹100 per basket (at which no baskets are bought). The polynomial is positive for \(0 < x < 100\), so the seller's sensible price range is between ₹0 and ₹100 per basket.
Case Study 5 — Number Pattern
A pattern of floor tiles gives the number of tiles in the \(n\)-th figure as \(T(n) = 2n^2 - 3n - 2\).
Question 1
Verify that the pattern gives \(T(1) = -3\), \(T(2) = 0\) and \(T(3) = 7\).
Question 2
Find the zeroes of \(T(n)\). Which one is meaningful for a pattern of tiles?
Question 3
Calculate the sum and product of the zeroes and verify the coefficient relationship.
Solution. (1) \(T(1) = 2 - 3 - 2 = -3\) ✔; \(T(2) = 8 - 6 - 2 = 0\) ✔; \(T(3) = 18 - 9 - 2 = 7\) ✔.
(2) \(2n^2 - 3n - 2 = (2n + 1)(n - 2)\), so the zeroes are \(n = -\dfrac{1}{2}\) and \(n = 2\). Since \(n\) counts the figure number, only \(n = 2\) is meaningful.
(3) Sum \(= -\dfrac{1}{2} + 2 = \dfrac{3}{2} = -\dfrac{b}{a} = \dfrac{3}{2}\) ✔; product \(= \left(-\dfrac{1}{2}\right)(2) = -1 = \dfrac{c}{a} = \dfrac{-2}{2}\) ✔.
10. Competency-Based Examples
Competency questions test your ability to choose the right method, spot an error, interpret information, or compare two situations — not just to repeat a routine calculation.
A student says the zeroes of \(x^2 - 6x + 8\) are 2 and 3. Check and correct the claim.
Reasoning. If the zeroes were 2 and 3, the product would be \(6\), but \(\dfrac{c}{a} = 8\). The correct zeroes satisfy the product rule, so factorise: \(x^2 - 6x + 8 = (x - 2)(x - 4)\), giving zeroes \(2\) and \(4\). The student chose the wrong factors.
Without factorising \(x^2 - 7x + 12\), state the sum and product of its zeroes, and then write down the zeroes.
Reasoning. Sum \(= 7\), product \(= 12\). Two numbers with sum 7 and product 12 are 3 and 4, so the zeroes are 3 and 4.
A parabola opens upwards, lies completely above the x-axis, and never touches it. How many real zeroes does the corresponding polynomial have? Explain.
Reasoning. A real zero is a point where the graph meets the x-axis. The graph never touches it, so the polynomial has no real zeroes.
For a quadratic polynomial in which the sum of the zeroes is 3 and the product is 2, what is the polynomial of the form \(x^2 + bx + c\)? Also give one polynomial with these zeroes but leading coefficient 3.
Reasoning. \(x^2 - 3x + 2\) has sum 3 and product 2. Multiplying by 3 does not change the zeroes: \(3x^2 - 9x + 6\) has the same zeroes (1 and 2).
Which of \(x^2 - 10x + 25\) and \(x^2 - 6x + 8\) has two equal real zeroes? Justify.
Reasoning. \(x^2 - 10x + 25 = (x - 5)^2\) has the equal real zeroes \(5, 5\). The other factorises as \((x - 2)(x - 4)\), giving distinct zeroes \(2, 4\). So the first polynomial has two equal real zeroes.
11. Solved Examples
Twenty fully worked examples: five basic, six moderate, five application/competency-based and four case-based. Each solution follows Given → Method → Working → Answer → Interpretation wherever appropriate.
Basic
Given. \(p(x) = 2x^2 + 5x - 3\).
Working. \(p(1) = 2 + 5 - 3 = 4\); \(p(0) = -3\); \(p(-2) = 2(4) + 5(-2) - 3 = 8 - 10 - 3 = -5\).
Check for a zero. \(p\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{4}\right) + 5\left(\dfrac{1}{2}\right) - 3 = \dfrac{1}{2} + \dfrac{5}{2} - 3 = 0\). So \(\dfrac{1}{2}\) is a zero.
Answer. \(p(1) = 4\), \(p(0) = -3\), \(p(-2) = -5\), and the zero \(\dfrac{1}{2}\) verifies via \(p\left(\dfrac{1}{2}\right) = 0\).
Given. Polynomials \(4x^2 + 3x\), \(7x - 1\), \(9\); also \(2x + \dfrac{3}{x}\) and \(\sqrt{x} + 5\).
Method. Identify by degree; check that every exponent is a whole number.
Working/Answer. Degrees \(2\), \(1\), \(0\). \(2x + \dfrac{3}{x}\) has exponent \(-1\) and \(\sqrt{x} + 5\) has exponent \(\frac{1}{2}\), so neither is a polynomial.
Given. \(3x^2 - 5 + 2x\).
Working. Rearrange in descending powers: \(3x^2 + 2x - 5\).
Answer. Standard form \(3x^2 + 2x - 5\), with \(a = 3\), \(b = 2\), \(c = -5\).
Given. \(p(x) = x^2 + x - 6\).
Working. \(p(2) = 4 + 2 - 6 = 0\), so 2 is a zero. \(p(-3) = 9 - 3 - 6 = 0\), so \(-3\) is a zero.
Answer. Yes — both 2 and \(-3\) are zeroes of \(p(x)\). Factorised form \((x - 2)(x + 3)\) confirms it.
Given. \(p(x) = 2x + 3\).
Working. \(p(1) = 5 \neq 0\), so 1 is not a zero. Solving \(2x + 3 = 0\): \(x = -\dfrac{3}{2}\).
Answer. The zero is \(-\dfrac{3}{2}\) (a linear polynomial has exactly one zero).
Moderate
Given. \(x^2 - 5x + 6\).
Method. Split the middle term; verify with the relationships.
Working. \(x^2 - 2x - 3x + 6 = (x - 2)(x - 3)\). Zeroes 2, 3.
Answer + Interpretation. Sum 5 and product 6 match \(-\dfrac{b}{a} = 5\) and \(\dfrac{c}{a} = 6\), so the relationship is verified.
Given. \(2x^2 - 6x\).
Working. Factor 2x: \(2x(x - 3) = 0\), so the zeroes are 0 and 3.
Answer + Interpretation. Sum 3 = \(-\dfrac{-6}{2}\) ✔; product 0 = \(\dfrac{0}{2}\) ✔. A zero can be 0 — never confuse "zero of a polynomial" with "the number zero".
Given. \(x^2 + 5x + 6\).
Working. \((x + 2)(x + 3) = 0\), so the zeroes are \(-2\) and \(-3\).
Answer + Interpretation. Both zeroes are negative. Sum \(= -5 = -\dfrac{b}{a}\) ✔; product \(= 6 = \dfrac{c}{a}\) ✔.
Given. \(3x^2 - x - 4\).
Method. Product \(ac = -12\), sum \(-1\) → numbers \(-4\) and \(3\).
Working. \(3x^2 - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (3x - 4)(x + 1)\). Zeroes \(\dfrac{4}{3}\) and \(-1\).
Answer + Interpretation. Sum \(= \dfrac{1}{3}\) matches \(-\dfrac{b}{a} = \dfrac{1}{3}\) ✔; product \(= -\dfrac{4}{3} = \dfrac{c}{a}\) ✔.
Given. \(p(x) = x^2 + 7x + 12\).
Method. Use the relationships directly, then confirm by factorising.
Working. Sum \(= -7\), product \(= 12\). Factorising \(x^2 + 7x + 12 = (x + 3)(x + 4)\) gives zeroes \(-3, -4\), whose sum is \(-7\) and product is 12 — consistent. ✔
Answer. Sum \(= -7\), product \(= 12\).
Given. \(\alpha\), \(\beta\) are the zeroes of \(x^2 - 4x + 1\).
Working. Sum \(= 4\), product \(= 1\). Then \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 16 - 2 = 14\).
Answer. \(\alpha + \beta = 4\), \(\alpha\beta = 1\), \(\alpha^2 + \beta^2 = 14\).
Application / Competency
Given. Zeroes 3 and \(-5\).
Method. Form the polynomial from the zeroes.
Working. Sum \(= 3 + (-5) = -2\); product \(= 3(-5) = -15\). Polynomial \(= x^2 - (-2)x + (-15) = x^2 + 2x - 15\).
Answer + Interpretation. \(x^2 + 2x - 15 = (x - 3)(x + 5)\), whose zeroes are indeed 3 and \(-5\). ✔
Given. Area \(x^2 + 7x + 12\) with one side \((x + 3)\).
Method. Other side = area ÷ given side.
Working. \((x + 3)(x + 4) = x^2 + 7x + 12\).
Answer + Interpretation. The other side is \((x + 4)\); both sides are positive for \(x > 0\).
Given. Sum of zeroes \(\dfrac{5}{3}\), product \(\dfrac{2}{3}\).
Working. \(x^2 - \dfrac{5}{3}x + \dfrac{2}{3}\), then multiply by 3: \(3x^2 - 5x + 2\).
Answer + Interpretation. \((x - 1)(3x - 2)\) gives zeroes \(1\) and \(\dfrac{2}{3}\), whose sum and product match. ✔
Given. A student claims the zeroes of \(x^2 - 6x + 8\) are 2 and 3.
Working. Product of the claimed zeroes \(= 6\), but the product rule gives \(8\). Correct factorisation: \((x - 2)(x - 4)\), zeroes 2 and 4.
Answer. The claim is wrong — the zeroes are 2 and 4. The sum (6) matched, but the product rule caught the error.
Given. Product of two consecutive odd positive integers is 195.
Working. Let them be \(x\) and \(x + 2\). Then \(x(x + 2) = 195 \Rightarrow x^2 + 2x - 195 = 0 \Rightarrow (x + 15)(x - 13) = 0\). Positive integer solution: \(x = 13\).
Answer + Interpretation. The integers are 13 and 15. (Check \(13 \times 15 = 195\). ✔)
Case-based
Given. Rectangular garden, area \(x^2 + 5x + 6\), breadth \((x + 2)\).
Working. Length \(= x + 3\). At \(x = 5\): area \(= 25 + 25 + 6 = 56\) m² (length 8, breadth 7). Zeroes of the area polynomial: \((x + 2)(x + 3) = 0\) gives \(-2, -3\).
Answer + Interpretation. Both zeroes are negative, so the area cannot be zero for a real garden with \(x > 0\). Sum \(= -5 = -\dfrac{b}{a}\) ✔; product \(= 6 = \dfrac{c}{a}\) ✔.
Given. Rows \(= x\), benches per row \(= x + 4\), total \(x^2 + 4x\).
Working. Zeroes: \(x = 0, -4\). For 60 benches: \(x^2 + 4x = 60 \Rightarrow (x + 10)(x - 6) = 0 \Rightarrow x = 6\).
Answer + Interpretation. 6 rows with 10 benches per row gives 60 benches. The zero \(-4\) is not meaningful (negative rows).
Given. Length \((x + 5)\), breadth \((x + 2)\), area \(x^2 + 7x + 10\).
Working. Zeroes \(-5, -2\). At \(x = 8\): length 13, breadth 10, area \(= 64 + 56 + 10 = 130\) m².
Answer + Interpretation. Sum \(= -7 = -\dfrac{b}{a}\) ✔; product \(= 10 = \dfrac{c}{a}\) ✔. The negative zeroes signal that the model is only meaningful for positive \(x\).
Given. \(R(x) = 600x - 100x^2\).
Working. \(100x(6 - x) = 0\) gives zeroes 0 and 6. Here \(a = -100\), \(b = 600\), \(c = 0\). Sum \(= 6 = -\dfrac{600}{-100}\) ✔; product \(= 0\) ✔.
Answer + Interpretation. Revenue is zero at 0 and at 6 units; it is positive between them, which is the feasible business range.
12. Important Formulae
For a quadratic polynomial \(ax^2 + bx + c\) with zeroes \(\alpha\) and \(\beta\):
\[\alpha + \beta = -\frac{b}{a}\]
\[\alpha \beta = \frac{c}{a}\]
\[p(x) = x^2 - (\alpha + \beta)x + \alpha\beta\]
Always identify \(a\), \(b\), and \(c\) correctly before applying the formula. Put every polynomial in standard form \(ax^2 + bx + c\) first, keep the signs of \(b\) and \(c\), and never drop the \(a\) in the denominator.
13. Common Exam Mistakes
| Common mistake | Correct method |
|---|---|
| Writing the sum of zeroes as \(+\dfrac{b}{a}\) (sign error) | Sum of zeroes is always \(-\dfrac{b}{a}\) |
| Writing the sum without dividing by \(a\), e.g. sum \(= -7\) for \(2x^2 + 7x - 4\) | \(-\dfrac{b}{a} = -\dfrac{7}{2}\) |
| Writing product as \(c\) instead of \(\dfrac{c}{a}\), e.g. product \(= -4\) for \(2x^2 + 7x - 4\) | Product \(= \dfrac{c}{a} = \dfrac{-4}{2} = -2\) |
| Confusing sum and product when forming the polynomial | Use \(x^2 - (\text{sum})x + \text{product}\); double-check by expanding |
| Incorrect factorisation when splitting the middle term | Find two numbers with product \(ac\) and sum \(b\), never with product \(c\) alone |
| Incorrect expansion, e.g. \((x - 3)(x + 5) = x^2 - 15\) | Expand fully: \(x^2 + 5x - 3x - 15 = x^2 + 2x - 15\) |
| Calling a computed value like \(p(0)\) "the zero" | The zero is the input \(k\) for which \(p(k) = 0\) |
| Reading "graph touches the x-axis at one point" as "one zero" | It means two equal real zeroes, not one zero |
| Arithmetic slips with negatives, e.g. \(\alpha + \beta\) for zeroes \(-2, -5\) | Compute carefully: \(-2 + (-5) = -7\) |
| Forgetting to verify zeroes | Always check \(p(k) = 0\) and the sum/product rules |
For \(2x^2 + 7x - 4\), the sum of zeroes is \(-\dfrac{7}{2}\) and the product is \(-2\) (not \(7/2\) and not \(-4\)). Take \(a = 2\), \(b = 7\), \(c = -4\) and apply both formulae with the signs intact.
14. CBSE Exam Tips
- Write the polynomial in standard form \(ax^2 + bx + c\) before identifying coefficients.
- Identify \(a\), \(b\), \(c\) carefully — especially the signs of \(b\) and \(c\), and the fact that \(a\) sits in the denominator.
- Show the factorisation steps. Marker-wise, the splitting of the middle term earns marks.
- Verify answers where possible: substitute the zeroes and check the sum and product.
- Use correct notation — \(p(x)\), "zeroes of \(p(x)\)", and the set of zeroes like \(\{2, 3\}\).
- For case-based questions, read the situation carefully before calculating.
- Do not assume every quadratic has two distinct real zeroes.
- In application questions, interpret the mathematical answer in the context — discard negative or impossible values (rows, lengths, prices).
15. Quick Revision
| Concept | Key point |
|---|---|
| Zero of a polynomial | \(p(k) = 0\) |
| Linear polynomial | One zero |
| Quadratic polynomial | At most two real zeroes |
| Sum of zeroes | \(-\dfrac{b}{a}\) |
| Product of zeroes | \(\dfrac{c}{a}\) |
| Polynomial from zeroes | \(x^2 - (\alpha + \beta)x + \alpha\beta\) |
16. Self-Practice Questions
Twenty-five original questions. Answers are at the end of the page.
Conceptual
Q1.
Write \(x^2 + 5 - 3x\) in standard form and state its degree.
Q2.
Which of the following are not polynomials? \(2x + \dfrac{1}{2}\); \(\dfrac{3}{\sqrt{x}}\); \(x^2 - \sqrt{5}x + 2\); \(x^{1/2} + 3\). Give one-line reasons.
Q3.
If \(p(x) = 3x^2 + 2x - 5\), find \(p(0)\) and \(p(-1)\).
Q4.
In the polynomial \(4x^2 - 7x + 9\), write the coefficient of \(x^2\) and the constant term.
Q5.
How many real zeroes does a linear polynomial have?
Moderate
Q6.
Find the zeroes of \(x^2 - 9\).
Q7.
Find the zeroes of \(x^2 + 7x + 10\).
Q8.
Find the zeroes of \(6x^2 - x - 2\).
Q9.
For \(2x^2 - 8x + 6\), state the sum and product of the zeroes, then find the zeroes.
Q10.
Find the zeroes of \(4x^2 - 4x + 1\) and verify the relationship with the coefficients.
Q11.
If one zero of \(x^2 - 5x + k\) is 2, find \(k\) and the other zero.
Q12.
Find a quadratic polynomial whose zeroes are 4 and \(-1\).
Application-Based
Q13.
A rectangular plot has area \(x^2 + 9x + 14\) sq m and breadth \((x + 2)\) m. Find the polynomial for its length.
Q14.
Rows of chairs number \(x\), with \((x + 5)\) chairs in each row; the total is 36. Form a quadratic and find \(x\).
Q15.
The revenue model \(R(x) = 400x - 2x^2\) becomes zero at what values of \(x\)?
Q16.
The product of two consecutive even positive integers is 168. Form a quadratic polynomial and find the integers.
Q17.
The side of a square is \((2x + 3)\) m. Express its area as a polynomial and find the area when \(x = 2\).
Q18.
A number exceeds another by 4 and their product is 12. Form a quadratic and find the numbers.
Competency / Reasoning
Q19.
A student says the zeroes of \(x^2 - 5x + 6\) are \(-2\) and \(-3\). Is the claim correct? Explain using the sum of zeroes.
Q20.
The graph of a quadratic polynomial touches the x-axis at exactly one point. How many real zeroes does it have?
Q21.
A student wrote the sum of the zeroes of \(2x^2 + 7x - 4\) as \(7/2\) and the product as \(4\). Identify both errors.
Q22.
Which polynomial has equal real zeroes: \(x^2 - 8x + 16\) or \(x^2 - 5x + 6\)? Justify.
Case-Based
Q23 — Rectangular garden.
A garden has length \((x + 5)\) m and breadth \((x + 1)\) m, so its area is \(x^2 + 6x + 5\). (i) Find the zeroes of the area polynomial and verify the coefficient relationship. (ii) If the area is 21 m², find \(x\) and the dimensions.
Q24 — Parking lot.
A parking area has area \(x^2 + 10x + 24\) sq m and length \((x + 6)\) m. (i) Find the breadth. (ii) Find the zeroes of the area polynomial and explain why neither has a physical meaning here.
Q25 — Tile pattern.
The number of tiles in the \(n\)-th figure is \(f(n) = 2n^2 - 5n - 3\). (i) Find the zeroes of \(f(n)\). (ii) Which zero is meaningful for a pattern? (iii) Verify the sum and product of the zeroes.
17. Answers
A1.
Standard form: \(x^2 - 3x + 5\). Degree 2.
A2.
\(\dfrac{3}{\sqrt{x}}\) (exponent \(-\frac{1}{2}\)) and \(x^{1/2} + 3\) (exponent \(\frac{1}{2}\)) are not polynomials. \(2x + \dfrac{1}{2}\) is linear and \(x^2 - \sqrt{5}x + 2\) is quadratic.
A3.
\(p(0) = -5\); \(p(-1) = 3 - 2 - 5 = -4\).
A4.
Coefficient of \(x^2\) is 4; constant term is 9.
A5.
Exactly one real zero.
A6.
\(x^2 - 9 = (x - 3)(x + 3)\), so zeroes are 3 and \(-3\).
A7.
\((x + 5)(x + 2) = 0\), so zeroes are \(-5\) and \(-2\).
A8.
\(6x^2 - x - 2 = (3x - 2)(2x + 1)\), so zeroes are \(\dfrac{2}{3}\) and \(-\dfrac{1}{2}\).
A9.
Sum 4, product 3. \(2x^2 - 8x + 6 = 2(x - 1)(x - 3)\), zeroes 1 and 3.
A10.
\((2x - 1)^2 = 0\), so the two equal real zeroes are \(\dfrac{1}{2}, \dfrac{1}{2}\). Sum 1 = \(-\dfrac{b}{a}\) ✔; product \(\dfrac{1}{4} = \dfrac{c}{a}\) ✔.
A11.
\(p(2) = 0 \Rightarrow 4 - 10 + k = 0 \Rightarrow k = 6\). The polynomial is \(x^2 - 5x + 6\), other zero 3.
A12.
Sum 3, product \(-4\), so \(x^2 - 3x - 4\). Check: \((x - 4)(x + 1)\). ✔
A13.
Length polynomial \(= x + 7\) (since \((x + 2)(x + 7) = x^2 + 9x + 14\)).
A14.
\(x(x + 5) = 36 \Rightarrow x^2 + 5x - 36 = 0 \Rightarrow (x + 9)(x - 4) = 0\). So \(x = 4\) rows (with 9 chairs per row).
A15.
\(400x - 2x^2 = 2x(200 - x) = 0\), so \(x = 0\) or \(x = 200\).
A16.
\(x(x + 2) = 168 \Rightarrow x^2 + 2x - 168 = 0 \Rightarrow (x + 14)(x - 12) = 0\). Positive \(x = 12\); the integers are 12 and 14.
A17.
Area \(= (2x + 3)^2 = 4x^2 + 12x + 9\). At \(x = 2\): \(4(4) + 24 + 9 = 49\) m² (side 7 m).
A18.
\(x(x + 4) = 12 \Rightarrow x^2 + 4x - 12 = 0 \Rightarrow (x + 6)(x - 2) = 0\). Numbers are 2 and 6.
A19.
No. If the zeroes were \(-2\) and \(-3\), their sum would be \(-5\), but \(-\dfrac{b}{a} = 5\). The correct zeroes are 2 and 3.
A20.
Two equal real zeroes — the graph meets the x-axis at one point, so the two zeroes coincide.
A21.
Sum should be \(-\dfrac{7}{2}\) (the student missed the negative sign). Product should be \(\dfrac{c}{a} = \dfrac{-4}{2} = -2\) (the student used \(c = 4\) instead of \(\dfrac{c}{a} = -2\)).
A22.
\(x^2 - 8x + 16 = (x - 4)^2\) has the equal real zeroes \(4, 4\). The other factorises as \((x - 2)(x - 3)\) with distinct zeroes 2 and 3.
A23.
(i) \(x^2 + 6x + 5 = (x + 5)(x + 1)\), zeroes \(-5\) and \(-1\). Sum \(-6 = -\dfrac{b}{a}\) ✔; product 5 = \(\dfrac{c}{a}\) ✔. (ii) \(x^2 + 6x + 5 = 21 \Rightarrow (x + 8)(x - 2) = 0\), so \(x = 2\); length 7 m, breadth 3 m, area 21 m².
A24.
(i) Breadth \(= x + 4\), since \((x + 6)(x + 4) = x^2 + 10x + 24\). (ii) Zeroes of the area polynomial are \(-6\) and \(-4\); both are negative, while a length or breadth must be positive, so neither zero has physical meaning. The model is valid only for \(x > 0\), where the area is always positive.
A25.
(i) \(2n^2 - 5n - 3 = (2n + 1)(n - 3)\), zeroes \(n = -\dfrac{1}{2}\) and \(n = 3\). (ii) \(n = 3\) is meaningful (a figure number must be a positive integer). (iii) Sum \(= -\dfrac{1}{2} + 3 = \dfrac{5}{2} = -\dfrac{b}{a}\) ✔; product \(= -\dfrac{3}{2} = \dfrac{c}{a}\) ✔.