Polynomials extends the Class 9 idea of degree to a new question: how are the zeroes of a polynomial linked to its coefficients? The chapter covers linear, quadratic and cubic polynomials, the value p(k) and zeroes, the graph meaning of zeroes, the sum and product of zeroes of a quadratic, and forming a quadratic from a given sum and product. It opens the Algebra unit and feeds straight into Quadratic Equations, and board papers test it through graph readings, verify-the-relationship questions and expressions in α and β.
Key Concepts
1. Polynomials, Degree and Types
A polynomial in one variable x is an expression made of terms like axn, where each coefficient is a real number and each power n is a whole number (0, 1, 2, 3, …). If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of p(x).
- 4x + 2 is a polynomial in x of degree 1.
- 2y2 − 3y + 4 is a polynomial in y of degree 2.
- 5x3 − 4x2 + x + 7 is a polynomial in x of degree 3.
- 7u6 − 4u2 + 8 is a polynomial in u of degree 6.
Expressions such as 1/(x − 1), √x + 2 and 1/x + x are not polynomials. They have x in a denominator or a power of x that is not a whole number.
The chapter names polynomials by their degree:
| Degree | Name | General form | Examples |
|---|---|---|---|
| 1 | Linear polynomial | ax + b, a ≠ 0 | 2x − 3, 3z + 4, √2 y |
| 2 | Quadratic polynomial | ax2 + bx + c, a ≠ 0 | x2 − 3x − 4, y2 − 2, 2x2 + 3x − 2/5 |
| 3 | Cubic polynomial | ax3 + bx2 + cx + d, a ≠ 0 | 2 − x3, x3, 3 − x2 + x3, 3x3 − 2x2 + x − 1 |
In every general form, a, b, c and d are real numbers and the leading coefficient a is non-zero. The condition a ≠ 0 matters: if a = 0 in ax2 + bx + c, the x2 term disappears and the polynomial is no longer quadratic. The word “quadratic” comes from “quadrate”, which means “square”.
Reading coefficients correctly. Before using any formula, rewrite the polynomial in standard form (descending powers) and read the coefficients with their signs.
| Polynomial as given | Standard form | a | b | c |
|---|---|---|---|---|
| 6x2 − 3 − 7x | 6x2 − 7x − 3 | 6 | −7 | −3 |
| 4u2 + 8u | 4u2 + 8u + 0 | 4 | 8 | 0 |
| t2 − 15 | t2 + 0t − 15 | 1 | 0 | −15 |
2. Value of a Polynomial and Zero of a Polynomial
If p(x) is a polynomial in x and k is any real number, the value obtained by replacing x by k in p(x) is called the value of p(x) at x = k. It is written p(k).
Take p(x) = x2 − 3x − 4.
- p(2) = 22 − 3 × 2 − 4 = 4 − 6 − 4 = −6
- p(0) = 0 − 0 − 4 = −4
- p(−1) = (−1)2 − 3 × (−1) − 4 = 1 + 3 − 4 = 0
- p(4) = 42 − 3 × 4 − 4 = 16 − 12 − 4 = 0
Since p(−1) = 0 and p(4) = 0, the numbers −1 and 4 are called the zeroes of x2 − 3x − 4.
Definition: a real number k is a zero of a polynomial p(x) if p(k) = 0.
Zero of a linear polynomial. If k is a zero of p(x) = 2x + 3, then 2k + 3 = 0, so k = −3/2. In general, if k is a zero of ax + b, then ak + b = 0, which gives k = −b/a. So
zero of ax + b = −b/a = −(constant term) / (coefficient of x)
Note that zero itself can be a zero of a polynomial. For x2 − 5x, p(0) = 0, so 0 is a zero. This happens exactly when the constant term is 0.
3. Geometrical Meaning of the Zeroes
Linear polynomial. The graph of y = ax + b (a ≠ 0) is a straight line. For y = 2x + 3, the line passes through (−2, −1) and (2, 7). It crosses the x-axis midway between x = −1 and x = −2, at the point (−3/2, 0). This x-coordinate is the zero of 2x + 3. In general, the line y = ax + b meets the x-axis at exactly one point, (−b/a, 0). So a linear polynomial has exactly one zero.
Quadratic polynomial. Take y = x2 − 3x − 4 and make a table of values (NCERT Table 2.1):
| x | −2 | −1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|---|
| y = x2 − 3x − 4 | 6 | 0 | −4 | −6 | −6 | −4 | 0 | 6 |
Plotting these points and joining them gives a U-shaped curve. The curve cuts the x-axis where y = 0, at x = −1 and x = 4, which are the zeroes of the polynomial. The graph of any quadratic y = ax2 + bx + c (a ≠ 0) is a curve called a parabola. It opens upwards when a > 0 and opens downwards when a < 0.
The zeroes of ax2 + bx + c are precisely the x-coordinates of the points where the parabola y = ax2 + bx + c meets the x-axis. Three cases can happen:
| Case | What the parabola does | Number of zeroes |
|---|---|---|
| (i) | Cuts the x-axis at two distinct points A and A′ | Two distinct zeroes (the x-coordinates of A and A′) |
| (ii) | Touches the x-axis at exactly one point A (the two points coincide) | Two equal zeroes, that is, one zero |
| (iii) | Lies completely above or completely below the x-axis | No zero |
So a polynomial of degree 2 has at most two zeroes. For example, x2 + 1 is at least 1 for every real x, so its graph stays above the x-axis and it has no zero (case iii). x2 − 4x + 4 = (x − 2)2 is 0 only at x = 2, so its graph touches the x-axis at (2, 0) (case ii).
Cubic polynomial. Take y = x3 − 4x (NCERT Table 2.2):
| x | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| y = x3 − 4x | 0 | 3 | 0 | −3 | 0 |
The zeroes are −2, 0 and 2, and these are the x-coordinates of the only points where the graph meets the x-axis. You can confirm this by factorising: x3 − 4x = x(x − 2)(x + 2). Two more cubics from the chapter:
- y = x3: the only zero is 0, and the graph meets the x-axis at the origin only.
- y = x3 − x2 = x2(x − 1): the only zeroes are 0 and 1, and the graph meets the x-axis at x = 0 and x = 1 only.
So a cubic polynomial has at most three zeroes.
General rule (Remark in NCERT): for a polynomial p(x) of degree n, the graph of y = p(x) meets the x-axis at most at n points. So a polynomial of degree n has at most n zeroes.
Counting zeroes from a graph. Count the number of distinct points where the curve meets the x-axis. A point where the curve only touches the axis and turns back still counts as a meeting point. Points where the curve meets the y-axis do not count. If the curve never meets the x-axis, the polynomial has no zero. The NCERT footnote says students are not meant to plot graphs of quadratic or cubic polynomials, and such plotting is not evaluated. Exercise 2.1 asks you to read given graphs and count zeroes.
4. Relationship between Zeroes and Coefficients of a Quadratic
Start with p(x) = 2x2 − 8x + 6. Split the middle term: we need two terms whose sum is −8x and whose product is 6 × 2x2 = 12x2. These are −6x and −2x.
2x2 − 8x + 6 = 2x2 − 6x − 2x + 6 = 2x(x − 3) − 2(x − 3) = (2x − 2)(x − 3) = 2(x − 1)(x − 3)
So the zeroes are 1 and 3. Now compare with the coefficients a = 2, b = −8, c = 6:
- Sum of zeroes = 1 + 3 = 4 = −(−8)/2 = −(coefficient of x)/(coefficient of x2)
- Product of zeroes = 1 × 3 = 3 = 6/2 = (constant term)/(coefficient of x2)
A second check: 3x2 + 5x − 2 = 3x2 + 6x − x − 2 = 3x(x + 2) − 1(x + 2) = (3x − 1)(x + 2). The zeroes are 1/3 and −2. Sum = 1/3 + (−2) = −5/3 = −b/a. Product = (1/3)(−2) = −2/3 = c/a.
The general proof. Let α and β (alpha and beta) be the zeroes of p(x) = ax2 + bx + c, a ≠ 0. Then (x − α) and (x − β) are factors of p(x), so
ax2 + bx + c = k(x − α)(x − β), where k is a constant
= k[x2 − (α + β)x + αβ]
= kx2 − k(α + β)x + kαβ
Comparing the coefficients of x2, x and the constant terms on both sides: a = k, b = −k(α + β) and c = kαβ. This gives
α + β = −b/a = −(coefficient of x) / (coefficient of x2)
αβ = c/a = (constant term) / (coefficient of x2)
5. Finding the Zeroes of a Quadratic and Verifying the Relationship
NCERT Exercise 2.2 asks you to find the zeroes and then verify the relationship. The routine has four steps.
- Write the polynomial in standard form ax2 + bx + c and note a, b, c with signs.
- Factorise it. Choose the method that fits.
- Set each factor equal to zero to get the two zeroes.
- Compute the sum and product of the zeroes, then compute −b/a and c/a separately, and show that they match.
| Form of the polynomial | Method | Example |
|---|---|---|
| ax2 + bx + c, all non-zero | Split the middle term: two numbers with product ac and sum b | x2 + 7x + 10 = (x + 2)(x + 5) |
| ax2 + bx (c = 0) | Take x (and any common number) out | 4u2 + 8u = 4u(u + 2) |
| x2 − k (b = 0, k > 0) | Use a2 − b2 = (a − b)(a + b) | x2 − 3 = (x − √3)(x + √3) |
| Perfect square | Use (a − b)2 or (a + b)2 | 4s2 − 4s + 1 = (2s − 1)2 |
Worked illustration (NCERT Example 3). x2 − 3 = (x − √3)(x + √3). The zeroes are √3 and −√3. Sum = √3 + (−√3) = 0 = −0/1 = −(coefficient of x)/(coefficient of x2). Product = (√3)(−√3) = −3 = −3/1 = (constant term)/(coefficient of x2).
Presentation tip. Write both sides of each check separately, then state “Hence verified”.
6. Forming a Quadratic from the Sum and Product of Its Zeroes
If the zeroes of a quadratic are α and β, then from the proof above the quadratic is k(x − α)(x − β) = k[x2 − (α + β)x + αβ]. So if the sum of zeroes is S and the product is P,
required quadratic = k(x2 − Sx + P), where k is any non-zero real number
NCERT Example 4. Find a quadratic polynomial, the sum and product of whose zeroes are −3 and 2, respectively. Let the polynomial be ax2 + bx + c with zeroes α and β. Then α + β = −3 = −b/a and αβ = 2 = c/a. Taking a = 1 gives b = 3 and c = 2. One such polynomial is x2 + 3x + 2. Every other quadratic that fits these conditions has the form k(x2 + 3x + 2), where k is real (and non-zero, so that it stays quadratic).
Why the answer is not unique. Multiplying a polynomial by a non-zero constant does not change where it equals zero. In the exam, one correct polynomial is enough.
Clearing fractions. When S or P is a fraction, choose k to remove the denominators. Sum 1/4 and product −1 give x2 − (1/4)x − 1. Taking k = 4 gives the neater answer 4x2 − x − 4. Check: −b/a = 1/4 and c/a = −4/4 = −1.
Sign trap. The middle coefficient is the negative of the sum. A sum of 5 gives −5x, and a sum of −5 gives +5x. The product goes in with its own sign.
7. Using the Relations: Unknown Coefficients and Expressions in α and β
Many questions give a condition on the zeroes and ask for a missing coefficient, or ask for the value of an expression in α and β. The relations α + β = −b/a and αβ = c/a do all the work, even when the zeroes are irrational.
Standard identities. Rewrite the expression in terms of (α + β) and αβ, then substitute.
| Expression | Rewrite as |
|---|---|
| α2 + β2 | (α + β)2 − 2αβ |
| (α − β)2 | (α + β)2 − 4αβ |
| 1/α + 1/β | (α + β)/αβ |
| α/β + β/α | (α2 + β2)/αβ |
| α2β + αβ2 | αβ(α + β) |
| α3 + β3 | (α + β)3 − 3αβ(α + β) |
Illustration. Let α, β be the zeroes of x2 − 5x + 3. Then α + β = 5 and αβ = 3.
- α2 + β2 = 25 − 6 = 19
- 1/α + 1/β = 5/3
- α3 + β3 = 125 − 3 × 3 × 5 = 125 − 45 = 80
Conditions that fix a coefficient.
- One zero is the reciprocal of the other: αβ = 1, so c/a = 1, that is c = a.
- One zero is the negative of the other: α + β = 0, so b = 0.
- The sum of zeroes equals their product: −b/a = c/a, so c = −b.
- One zero is known: substitute it into p(x) = 0 to find the unknown, then use the sum or product to get the other zero.
8. Zeroes and Coefficients of a Cubic Polynomial
A similar relationship holds for a cubic. Take p(x) = 2x3 − 5x2 − 14x + 8. You can check that p(4) = 0, p(−2) = 0 and p(1/2) = 0. A cubic has at most three zeroes, so 4, −2 and 1/2 are its zeroes. Here a = 2, b = −5, c = −14, d = 8.
- Sum of zeroes = 4 + (−2) + 1/2 = 5/2 = −(−5)/2 = −b/a
- Sum of products of zeroes taken two at a time = 4(−2) + (−2)(1/2) + (1/2)(4) = −8 − 1 + 2 = −7 = −14/2 = c/a
- Product of zeroes = 4 × (−2) × 1/2 = −4 = −8/2 = −d/a
In general, if α, β, γ (gamma) are the zeroes of ax3 + bx2 + cx + d, a ≠ 0, then
α + β + γ = −b/a
αβ + βγ + γα = c/a
αβγ = −d/a
Memory pattern. The signs alternate: minus, plus, minus. Divide the next coefficient after a by a each time: −b/a, then +c/a, then −d/a.
NCERT marks its cubic verification example (Example 5) with the note “Not from the examination point of view”. The three relations do appear in the chapter summary, so know them, but keep the focus of your revision on the quadratic relations.
Formula and Theorem Sheet
| Result | Statement | Remember |
|---|---|---|
| Linear, quadratic, cubic | Degree 1, 2, 3 | Leading coefficient a ≠ 0 |
| Zero of p(x) | k is a zero if p(k) = 0 | A zero is a real number |
| Zero of ax + b | −b/a | Exactly one zero |
| Geometrical meaning | Zeroes = x-coordinates where y = p(x) meets the x-axis | Ignore y-axis crossings |
| Shape of y = ax2 + bx + c | Parabola: opens upwards if a > 0, downwards if a < 0 | Two, one or no zeroes |
| Maximum number of zeroes | Degree n ⇒ at most n zeroes | Quadratic ≤ 2, cubic ≤ 3 |
| Sum of zeroes, quadratic | α + β = −b/a | Note the minus sign |
| Product of zeroes, quadratic | αβ = c/a | No minus sign |
| Quadratic with given S and P | k(x2 − Sx + P), k ≠ 0 | Choose k to clear fractions |
| Cubic: sum | α + β + γ = −b/a | Pattern: −, +, − |
| Cubic: pair sum | αβ + βγ + γα = c/a | Three products of two zeroes |
| Cubic: product | αβγ = −d/a | Note the minus sign |
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Important Definitions
- Degree of a polynomial: the highest power of the variable in the polynomial.
- Linear polynomial: a polynomial of degree 1, of the form ax + b with a ≠ 0.
- Quadratic polynomial: a polynomial of degree 2. Any quadratic polynomial in x is of the form ax2 + bx + c, where a, b, c are real numbers and a ≠ 0.
- Cubic polynomial: a polynomial of degree 3. Its most general form is ax3 + bx2 + cx + d, where a, b, c, d are real numbers and a ≠ 0.
- Value of a polynomial at x = k: the value obtained by replacing x by k in p(x), written p(k).
- Zero of a polynomial: a real number k is a zero of p(x) if p(k) = 0.
- Parabola: the curve that is the graph of y = ax2 + bx + c, a ≠ 0. It opens upwards if a > 0 and downwards if a < 0.
- Geometrical meaning of zeroes: the zeroes of p(x) are precisely the x-coordinates of the points where the graph of y = p(x) intersects the x-axis.
- α, β, γ: Greek letters read as alpha, beta and gamma, used to name the zeroes.
Solved Examples (NCERT-Based)
Example 1: Zeroes of x2 − 2x − 8 (NCERT Exercise 2.2, Q1 (i))
Question: Find the zeroes of the quadratic polynomial x2 − 2x − 8 and verify the relationship between the zeroes and the coefficients.
Solution: Two numbers with product −8 and sum −2 are −4 and 2.
x2 − 2x − 8 = x2 − 4x + 2x − 8 = x(x − 4) + 2(x − 4) = (x − 4)(x + 2)
So the zeroes are 4 and −2. Here a = 1, b = −2, c = −8.
Sum of zeroes = 4 + (−2) = 2, and −b/a = −(−2)/1 = 2.
Product of zeroes = 4 × (−2) = −8, and c/a = −8/1 = −8. Hence verified.
Example 2: Zeroes of 4s2 − 4s + 1 (NCERT Exercise 2.2, Q1 (ii))
Question: Find the zeroes of 4s2 − 4s + 1 and verify the relationship between the zeroes and the coefficients.
Solution: 4s2 − 4s + 1 = (2s)2 − 2(2s)(1) + 12 = (2s − 1)2 = (2s − 1)(2s − 1).
So the zeroes are 1/2 and 1/2 (two equal zeroes). Here a = 4, b = −4, c = 1.
Sum = 1/2 + 1/2 = 1, and −b/a = 4/4 = 1.
Product = 1/2 × 1/2 = 1/4, and c/a = 1/4. Hence verified.
The graph of y = 4s2 − 4s + 1 would touch the axis at one point, s = 1/2. This is case (ii).
Example 3: Zeroes of 6x2 − 3 − 7x (NCERT Exercise 2.2, Q1 (iii))
Question: Find the zeroes of 6x2 − 3 − 7x and verify the relationship between the zeroes and the coefficients.
Solution: Standard form: 6x2 − 7x − 3, so a = 6, b = −7, c = −3. We need two numbers with product 6 × (−3) = −18 and sum −7: these are −9 and 2.
6x2 − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3)
So the zeroes are −1/3 and 3/2.
Sum = 3/2 − 1/3 = 9/6 − 2/6 = 7/6, and −b/a = 7/6.
Product = (3/2)(−1/3) = −1/2, and c/a = −3/6 = −1/2. Hence verified.
Example 4: Zeroes of 4u2 + 8u (NCERT Exercise 2.2, Q1 (iv))
Question: Find the zeroes of 4u2 + 8u and verify the relationship between the zeroes and the coefficients.
Solution: 4u2 + 8u = 4u(u + 2). The zeroes are 0 and −2. Here a = 4, b = 8, c = 0.
Sum = 0 + (−2) = −2, and −b/a = −8/4 = −2.
Product = 0 × (−2) = 0, and c/a = 0/4 = 0. Hence verified.
Example 5: Zeroes of t2 − 15 (NCERT Exercise 2.2, Q1 (v))
Question: Find the zeroes of t2 − 15 and verify the relationship between the zeroes and the coefficients.
Solution: t2 − 15 = t2 − (√15)2 = (t − √15)(t + √15). The zeroes are √15 and −√15. Here a = 1, b = 0, c = −15.
Sum = √15 + (−√15) = 0, and −b/a = −0/1 = 0.
Product = (√15)(−√15) = −15, and c/a = −15. Hence verified.
Example 6: Zeroes of 3x2 − x − 4 (NCERT Exercise 2.2, Q1 (vi))
Question: Find the zeroes of 3x2 − x − 4 and verify the relationship between the zeroes and the coefficients.
Solution: Product ac = 3 × (−4) = −12 and sum b = −1: the numbers are −4 and 3.
3x2 − 4x + 3x − 4 = x(3x − 4) + 1(3x − 4) = (x + 1)(3x − 4)
So the zeroes are −1 and 4/3.
Sum = −1 + 4/3 = 1/3, and −b/a = −(−1)/3 = 1/3.
Product = (−1)(4/3) = −4/3, and c/a = −4/3. Hence verified.
Example 7: Quadratics from sum and product, part A (NCERT Exercise 2.2, Q2 (i) to (iii))
Question: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively: (i) 1/4, −1 (ii) √2, 1/3 (iii) 0, √5.
Solution: Use k(x2 − Sx + P).
(i) x2 − (1/4)x − 1. Taking k = 4: 4x2 − x − 4. Check: −b/a = 1/4, c/a = −4/4 = −1.
(ii) x2 − √2 x + 1/3. Taking k = 3: 3x2 − 3√2 x + 1. Check: −b/a = 3√2/3 = √2, c/a = 1/3.
(iii) x2 − 0 × x + √5 = x2 + √5. Check: −b/a = 0, c/a = √5.
Example 8: Quadratics from sum and product, part B (NCERT Exercise 2.2, Q2 (iv) to (vi))
Question: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively: (iv) 1, 1 (v) −1/4, 1/4 (vi) 4, 1.
Solution:
(iv) x2 − x + 1. Check: −b/a = 1, c/a = 1. (x2 − x + 1 has no real zeroes, since its graph stays above the x-axis. The question only fixes −b/a and c/a, and both equal 1 here.)
(v) x2 − (−1/4)x + 1/4 = x2 + (1/4)x + 1/4. Taking k = 4: 4x2 + x + 1. Check: −b/a = −1/4, c/a = 1/4.
(vi) x2 − 4x + 1. Check: −b/a = 4, c/a = 1.
Example 9: Finding k from a condition on the zeroes
Question: Find k if the sum of the zeroes of x2 − (k + 6)x + 2(2k − 1) is equal to half of their product.
Solution: Here a = 1, b = −(k + 6), c = 2(2k − 1).
Sum = −b/a = k + 6. Product = c/a = 2(2k − 1) = 4k − 2, so half of the product = 2k − 1.
k + 6 = 2k − 1 ⇒ k = 7.
Check: with k = 7, the polynomial is x2 − 13x + 26. Sum = 13 and product = 26, and 13 is half of 26.
Example 10: Expressions in α and β, and a new quadratic
Question: If α and β are the zeroes of 2x2 − 4x + 1, find (i) α2 + β2 (ii) 1/α + 1/β (iii) a quadratic polynomial whose zeroes are 1/α and 1/β.
Solution: α + β = −(−4)/2 = 2 and αβ = 1/2.
(i) α2 + β2 = (α + β)2 − 2αβ = 4 − 1 = 3.
(ii) 1/α + 1/β = (α + β)/αβ = 2 ÷ (1/2) = 4.
(iii) New sum S = 1/α + 1/β = 4. New product P = (1/α)(1/β) = 1/αβ = 2. Required polynomial: x2 − 4x + 2 (or any non-zero multiple).
No square roots were needed, even though the zeroes of 2x2 − 4x + 1 are irrational.
Example 11: Verifying the zeroes of a cubic (NCERT Example 5)
Question: Verify that 3, −1, −1/3 are the zeroes of the cubic polynomial p(x) = 3x3 − 5x2 − 11x − 3, and then verify the relationship between the zeroes and the coefficients.
Solution: Comparing with ax3 + bx2 + cx + d: a = 3, b = −5, c = −11, d = −3.
p(3) = 3 × 27 − 5 × 9 − 11 × 3 − 3 = 81 − 45 − 33 − 3 = 0
p(−1) = 3(−1) − 5(1) − 11(−1) − 3 = −3 − 5 + 11 − 3 = 0
p(−1/3) = 3(−1/27) − 5(1/9) − 11(−1/3) − 3 = −1/9 − 5/9 + 11/3 − 3 = −2/3 + 11/3 − 9/3 = 0
So α = 3, β = −1, γ = −1/3.
α + β + γ = 3 − 1 − 1/3 = 5/3 = −(−5)/3 = −b/a
αβ + βγ + γα = (3)(−1) + (−1)(−1/3) + (−1/3)(3) = −3 + 1/3 − 1 = −11/3 = c/a
αβγ = 3 × (−1) × (−1/3) = 1 = −(−3)/3 = −d/a
All three relations hold. (NCERT marks this example “Not from the examination point of view”.)
Competency-Based Questions (with answers)
1. Case-based: The fountain jet
In a garden, the path of a water jet is modelled by y = −x2 + 6x − 8, where x is the horizontal distance in metres and y is the height in metres above the water surface of the pool (the x-axis). The jet leaves the water surface and falls back into it.
(a) Does the curve open upwards or downwards? Why? Downwards, because the coefficient of x2 is −1, which is less than 0.
(b) At what horizontal distances does the jet meet the water surface? Solve −x2 + 6x − 8 = 0, that is x2 − 6x + 8 = 0, (x − 2)(x − 4) = 0. The jet meets the surface at x = 2 m and x = 4 m. These are the zeroes of the polynomial.
(c) Verify the sum and product of these zeroes from the coefficients. a = −1, b = 6, c = −8. −b/a = −6/(−1) = 6 = 2 + 4. c/a = −8/(−1) = 8 = 2 × 4.
2. Case-based: The rectangular plot
A rectangular plot has sides (x + 3) m and (2x − 1) m.
(a) Write the area A(x) as a polynomial and name its type. A(x) = (x + 3)(2x − 1) = 2x2 − x + 6x − 3 = 2x2 + 5x − 3. It is a quadratic polynomial.
(b) Find the zeroes of A(x). From the factors, x = −3 and x = 1/2.
(c) Verify the relationship. Sum = −3 + 1/2 = −5/2 = −b/a. Product = (−3)(1/2) = −3/2 = c/a.
(d) For which values of x does the model describe a real plot? Both sides must be positive: x + 3 > 0 and 2x − 1 > 0. So x > 1/2. At the zero x = 1/2 one side has length 0.
3. Source-based: Reading a table of values
A student makes this table for y = x3 − 4x: at x = −2, −1, 0, 1, 2 the values of y are 0, 3, 0, −3, 0.
(a) Name the zeroes shown by the table. −2, 0 and 2, since y = 0 there.
(b) Can the polynomial have any other zero? Give a reason. No. It is a cubic, so it has at most three zeroes, and three have been found.
(c) Verify the cubic relations (practice only: NCERT marks cubic verification as not from the examination point of view). a = 1, b = 0, c = −4, d = 0. Sum = −2 + 0 + 2 = 0 = −b/a. Pair sum = (−2)(0) + (0)(2) + (2)(−2) = −4 = c/a. Product = 0 = −d/a.
4. Assertion-Reason
Assertion (A): The polynomial x2 + 1 has no zero.
Reason (R): The graph of y = x2 + 1 lies completely above the x-axis.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a). x2 ≥ 0, so x2 + 1 ≥ 1 for every real x. The graph never meets the x-axis, which is exactly why there is no zero.
5. Assertion-Reason
Assertion (A): The polynomial x3 − x has four zeroes.
Reason (R): A polynomial of degree n has at most n zeroes.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d). x3 − x = x(x − 1)(x + 1) has exactly three zeroes: −1, 0, 1. A cubic cannot have four zeroes, so A is false. R is the correct general rule.
6. Assertion-Reason
Assertion (A): x2 − 5x + 6 and 3x2 − 15x + 18 have the same zeroes.
Reason (R): Multiplying a polynomial by a non-zero constant does not change its zeroes.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a). 3x2 − 15x + 18 = 3(x2 − 5x + 6), and both have zeroes 2 and 3. R gives the reason: if p(k) = 0 then 3p(k) = 0 too.
7. Error analysis: The wrong sign
Asked for a quadratic polynomial whose zeroes have sum 3 and product −10, a student writes x2 + 3x − 10. Find the error and correct it.
Answer: The middle coefficient must be the negative of the sum. The student’s polynomial has −b/a = −3, so its zeroes (−5 and 2) add up to −3. The correct answer is x2 − 3x − 10 = (x − 5)(x + 2), with zeroes 5 and −2: sum 3, product −10.
8. Competency MCQ: Matching zeroes to a polynomial
Which polynomial has zeroes −2 and 5?
(a) x2 − 3x − 10 (b) x2 + 3x − 10 (c) x2 − 7x + 10 (d) x2 + 7x + 10
Answer: (a). Sum = 3 and product = −10, so the polynomial is x2 − 3x − 10. Check: (−2)2 − 3(−2) − 10 = 4 + 6 − 10 = 0.
Important Questions for Board Exams
1-Mark Questions
Q1. If one zero of 5x2 + 13x + k is the reciprocal of the other, find k.
Product of zeroes = 1, so k/5 = 1 and k = 5.
Q2. The graph of y = p(x), where p(x) is a quadratic, touches the x-axis at exactly one point. How many zeroes does p(x) have?
One zero (the two zeroes are equal). The x-coordinate of the touching point is that zero.
Q3. If α and β are the zeroes of x2 − x − 6, find α2β + αβ2.
α + β = 1, αβ = −6. α2β + αβ2 = αβ(α + β) = −6 × 1 = −6.
Q4. What is the maximum number of zeroes a cubic polynomial can have?
Three, because a polynomial of degree n has at most n zeroes.
Q5. Write a quadratic polynomial whose zeroes are −3 and 4.
S = 1, P = −12, so x2 − x − 12.
2-Mark Questions
Q6. Find the zeroes of x2 + 2√2 x − 6 and verify the relationship between the zeroes and the coefficients.
Split: product −6, sum 2√2: the terms are 3√2 x and −√2 x.
x2 + 3√2 x − √2 x − 6 = x(x + 3√2) − √2(x + 3√2) = (x − √2)(x + 3√2).
Zeroes: √2 and −3√2. Sum = −2√2 = −b/a. Product = −3 × 2 = −6 = c/a. Hence verified.
Q7. Find a quadratic polynomial whose zeroes have sum 2 and product −1/3.
x2 − 2x − 1/3. Multiply by 3: 3x2 − 6x − 1. Check: −b/a = 6/3 = 2, c/a = −1/3.
Q8. If 2 is a zero of x2 + 3x + k, find k and the other zero.
p(2) = 4 + 6 + k = 0, so k = −10. Product = −10, so the other zero is −10/2 = −5. Check: sum 2 + (−5) = −3 = −b/a.
Q9. α and β are the zeroes of x2 − 6x + k, and 3α + 2β = 20. Find k.
α + β = 6. Then 3α + 2β = 2(α + β) + α = 12 + α = 20, so α = 8 and β = −2. k = αβ = −16.
3-Mark Questions
Q10. Find the zeroes of 4√3 x2 + 5x − 2√3 and verify the relationship between the zeroes and the coefficients.
ac = 4√3 × (−2√3) = −24 and b = 5, so split 5x as 8x − 3x.
4√3 x2 + 8x − 3x − 2√3 = 4x(√3 x + 2) − √3(√3 x + 2) = (4x − √3)(√3 x + 2).
Zeroes: √3/4 and −2/√3 = −2√3/3.
Sum = √3/4 − 2√3/3 = (3√3 − 8√3)/12 = −5√3/12, and −b/a = −5/(4√3) = −5√3/12.
Product = (√3/4)(−2/√3) = −1/2, and c/a = −2√3/(4√3) = −1/2. Hence verified.
Q11. If α and β are the zeroes of x2 − 3x + 1, find a quadratic polynomial whose zeroes are α2 and β2.
α + β = 3, αβ = 1.
New sum = α2 + β2 = 9 − 2 = 7. New product = α2β2 = (αβ)2 = 1.
Required polynomial: x2 − 7x + 1.
Q12. The difference of the zeroes of x2 − 7x + k is 3. Find k and the zeroes.
Let the zeroes be α and β with α > β. α + β = 7 and α − β = 3. Adding: 2α = 10, so α = 5 and β = 2. k = αβ = 10. The zeroes are 5 and 2.
5-Mark Questions
Q13. If α and β are the zeroes of p(x) = 3x2 − 5x − 2, (i) find the zeroes and verify the relationship between the zeroes and the coefficients, (ii) find α/β + β/α, (iii) find a quadratic polynomial whose zeroes are 2α and 2β.
(i) a = 3, b = −5, c = −2. Two numbers with product 3 × (−2) = −6 and sum −5 are −6 and 1.
3x2 − 6x + x − 2 = 3x(x − 2) + 1(x − 2) = (3x + 1)(x − 2). The zeroes are 2 and −1/3.
Sum = 2 − 1/3 = 5/3, and −b/a = −(−5)/3 = 5/3.
Product = 2 × (−1/3) = −2/3, and c/a = −2/3. Hence verified.
(ii) α2 + β2 = (α + β)2 − 2αβ = 25/9 + 4/3 = 25/9 + 12/9 = 37/9.
α/β + β/α = (α2 + β2)/αβ = (37/9) ÷ (−2/3) = (37/9) × (−3/2) = −37/6.
Check with the zeroes: 2 ÷ (−1/3) + (−1/3) ÷ 2 = −6 − 1/6 = −37/6.
(iii) New sum = 2(α + β) = 10/3. New product = 4αβ = −8/3.
x2 − (10/3)x − 8/3. Multiply by 3: 3x2 − 10x − 8. Check: 3x2 − 10x − 8 = (3x + 2)(x − 4), with zeroes 4 = 2 × 2 and −2/3 = 2 × (−1/3).
Q14. If α and β are the zeroes of 2x2 − 5x + 1, find a quadratic polynomial whose zeroes are 2α + 3β and 3α + 2β.
α + β = 5/2, αβ = 1/2.
New sum = (2α + 3β) + (3α + 2β) = 5(α + β) = 25/2.
New product = (2α + 3β)(3α + 2β) = 6α2 + 4αβ + 9αβ + 6β2 = 6(α2 + β2) + 13αβ.
Now 6(α2 + β2) = 6[(α + β)2 − 2αβ] = 6(α + β)2 − 12αβ, so the product = 6(α + β)2 + αβ = 6 × 25/4 + 1/2 = 75/2 + 1/2 = 38.
Polynomial: x2 − (25/2)x + 38. Multiply by 2: 2x2 − 25x + 76.
Common Mistakes and Examiner Tips
- Reading coefficients from the unsorted form. In 6x2 − 3 − 7x, students take b = −3. Fix: rewrite as 6x2 − 7x − 3 first, so b = −7 and c = −3.
- Dropping the minus in the sum. Writing α + β = b/a. Fix: the sum is −b/a. For x2 − 2x − 8 the sum is +2.
- Adding a minus to the product. Writing αβ = −c/a for a quadratic. Fix: the product is c/a. The minus sign on the product belongs to the cubic, αβγ = −d/a.
- Wrong sign in the formed quadratic. Writing x2 + Sx + P. Fix: use x2 − Sx + P and check the answer with −b/a.
- Verifying only one side. Writing “sum = 4” and stopping. Fix: compute the sum from the zeroes and −b/a from the coefficients separately, show both, then write “Hence verified”.
- Losing the zero 0. For 4u2 + 8u, dividing by u gives only u = −2. Fix: factorise as 4u(u + 2), which gives both zeroes 0 and −2.
- Giving only the positive root. For t2 − 15, writing only √15. Fix: t2 − 15 = (t − √15)(t + √15), so −√15 is also a zero.
- Counting y-axis crossings as zeroes. Fix: zeroes are only where the graph meets the x-axis. A point where the curve touches the x-axis and turns back still counts once.
- Claiming more zeroes than the degree allows. Fix: a quadratic has at most two zeroes and a cubic at most three. If your count from a sketch exceeds the degree, recheck.
- Finding irrational zeroes when you do not need them. For α2 + β2, students solve the quadratic and square surds. Fix: use (α + β)2 − 2αβ. It is faster and has fewer places to slip.
- Sign slips when substituting negatives. p(−1) for 3x3 − 5x2 gives −3 − 5, not −3 + 5. Fix: always put negative numbers and fractions in brackets before you square or cube them.
Quick Revision Points
- The degree of a polynomial is the highest power of its variable.
- Degrees 1, 2 and 3 give linear, quadratic and cubic polynomials.
- General quadratic: ax2 + bx + c, with a, b, c real and a ≠ 0.
- General cubic: ax3 + bx2 + cx + d, with a, b, c, d real and a ≠ 0.
- k is a zero of p(x) if p(k) = 0.
- The zero of ax + b is −b/a.
- Zeroes are the x-coordinates of the points where y = p(x) meets the x-axis.
- The graph of a quadratic is a parabola: upwards if a > 0, downwards if a < 0.
- A quadratic has two distinct zeroes, two equal zeroes, or no zero.
- A polynomial of degree n has at most n zeroes.
- x3 − 4x has zeroes −2, 0, 2; x3 has only 0; x3 − x2 has 0 and 1.
- For ax2 + bx + c: α + β = −b/a and αβ = c/a.
- Quadratic with zero-sum S and zero-product P: k(x2 − Sx + P), k ≠ 0.
- Multiplying by a non-zero constant does not change the zeroes.
- For ax3 + bx2 + cx + d: α + β + γ = −b/a, αβ + βγ + γα = c/a, αβγ = −d/a.
- α2 + β2 = (α + β)2 − 2αβ and 1/α + 1/β = (α + β)/αβ.
- Reciprocal zeroes: c = a. Zeroes that are negatives of each other: b = 0. One zero equal to 0: c = 0.
- Plotting graphs of quadratic or cubic polynomials is not evaluated; Exercise 2.1 practises reading given graphs.
Weightage in Board Exams
Polynomials belongs to the Algebra unit of Class 10 Maths, together with Pair of Linear Equations in Two Variables, Quadratic Equations and Arithmetic Progressions. Check the current CBSE course structure for the marks given to the unit. Within it, this chapter is tested in a predictable way.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Number of zeroes from a described or drawn graph; sum or product of zeroes; the polynomial with given zeroes; k from a reciprocal or given-zero condition |
| Short answers | Find the zeroes of a quadratic and verify the relationship (including surd coefficients); form a quadratic from a given sum and product |
| Longer answers | Expressions in α and β; a new quadratic whose zeroes are built from α and β; finding unknown coefficients from conditions on the zeroes |
| Case-based | A curved path or an area written as a quadratic, split into parts on zeroes, shape and the relationship with coefficients |
Practise every NCERT exercise question in full, then the α and β identities and the sign rules.
Class 10 Maths · Chapter 2 – swipe through all 9 cards to understand the whole chapter.
Linear, quadratic, cubic
A polynomial is named by its degree, the highest power of the variable.
Rewrite in descending powers before reading a, b, c.
- Degree 1 is linear, degree 2 is quadratic, degree 3 is cubic.
- 2x + 5 − x2 is quadratic because its highest power is 2.
- 1/(x − 1) and √x + 2 are not polynomials.
When p(k) = 0
A real number k is a zero of p(x) if the value of p(x) at x = k is 0.
Put negatives and fractions in brackets when substituting.
- For x2 − 3x − 4: p(2) = −6 but p(−1) = 0 and p(4) = 0.
- So −1 and 4 are the zeroes of x2 − 3x − 4.
- The zero of 2x + 3 is −3/2.
Zeroes live on the x-axis
Zeroes are the x-coordinates of the points where y = p(x) meets the x-axis.
Plotting quadratic or cubic graphs is not evaluated; Exercise 2.1 practises reading given graphs.
- A line y = ax + b meets the x-axis once, at (−b/a, 0).
- y = x3 − 4x meets the x-axis at −2, 0 and 2.
- Crossings of the y-axis are never zeroes.
Two, one or no zero
The graph of ax2 + bx + c is a parabola: it opens up for positive a and down for negative a.
A quadratic has at most two zeroes.
- Cuts the x-axis at two points: two distinct zeroes.
- Touches the x-axis at one point: two equal zeroes.
- Stays above or below the x-axis: no zero, as for x2 + 1.
Zeroes vs coefficients
For ax2 + bx + c with zeroes α and β, the coefficients give the sum and product.
Minus sign on the sum only, never on the product.
- 2x2 − 8x + 6 = 2(x − 1)(x − 3): sum 4 = 8/2, product 3 = 6/2.
- 3x2 + 5x − 2 has zeroes 1/3 and −2: sum −5/3, product −2/3.
- Proof: compare ax2 + bx + c with k(x − α)(x − β).
The Exercise 2.2 routine
Factorise, read the zeroes, then check the sum and product against the coefficients.
Show both sides of each check, then write Hence verified.
- 6x2 − 3 − 7x becomes 6x2 − 7x − 3 = (3x + 1)(2x − 3).
- 4u2 + 8u = 4u(u + 2) gives zeroes 0 and −2.
- t2 − 15 has zeroes √15 and −√15, sum 0, product −15.
From sum and product
Knowing the sum S and product P of the zeroes fixes the quadratic up to a constant.
Choose k to clear fractions from the answer.
- Sum −3, product 2 gives x2 + 3x + 2.
- Sum 1/4, product −1 gives 4x2 − x − 4.
- Sum √2, product 1/3 gives 3x2 − 3√2x + 1.
Work with α + β and αβ
Rewrite any symmetric expression in α and β using only their sum and product.
No need to find irrational zeroes.
- 1/α + 1/β = (α + β)/αβ.
- For x2 − 5x + 3: α2 + β2 = 25 − 6 = 19.
- Reciprocal zeroes mean αβ = 1, so c = a.
Three zeroes, three relations
For ax3 + bx2 + cx + d with zeroes α, β, γ the signs alternate minus, plus, minus.
NCERT marks its cubic verification example not from the examination point of view, so keep this light.
- 2x3 − 5x2 − 14x + 8 has zeroes 4, −2 and 1/2.
- Sum 5/2 = 5/2 and pair sum −7 = −14/2.
- Product −4 = −8/2.
📝 Practice Polynomials - 10 board questions
CBSE previous-year and competency-based · with answers & explanations
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Why B: y² + 4y + 3 = (y + 1)(y + 3), so its zeroes are −1 and −3; the Assertion is true. The Reason is also true, since a polynomial of degree 2 has at most two zeroes. But “at most two” only sets an upper limit; it does not tell us that this polynomial actually has two. That comes from factorising it (or from b² − 4ac = 16 − 12 = 4 > 0). So R does not explain A.
Why not A: R allows zero, one or two zeroes, so it cannot be the reason this one has exactly two.
Why not C: R is a true statement about every quadratic.
Why not D: A is true, since −1 and −3 both make p(y) = 0.
Remember: An upper limit (at most two) never proves that a particular quadratic has two zeroes.
Why B: (x + 3)(−x + 8) = −(x + 3)(x − 8). The factor x + 3 is 0 at x = −3 and −x + 8 is 0 at x = 8, so the zeroes are exactly −3 and 8. The constant multiple −1 does not change the zeroes.
Why not A: x² + 5x − 4 has product of zeroes −4, but (−3) × 8 = −24.
Why not C: a(x² + 5x − 24) = a(x + 8)(x − 3) has zeroes −8 and 3; the sum was used with the wrong sign, x² + Sx + P instead of x² − Sx + P.
Why not D: x² − 24 has zeroes ±√24, and its sum of zeroes is 0, not 5.
Remember: Zeroes α and β give k(x − α)(x − β), which expands to k(x² − (α + β)x + αβ).
Why B: Step 1 (sum of zeroes = −b/a): α + β = 3√2. Step 2 (product of zeroes = c/a): αβ = 4. Step 3 (find the pair): 2√2 + √2 = 3√2 and 2√2 × √2 = 2 × 2 = 4, so the zeroes are 2√2 and √2. In factor form, x² − 3√2x + 4 = (x − 2√2)(x − √2).
Why not A: 2 and √2 multiply to 2√2, not 4, and add to 2 + √2, not 3√2.
Why not C: 4√2 and −√2 add to 3√2 but multiply to −8, not 4.
Why not D: √2 and 2 is the same wrong pair as A in a different order.
Remember: Always check both the sum and the product; one match is not enough.
Why D: Step 1 (sum of zeroes = −b/a): α + β = −6/3 = −2. Step 2 (product of zeroes = c/a): αβ = k/3. Step 3 (solve the given condition): −2 + k/3 = −2/3, so k/3 = 2 − 2/3 = 4/3, and k = 4.
Why not A: −8 comes from taking α + β = +2, giving 2 + k/3 = −2/3 and k = −8.
Why not B: 8 is −8 with its sign also lost; check: with k = 8, −2 + 8/3 = 2/3, not −2/3.
Why not C: −4 is the right size with the wrong sign, a slip when moving −2 to the other side.
Remember: Write α + β and αβ first, substitute them into the given condition, then solve for k.
Why B: Step 1 (zero means p(−4) = 0): (−4)² − (−4) − (2 + 2k) = 0. Step 2 (simplify): 16 + 4 − 2 − 2k = 0, so 18 − 2k = 0. Step 3 (solve): k = 9.
Why not A: with k = 3, p(−4) = 18 − 6 = 12, not 0, so −4 would not be a zero.
Why not C: with k = 6, p(−4) = 18 − 12 = 6, not 0; a common slip is reading −(−4) as −4.
Why not D: −9 is the right size but with the sign reversed while solving 18 − 2k = 0.
Remember: Substitute the zero into p(x), set it equal to 0, and watch each minus sign.
Why C: For x² − ax − b, the coefficients are 1, −a and −b. Sum of zeroes = −(−a)/1 = a. Product of zeroes = −b/1 = −b. So α + β + αβ = a + (−b) = a − b.
Why not A: a + b takes the product as +b, dropping the minus sign of the constant term.
Why not B: −a − b takes the sum as −a, forgetting the minus in −b/a.
Why not D: −a + b reverses both signs, sum as −a and product as +b.
Remember: Sum = −(coefficient of x)/(coefficient of x²); product = constant/(coefficient of x²).
Why B: Step 1 (substitute x = 1): p(1) = 1 − 5 + 6 = 2. Step 2 (substitute x = 4): p(4) = 16 − 20 + 6 = 2. Step 3 (add): p(1) + p(4) = 2 + 2 = 4.
Why not A: 0 would need both values to be zeroes, but the zeroes of this polynomial are 2 and 3, not 1 and 4.
Why not C: 2 is only one of the two values; the second one was not added.
Why not D: −4 comes from a sign slip, such as taking p(1) = −2 and p(4) = −2.
Remember: Substitute the number in every term, keep every sign, then simplify.
Why A: Step 1 (split the middle term): ac = 2 × (−9) = −18 = −6 × 3 and −6 + 3 = −3, so 2x² − 6x + 3x − 9 = 2x(x − 3) + 3(x − 3) = (x − 3)(2x + 3). Step 2 (zero means p(x) = 0): x − 3 = 0 gives 3, and 2x + 3 = 0 gives −3/2. Check: sum 3 − 3/2 = 3/2 = −b/a.
Why not B: −3 and −3/2 changes the sign of the first zero; p(−3) = 18 + 9 − 9 = 18, not 0.
Why not C: −3 and 3/2 has both signs reversed; these are zeroes of 2x² + 3x − 9.
Why not D: 3 and 3/2 has a product of 9/2, but the product must be c/a = −9/2.
Remember: After factorising, set each factor to 0 and check the sum and product.
Why D: For x² + 3x + 3, b² − 4ac = 9 − 12 = −3, which is negative, so p(x) = 0 has no real solution; also p(x) = (x + 3/2)² + 3/4 is always positive, so its graph never meets the x-axis. The Assertion is false. The Reason is a true general fact: a polynomial of degree 2 can have at most 2 zeroes.
Why not A: A is false, so R cannot explain it.
Why not B: B needs both statements true, but A is false.
Why not C: R is true; “at most two” is exactly what the degree rule says.
Remember: At most two zeroes for a quadratic, and it can have two, one or none.
Why B: A polynomial with zeroes −3 and 5 is k(x + 3)(x − 5) = k(x² − 2x − 15). For every non-zero real k, such as 1, 2, −1 or 1/2, this gives a different polynomial with the same zeroes −3 and 5. There are infinitely many choices of k, so infinitely many polynomials. Polynomials of higher degree such as (x + 3)(x − 5)(x² + 1) also work.
Why not A: x² − 2x − 15 is only the simplest one, with k = 1.
Why not C: Two would count only k = 1 and k = −1, but any non-zero k works.
Why not D: The zeroes fix the factors, not the number of polynomials; there is no upper limit of two.
Remember: Zeroes decide the polynomial only up to a non-zero constant multiple k.
Chapter Navigation
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Related Chapters in Class 10 Maths
- Real Numbers Class 10 Notes
- Pair of Linear Equations in Two Variables Class 10 Notes
- Quadratic Equations Class 10 Notes
- Arithmetic Progressions Class 10 Notes
- Triangles Class 10 Notes
- Coordinate Geometry Class 10 Notes
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Frequently Asked Questions
A real number k is a zero of a polynomial p(x) if p(k) = 0. For p(x) = x² − 3x − 4, p(−1) = 1 + 3 − 4 = 0 and p(4) = 16 − 12 − 4 = 0, so −1 and 4 are its zeroes. The zero of a linear polynomial ax + b is −b/a.
The zeroes of p(x) are exactly the x-coordinates of the points where the graph of y = p(x) meets the x-axis. So to count zeroes from a graph, count the distinct points where the curve meets the x-axis, including points where it only touches and turns back. Crossings of the y-axis do not count.
At most two. Its graph is a parabola, which can cut the x-axis at two distinct points (two zeroes), touch it at one point (two equal zeroes, that is one zero), or stay completely above or below it (no zero). In general, a polynomial of degree n has at most n zeroes, so a cubic has at most three.
If α and β are the zeroes of ax² + bx + c, a ≠ 0, then α + β = −b/a and αβ = c/a. In words, the sum of zeroes is minus the coefficient of x divided by the coefficient of x², and the product is the constant term divided by the coefficient of x². For 2x² − 8x + 6 the zeroes 1 and 3 give sum 4 = 8/2 and product 3 = 6/2.
Use k(x² − Sx + P), where S is the sum, P is the product and k is any non-zero real number. For sum −3 and product 2 the answer is x² + 3x + 2. If S or P is a fraction, choose k to clear it: sum 1/4 and product −1 give 4x² − x − 4. Note the minus sign in front of S.
Multiplying a polynomial by a non-zero constant does not change the values of x that make it zero. So x² + 3x + 2, 2x² + 6x + 4 and −x² − 3x − 2 all have the zeroes −1 and −2. Every quadratic with those zeroes is of the form k(x² + 3x + 2). Giving any one correct polynomial is enough.
If α, β, γ are the zeroes of ax³ + bx² + cx + d, a ≠ 0, then α + β + γ = −b/a, αβ + βγ + γα = c/a and αβγ = −d/a. The signs alternate minus, plus, minus. For 2x³ − 5x² − 14x + 8 with zeroes 4, −2 and 1/2: sum 5/2, pair sum −7 and product −4. NCERT marks its cubic verification example as not from the examination point of view.
Use α² + β² = (α + β)² − 2αβ and substitute α + β = −b/a and αβ = c/a. For 2x² − 4x + 1, α + β = 2 and αβ = 1/2, so α² + β² = 4 − 1 = 3. Other useful forms are 1/α + 1/β = (α + β)/αβ and (α − β)² = (α + β)² − 4αβ.
