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Chapter 2

Polynomials

Class 10 Maths  ·  NCERT Solutions  ·  5 Examples + 3 Questions

Chapter 1: Real Numbers All Chapters Chapter 3: Pair of Linear Equations
5 Examples + 3 Questions Solved
Chapter 2: Polynomials — NCERT Solutions (rationalised syllabus 2026–27), with all 5 worked Examples from the chapter plus every Exercise question. This chapter covers zeroes of a polynomial and their link to the graph of y = p(x) (a degree-n polynomial has at most n zeroes), and the relationship between zeroes and coefficients: for a quadratic \(ax^2+bx+c\) with zeroes \(\alpha,\beta\): \(\alpha+\beta=\dfrac{-b}{a}\), \(\alpha\beta=\dfrac{c}{a}\); for a cubic \(ax^3+bx^2+cx+d\) with zeroes \(\alpha,\beta,\gamma\): \(\alpha+\beta+\gamma=\dfrac{-b}{a}\), \(\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{c}{a}\), \(\alpha\beta\gamma=\dfrac{-d}{a}\).

Worked Examples (Textbook)

Example 1

Look at the graphs below. Each is the graph of y = p(x), where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x).

(i)(ii)(iii)(iv)(v)(vi)
Fig. 2.9 — each graph is y = p(x) for some polynomial p(x)
Solution

The number of zeroes of p(x) equals the number of points where its graph intersects the x-axis (a point where the curve only touches the axis, as in (v), still counts as one zero, a repeated one).

(i)1 zero — the graph crosses the x-axis at exactly one point.
(ii)2 zeroes — the graph crosses the x-axis at two points.
(iii)3 zeroes — the graph crosses the x-axis at three points.
(iv)1 zero — a straight-line graph crosses the x-axis exactly once.
(v)1 zero — the graph only touches the x-axis (a repeated zero), it does not cross it.
(vi)4 zeroes — the graph crosses the x-axis at four points.
Example 2

Find the zeroes of the quadratic polynomial \(x^{2}+7x+10\), and verify the relationship between the zeroes and the coefficients.

Solution

Splitting the middle term:

\(x^{2}+7x+10=(x+2)(x+5)\)

So p(x) = 0 when x + 2 = 0 or x + 5 = 0, i.e. x = −2 or x = −5. The zeroes are −2 and −5.

\(\text{sum of zeroes} = -2+(-5) = -7 = \dfrac{-(7)}{1} = \dfrac{-(\text{coeff. of }x)}{\text{coeff. of }x^{2}}\)
\(\text{product of zeroes} = (-2)\times(-5) = 10 = \dfrac{10}{1} = \dfrac{\text{constant term}}{\text{coeff. of }x^{2}}\)
Example 3

Find the zeroes of the polynomial \(x^{2}-3\) and verify the relationship between the zeroes and the coefficients.

Solution

Using the identity \(a^{2}-b^{2}=(a-b)(a+b)\):

\(x^{2}-3 = \left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)

So p(x) = 0 when \(x=\sqrt{3}\) or \(x=-\sqrt{3}\). The zeroes are \(\sqrt{3}\) and \(-\sqrt{3}\).

\(\text{sum of zeroes} = \sqrt{3}-\sqrt{3} = 0 = \dfrac{-(\text{coeff. of }x)}{\text{coeff. of }x^{2}}\)
\(\text{product of zeroes} = \left(\sqrt{3}\right)\left(-\sqrt{3}\right) = -3 = \dfrac{-3}{1} = \dfrac{\text{constant term}}{\text{coeff. of }x^{2}}\)
Example 4

Find a quadratic polynomial, the sum and product of whose zeroes are −3 and 2, respectively.

Solution

Let the quadratic polynomial be \(ax^{2}+bx+c\) with zeroes \(\alpha,\beta\). We have:

\(\alpha+\beta = -3 = \dfrac{-b}{a}\)
\(\alpha\beta = 2 = \dfrac{c}{a}\)

Taking a = 1 gives b = 3 and c = 2. So one quadratic polynomial that fits is:

\(x^{2}+3x+2\)

Any other polynomial that fits will be of the form \(k(x^{2}+3x+2)\), where k is a non-zero real number.

Example 5

Verify that 3, −1, \(-\dfrac{1}{3}\) are the zeroes of the cubic polynomial \(p(x)=3x^{3}-5x^{2}-11x-3\), and then verify the relationship between the zeroes and the coefficients.
* Not from the examination point of view (NCERT).

Solution

Comparing with \(ax^{3}+bx^{2}+cx+d\): a = 3, b = −5, c = −11, d = −3. Checking each value:

\(p(3) = 3(3)^{3}-5(3)^{2}-11(3)-3 = 81-45-33-3 = 0\)
\(p(-1) = 3(-1)^{3}-5(-1)^{2}-11(-1)-3 = -3-5+11-3 = 0\)
\(p\!\left(-\dfrac{1}{3}\right) = -\dfrac{1}{9}-\dfrac{5}{9}+\dfrac{11}{3}-3 = -\dfrac{2}{3}+\dfrac{2}{3} = 0\)

So 3, −1 and \(-\tfrac{1}{3}\) are indeed the zeroes. Take \(\alpha=3,\ \beta=-1,\ \gamma=-\dfrac{1}{3}\). Now:

\(\alpha+\beta+\gamma = 3+(-1)+\left(-\dfrac{1}{3}\right) = \dfrac{5}{3} = \dfrac{-(-5)}{3} = \dfrac{-b}{a}\)
\(\alpha\beta+\beta\gamma+\gamma\alpha = 3(-1)+(-1)\!\left(-\dfrac{1}{3}\right)+\left(-\dfrac{1}{3}\right)3 = -\dfrac{11}{3} = \dfrac{c}{a}\)
\(\alpha\beta\gamma = 3(-1)\!\left(-\dfrac{1}{3}\right) = 1 = \dfrac{-(-3)}{3} = \dfrac{-d}{a}\)

Exercise 2.1

Q1

The graphs of y = p(x) are given below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

(i)(ii)(iii)(iv)(v)(vi)
Fig. 2.10 — find the number of zeroes of p(x) in each case
Answer

Counting the points where each graph meets the x-axis:

(i)0 zeroes — this graph is a horizontal line that never touches the x-axis.
(ii)1 zero — the graph crosses the x-axis exactly once.
(iii)3 zeroes — the graph crosses the x-axis at three points.
(iv)2 zeroes — the graph crosses the x-axis at two points.
(v)4 zeroes — the graph crosses the x-axis at four points.
(vi)3 zeroes — the graph crosses the x-axis at three points.

Exercise 2.2

Q1

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:

(i) \(x^{2}-2x-8\)   (ii) \(4s^{2}-4s+1\)   (iii) \(6x^{2}-3-7x\)   (iv) \(4u^{2}+8u\)   (v) \(t^{2}-15\)   (vi) \(3x^{2}-x-4\)

Answer

(i) Split the middle term: we need two numbers that multiply to −8 and add to −2, which are −4 and 2.

\(x^{2}-2x-8 = x^{2}-4x+2x-8 = x(x-4)+2(x-4) = (x-4)(x+2)\)

So p(x) = 0 when x = 4 or x = −2. Zeroes: 4, −2.

\(\text{sum} = 4+(-2) = 2 = \dfrac{-(-2)}{1}\ \checkmark\)
\(\text{product} = 4\times(-2) = -8 = \dfrac{-8}{1}\ \checkmark\)

(ii) Recognise the perfect square: \((2s)^{2}-2(2s)(1)+1^{2}\).

\(4s^{2}-4s+1 = (2s-1)^{2}\)

So p(x) = 0 only when 2s − 1 = 0, i.e. s = 1/2. This is a repeated (double) zero: 1/2, 1/2.

\(\text{sum} = \dfrac{1}{2}+\dfrac{1}{2} = 1 = \dfrac{-(-4)}{4}\ \checkmark\)
\(\text{product} = \dfrac{1}{2}\times\dfrac{1}{2} = \dfrac{1}{4} = \dfrac{1}{4}\ \checkmark\)

(iii) Rewrite as \(6x^{2}-7x-3\). We need two numbers multiplying to \(6\times(-3)=-18\) and adding to −7, which are −9 and 2.

\(6x^{2}-7x-3 = 6x^{2}-9x+2x-3 = 3x(2x-3)+1(2x-3) = (3x+1)(2x-3)\)

So p(x) = 0 when 3x + 1 = 0 or 2x − 3 = 0, i.e. x = −1/3 or x = 3/2. Zeroes: 3/2, −1/3.

\(\text{sum} = \dfrac{3}{2}+\left(-\dfrac{1}{3}\right) = \dfrac{7}{6} = \dfrac{-(-7)}{6}\ \checkmark\)
\(\text{product} = \dfrac{3}{2}\times\left(-\dfrac{1}{3}\right) = -\dfrac{1}{2} = \dfrac{-3}{6}\ \checkmark\)

(iv) Take out the common factor 4u:

\(4u^{2}+8u = 4u(u+2)\)

So p(x) = 0 when 4u = 0 or u + 2 = 0, i.e. u = 0 or u = −2. Zeroes: 0, −2.

\(\text{sum} = 0+(-2) = -2 = \dfrac{-8}{4}\ \checkmark\)
\(\text{product} = 0\times(-2) = 0 = \dfrac{0}{4}\ \checkmark\)

(v) Use the identity \(a^{2}-b^{2}=(a-b)(a+b)\):

\(t^{2}-15 = \left(t-\sqrt{15}\right)\left(t+\sqrt{15}\right)\)

So p(x) = 0 when \(t=\sqrt{15}\) or \(t=-\sqrt{15}\). Zeroes: \(\sqrt{15},\,-\sqrt{15}\).

\(\text{sum} = \sqrt{15}-\sqrt{15} = 0\ \checkmark\)
\(\text{product} = \left(\sqrt{15}\right)\left(-\sqrt{15}\right) = -15 = \dfrac{-15}{1}\ \checkmark\)

(vi) We need two numbers multiplying to \(3\times(-4)=-12\) and adding to −1, which are −4 and 3.

\(3x^{2}-x-4 = 3x^{2}-4x+3x-4 = x(3x-4)+1(3x-4) = (3x-4)(x+1)\)

So p(x) = 0 when 3x − 4 = 0 or x + 1 = 0, i.e. x = 4/3 or x = −1. Zeroes: 4/3, −1.

\(\text{sum} = \dfrac{4}{3}+(-1) = \dfrac{1}{3} = \dfrac{-(-1)}{3}\ \checkmark\)
\(\text{product} = \dfrac{4}{3}\times(-1) = -\dfrac{4}{3} = \dfrac{-4}{3}\ \checkmark\)
Q2

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:

(i) \(\dfrac{1}{4},\,-1\)   (ii) \(\sqrt{2},\,\dfrac{1}{3}\)   (iii) \(0,\,\sqrt{5}\)   (iv) \(1,\,1\)   (v) \(-\dfrac{1}{4},\,\dfrac{1}{4}\)   (vi) \(4,\,1\)

Answer

For every part we substitute the given sum and product into \(p(x) = x^{2} - (\text{sum})x + (\text{product})\), then clear any fractions by multiplying through by a suitable constant k (this does not change the zeroes).

(i) Sum \(=\dfrac{1}{4}\), product \(=-1\):

\(x^{2}-\dfrac{1}{4}x+(-1) = x^{2}-\dfrac{1}{4}x-1\)

Multiplying every term by \(k=4\) to clear the fraction:

\(4x^{2}-x-4\)

(ii) Sum \(=\sqrt{2}\), product \(=\dfrac{1}{3}\):

\(x^{2}-\sqrt{2}\,x+\dfrac{1}{3}\)

Multiplying every term by \(k=3\) to clear the fraction:

\(3x^{2}-3\sqrt{2}\,x+1\)

(iii) Sum \(=0\), product \(=\sqrt{5}\):

\(x^{2}-0\cdot x+\sqrt{5} = x^{2}+\sqrt{5}\)

No fraction to clear here, so \(k=1\) and the polynomial is already in simplest form.

(iv) Sum \(=1\), product \(=1\):

\(x^{2}-x+1\)

Both coefficients are already integers, so \(k=1\).

(v) Sum \(=-\dfrac{1}{4}\), product \(=\dfrac{1}{4}\):

\(x^{2}-\left(-\dfrac{1}{4}\right)x+\dfrac{1}{4} = x^{2}+\dfrac{1}{4}x+\dfrac{1}{4}\)

Multiplying every term by \(k=4\) to clear the fractions:

\(4x^{2}+x+1\)

(vi) Sum \(=4\), product \(=1\):

\(x^{2}-4x+1\)

Both coefficients are already integers, so \(k=1\).

Final answers: (i) \(4x^{2}-x-4\)   (ii) \(3x^{2}-3\sqrt{2}\,x+1\)   (iii) \(x^{2}+\sqrt{5}\)   (iv) \(x^{2}-x+1\)   (v) \(4x^{2}+x+1\)   (vi) \(x^{2}-4x+1\)

Chapter 1: Real Numbers
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Chapter 3: Pair of Linear Equations