NCERT Exemplar Class 10 Maths Chapter 2 Polynomials Solutions (CBSE 2026-27)
In alignment with the official curriculum for Chapter 2 (Polynomials), this guide covers: geometric meaning of zeroes, finding zeroes of quadratic polynomials algebraically & graphically, and verifying the relationship between zeroes and coefficients. Deleted topics—specifically the Division Algorithm for polynomials and cubic polynomials—have been pruned so you practice strictly what is tested.
Section A: Multiple Choice Questions (MCQs)
Let \(p(x) = x^2 + 3x + k\). Since \(2\) is a zero of \(p(x)\), \(p(2) = 0\):
\[(2)^2 + 3(2) + k = 0\] \[4 + 6 + k = 0 \implies 10 + k = 0 \implies k = -10\]Let \(p(x) = (k – 1)x^2 + kx + 1\). Since \(-3\) is a zero of the polynomial, \(p(-3) = 0\):
\[(k – 1)(-3)^2 + k(-3) + 1 = 0\] \[9(k – 1) – 3k + 1 = 0\] \[9k – 9 – 3k + 1 = 0\] \[6k – 8 = 0 \implies 6k = 8 \implies k = \frac{8}{6} = \frac{4}{3}\]Let \(\alpha = -3\) and \(\beta = 4\). The sum and product of zeroes are:
\[\alpha + \beta = -3 + 4 = 1\] \[\alpha\beta = (-3)(4) = -12\]A family of quadratic polynomials with zeroes \(\alpha\) and \(\beta\) is given by \(k[x^2 – (\alpha + \beta)x + \alpha\beta]\) where \(k \neq 0\):
\[p(x) = k(x^2 – x – 12)\]Choosing \(k = \frac{1}{2}\):
\[p(x) = \frac{1}{2}(x^2 – x – 12) = \frac{x^2}{2} – \frac{x}{2} – 6\]For the polynomial \(p(x) = x^2 + (a + 1)x + b\), comparing with \(Ax^2 + Bx + C\), we have \(A = 1\), \(B = a + 1\), and \(C = b\).
Given zeroes are \(\alpha = 2\) and \(\beta = -3\):
\[\text{Sum of zeroes} = \alpha + \beta = -\frac{B}{A} \implies 2 + (-3) = -(a + 1)\] \[-1 = -a – 1 \implies a = 0\] \[\text{Product of zeroes} = \alpha\beta = \frac{C}{A} \implies (2)(-3) = b \implies b = -6\]Therefore, \(a = 0\) and \(b = -6\).
A polynomial having zeroes \(-2\) and \(5\) can be written in the form:
\[p(x) = k(x – (-2))(x – 5) = k(x + 2)(x – 5)\]where \(k\) is any non-zero real number. Since there are infinitely many real values for \(k\), there exist infinitely many polynomials having \(-2\) and \(5\) as zeroes. Hence, the answer is more than 3.
Comparing \(x^2 + 99x + 127\) with \(ax^2 + bx + c\), we have \(a = 1 > 0\), \(b = 99 > 0\), and \(c = 127 > 0\).
Let the zeroes be \(\alpha\) and \(\beta\):
\[\alpha + \beta = -\frac{b}{a} = -99 < 0\] \[\alpha\beta = \frac{c}{a} = 127 > 0\]Since the product \(\alpha\beta > 0\), both zeroes must have the same sign. Because the sum \(\alpha + \beta < 0\), both zeroes must be negative.
For \(p(x) = x^2 + kx + k\), let \(\alpha\) and \(\beta\) be the zeroes:
\[\alpha + \beta = -k \quad \text{and} \quad \alpha\beta = k\]If both zeroes were positive, then \(\alpha\beta > 0 \implies k > 0\). However, this would mean \(\alpha + \beta = -k < 0\), which is impossible for two positive numbers. Hence, the zeroes cannot both be positive.
For the zeroes to be equal, the discriminant \(D = b^2 – 4ac\) must equal zero:
\[b^2 – 4ac = 0 \implies b^2 = 4ac \implies ac = \frac{b^2}{4}\]Since \(b^2 > 0\) (as \(c \neq 0 \implies b \neq 0\)), the product \(ac > 0\). The product of two numbers is positive if and only if both numbers have the same sign. Thus, \(c\) and \(a\) must have the same sign.
Let the zeroes be \(\alpha\) and \(-\alpha\):
\[\text{Sum of zeroes} = \alpha + (-\alpha) = 0 \implies -a = 0 \implies a = 0\]Since \(a = 0\), the polynomial has no linear term (no \(x\) term).
\[\text{Product of zeroes} = \alpha(-\alpha) = -\alpha^2 = b\]For real distinct zeroes \(\alpha \neq 0\), \(\alpha^2 > 0\), which means \(b = -\alpha^2 < 0\). Thus, the polynomial has no linear term and the constant term is negative.
The graph of any quadratic polynomial \(p(x) = ax^2 + bx + c\) is a parabola that can intersect or touch the x-axis at at most \(2\) points (at most two real zeroes). The curve in option (D) intersects the x-axis at \(3\) distinct points, which corresponds to a polynomial of degree at least \(3\) (such as a cubic polynomial). Hence, (D) is not a quadratic polynomial graph.
Section B: Short Answer Questions with Reasoning
Let \(\alpha\) and \(\beta\) be the two negative zeroes (\(\alpha < 0, \beta < 0\)):
\[\text{Sum of zeroes} = \alpha + \beta < 0 \implies -\frac{b}{a} < 0 \implies \frac{b}{a} > 0\]\(\frac{b}{a} > 0\) implies that \(a\) and \(b\) have the same sign.
\[\text{Product of zeroes} = \alpha\beta > 0 \implies \frac{c}{a} > 0\]\(\frac{c}{a} > 0\) implies that \(a\) and \(c\) have the same sign.
Since \(b\) and \(c\) both share the same sign as \(a\), all three coefficients \(a, b,\) and \(c\) must have the same sign.
For the quadratic polynomial \(x^2 + kx + k\) to have equal zeroes, its discriminant must be zero:
\[D = b^2 – 4ac = 0\] \[k^2 – 4(1)(k) = 0 \implies k(k – 4) = 0\]This yields \(k = 0\) or \(k = 4\). Both values are even integers; neither is an odd integer greater than \(1\). Therefore, the polynomial cannot have equal zeroes for any odd integer \(k > 1\).
Let the zeroes be \(\alpha > 0\) and \(\beta > 0\):
\[\text{Sum of zeroes} = \alpha + \beta > 0 \implies -\frac{b}{a} > 0 \implies \frac{b}{a} < 0\]\(\frac{b}{a} < 0\) means \(a\) and \(b\) have opposite signs.
\[\text{Product of zeroes} = \alpha\beta > 0 \implies \frac{c}{a} > 0\]\(\frac{c}{a} > 0\) means \(a\) and \(c\) have the same sign.
Because \(b\) has an opposite sign to \(a\), the coefficients \(a, b,\) and \(c\) cannot all have the same sign.
A quadratic polynomial can have two equal real zeroes (for example, \(p(x) = (x – 2)^2 = x^2 – 4x + 4\)). In this case, its parabolic graph touches the x-axis at exactly one point \((2, 0)\). Hence, a graph intersecting or touching the x-axis at only one point can certainly be a quadratic polynomial.
A polynomial of degree higher than \(2\) can also intersect the x-axis at exactly two points. For example, consider the 4th-degree polynomial \(p(x) = x^2(x^2 + 1)\) or a cubic polynomial with one single real zero and one repeated zero. Thus, having exactly two x-intercepts does not restrict a polynomial strictly to being quadratic.
For equal zeroes, the discriminant must be zero:
\[D = b^2 – 4ac = 0 \implies (1)^2 – 4(k)(k) = 0\] \[1 – 4k^2 = 0 \implies 4k^2 = 1 \implies k^2 = \frac{1}{4} \implies k = \pm\frac{1}{2}\]Since \(k\) can be either \(\frac{1}{2}\) or \(-\frac{1}{2}\), \(\frac{1}{2}\) is not the only value.
Section C: Finding Zeroes & Verifying Coefficient Relations
Let \(p(x) = x^2 + \frac{1}{6}x – 2 = \frac{1}{6}(6x^2 + x – 12)\). Factoring by splitting the middle term (\(6 \times (-12) = -72 = 9 \times (-8)\)):
\[p(x) = \frac{1}{6}[6x^2 + 9x – 8x – 12] = \frac{1}{6}[3x(2x + 3) – 4(2x + 3)]\] \[= \frac{1}{6}(3x – 4)(2x + 3)\]Setting \(p(x) = 0\) gives \(3x – 4 = 0 \implies x = \frac{4}{3}\) or \(2x + 3 = 0 \implies x = -\frac{3}{2}\).
Zeroes: \(\alpha = \frac{4}{3}\) and \(\beta = -\frac{3}{2}\).
Verification:
\[\text{Sum of zeroes} = \frac{4}{3} + \left(-\frac{3}{2}\right) = \frac{8 – 9}{6} = -\frac{1}{6} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}\] \[\text{Product of zeroes} = \left(\frac{4}{3}\right)\left(-\frac{3}{2}\right) = -2 = \frac{\text{Constant term}}{\text{Coefficient of } x^2}\]The relations are verified.
Factoring \(4x^2 – 3x – 1\) by splitting the middle term (\(4 \times (-1) = -4 = -4 \times 1\)):
\[4x^2 – 4x + x – 1 = 4x(x – 1) + 1(x – 1) = (4x + 1)(x – 1)\]Setting to zero: \(4x + 1 = 0 \implies x = -\frac{1}{4}\) and \(x – 1 = 0 \implies x = 1\).
Zeroes: \(\alpha = 1\) and \(\beta = -\frac{1}{4}\).
Verification:
\[\alpha + \beta = 1 + \left(-\frac{1}{4}\right) = \frac{3}{4} = -\frac{(-3)}{4} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}\] \[\alpha\beta = (1)\left(-\frac{1}{4}\right) = -\frac{1}{4} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}\]Splitting the middle term (\(3 \times (-4) = -12 = 6 \times (-2)\)):
\[3x^2 + 6x – 2x – 4 = 3x(x + 2) – 2(x + 2) = (3x – 2)(x + 2)\]Setting to zero gives zeroes: \(\alpha = \frac{2}{3}\) and \(\beta = -2\).
Verification:
\[\alpha + \beta = \frac{2}{3} + (-2) = \frac{2 – 6}{3} = -\frac{4}{3} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}\] \[\alpha\beta = \left(\frac{2}{3}\right)(-2) = -\frac{4}{3} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}\]Splitting the middle term (\(5 \times 7 = 35 = 5 + 7\)):
\[5t^2 + 5t + 7t + 7 = 5t(t + 1) + 7(t + 1) = (5t + 7)(t + 1)\]Setting to zero: \(\alpha = -1\) and \(\beta = -\frac{7}{5}\).
Verification:
\[\alpha + \beta = -1 + \left(-\frac{7}{5}\right) = -\frac{12}{5} = -\frac{\text{Coefficient of } t}{\text{Coefficient of } t^2}\] \[\alpha\beta = (-1)\left(-\frac{7}{5}\right) = \frac{7}{5} = \frac{\text{Constant term}}{\text{Coefficient of } t^2}\]Factor out \(\frac{1}{4}\):
\[p(x) = \frac{1}{4}(8x^2 + 14x + 3)\]Splitting the middle term (\(8 \times 3 = 24 = 12 \times 2\)):
\[p(x) = \frac{1}{4}[8x^2 + 12x + 2x + 3] = \frac{1}{4}[4x(2x + 3) + 1(2x + 3)] = \frac{1}{4}(4x + 1)(2x + 3)\]Setting to zero yields zeroes: \(\alpha = -\frac{1}{4}\) and \(\beta = -\frac{3}{2}\).
Verification:
\[\alpha + \beta = -\frac{1}{4} + \left(-\frac{3}{2}\right) = -\frac{7}{4} = -\frac{7/2}{2} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}\] \[\alpha\beta = \left(-\frac{1}{4}\right)\left(-\frac{3}{2}\right) = \frac{3}{8} = \frac{3/4}{2} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}\]Splitting the middle term: product \(= 4 \times (-3) = -12\), sum \(= 5\sqrt{2}\). We choose \(6\sqrt{2}\) and \(-\sqrt{2}\) (since \(6\sqrt{2} \times (-\sqrt{2}) = -12\)):
\[4x^2 + 6\sqrt{2}x – \sqrt{2}x – 3 = 2\sqrt{2}x(\sqrt{2}x + 3) – 1(\sqrt{2}x + 3) = (2\sqrt{2}x – 1)(\sqrt{2}x + 3)\]Setting to zero: \(\alpha = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}\) and \(\beta = -\frac{3}{\sqrt{2}} = -\frac{3\sqrt{2}}{2}\).
Verification:
\[\alpha + \beta = \frac{\sqrt{2}}{4} – \frac{6\sqrt{2}}{4} = -\frac{5\sqrt{2}}{4} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}\] \[\alpha\beta = \left(\frac{1}{2\sqrt{2}}\right)\left(-\frac{3}{\sqrt{2}}\right) = -\frac{3}{4} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}\]Expanding the middle term:
\[2s^2 – s – 2\sqrt{2}s + \sqrt{2} = s(2s – 1) – \sqrt{2}(2s – 1) = (2s – 1)(s – \sqrt{2})\]Setting to zero: \(\alpha = \frac{1}{2}\) and \(\beta = \sqrt{2}\).
Verification:
\[\alpha + \beta = \frac{1}{2} + \sqrt{2} = \frac{1 + 2\sqrt{2}}{2} = -\frac{-(1 + 2\sqrt{2})}{2} = -\frac{\text{Coefficient of } s}{\text{Coefficient of } s^2}\] \[\alpha\beta = \left(\frac{1}{2}\right)(\sqrt{2}) = \frac{\sqrt{2}}{2} = \frac{\text{Constant term}}{\text{Coefficient of } s^2}\]Splitting the middle term: product \(= -15\), sum \(= 4\sqrt{3}\). We select \(5\sqrt{3}\) and \(-\sqrt{3}\) (since \(5\sqrt{3} \times (-\sqrt{3}) = -15\)):
\[v^2 + 5\sqrt{3}v – \sqrt{3}v – 15 = v(v + 5\sqrt{3}) – \sqrt{3}(v + 5\sqrt{3}) = (v – \sqrt{3})(v + 5\sqrt{3})\]Setting to zero: \(\alpha = \sqrt{3}\) and \(\beta = -5\sqrt{3}\).
Verification:
\[\alpha + \beta = \sqrt{3} + (-5\sqrt{3}) = -4\sqrt{3} = -\frac{\text{Coefficient of } v}{\text{Coefficient of } v^2}\] \[\alpha\beta = (\sqrt{3})(-5\sqrt{3}) = -15 = \frac{\text{Constant term}}{\text{Coefficient of } v^2}\]Multiply by \(\frac{1}{2}\):
\[p(y) = \frac{1}{2}(2y^2 + 3\sqrt{5}y – 10)\]For \(2y^2 + 3\sqrt{5}y – 10\): product \(= -20\), sum \(= 3\sqrt{5}\). Choose \(4\sqrt{5}\) and \(-\sqrt{5}\) (since \(4\sqrt{5} \times (-\sqrt{5}) = -20\)):
\[2y^2 + 4\sqrt{5}y – \sqrt{5}y – 10 = 2y(y + 2\sqrt{5}) – \sqrt{5}(y + 2\sqrt{5}) = (2y – \sqrt{5})(y + 2\sqrt{5})\]Setting to zero yields zeroes: \(\alpha = \frac{\sqrt{5}}{2}\) and \(\beta = -2\sqrt{5}\).
Verification:
\[\alpha + \beta = \frac{\sqrt{5}}{2} – 2\sqrt{5} = -\frac{3\sqrt{5}}{2} = -\frac{\text{Coefficient of } y}{\text{Coefficient of } y^2}\] \[\alpha\beta = \left(\frac{\sqrt{5}}{2}\right)(-2\sqrt{5}) = -5 = \frac{\text{Constant term}}{\text{Coefficient of } y^2}\]Factor out \(\frac{1}{3}\):
\[p(y) = \frac{1}{3}(21y^2 – 11y – 2)\]Splitting the middle term (\(21 \times (-2) = -42 = -14 \times 3\)):
\[p(y) = \frac{1}{3}[21y^2 – 14y + 3y – 2] = \frac{1}{3}[7y(3y – 2) + 1(3y – 2)] = \frac{1}{3}(7y + 1)(3y – 2)\]Zeroes are \(\alpha = \frac{2}{3}\) and \(\beta = -\frac{1}{7}\).
Verification:
\[\alpha + \beta = \frac{2}{3} – \frac{1}{7} = \frac{14 – 3}{21} = \frac{11}{21} = -\frac{-11/3}{7} = -\frac{\text{Coefficient of } y}{\text{Coefficient of } y^2}\] \[\alpha\beta = \left(\frac{2}{3}\right)\left(-\frac{1}{7}\right) = -\frac{2}{21} = \frac{-2/3}{7} = \frac{\text{Constant term}}{\text{Coefficient of } y^2}\]Section D: Forming Quadratic Polynomials Given Sum and Product of Zeroes
A quadratic polynomial with sum \(S = \sqrt{2}\) and product \(P = -\frac{3}{2}\) is:
\[p(x) = x^2 – Sx + P = x^2 – \sqrt{2}x – \frac{3}{2} = \frac{1}{2}(2x^2 – 2\sqrt{2}x – 3)\]To find the zeroes, factorise \(2x^2 – 2\sqrt{2}x – 3 = 0\). Product \(= -6\), sum \(= -2\sqrt{2}\). Choose \(-3\sqrt{2}\) and \(\sqrt{2}\):
\[2x^2 – 3\sqrt{2}x + \sqrt{2}x – 3 = \sqrt{2}x(\sqrt{2}x + 1) – 3(\sqrt{2}x + 1) = (\sqrt{2}x – 3)(\sqrt{2}x + 1) = 0\]Therefore, the zeroes are \(\frac{3}{\sqrt{2}}\) and \(-\frac{1}{\sqrt{2}}\). Rationalising denominators gives \(\frac{3\sqrt{2}}{2}\) and \(-\frac{\sqrt{2}}{2}\).
A quadratic polynomial is given by \(p(x) = x^2 – Sx + P\):
\[p(x) = x^2 – \left(-\frac{8}{3}\right)x + \frac{4}{3} = x^2 + \frac{8}{3}x + \frac{4}{3} = \frac{1}{3}(3x^2 + 8x + 4)\]Factoring \(3x^2 + 8x + 4 = 0\) (product \(= 12\), sum \(= 8 \implies 6\) and \(2\)):
\[3x^2 + 6x + 2x + 4 = 3x(x + 2) + 2(x + 2) = (3x + 2)(x + 2) = 0\]Zeroes: \(x = -\frac{2}{3}\) and \(x = -2\).
The polynomial is:
\[p(x) = x^2 – \frac{21}{8}x + \frac{5}{16} = \frac{1}{16}(16x^2 – 42x + 5)\]Factoring \(16x^2 – 42x + 5 = 0\) (product \(= 80\), sum \(= -42 \implies -40\) and \(-2\)):
\[16x^2 – 40x – 2x + 5 = 8x(2x – 5) – 1(2x – 5) = (8x – 1)(2x – 5) = 0\]Zeroes: \(x = \frac{1}{8}\) and \(x = \frac{5}{2}\).
The polynomial is:
\[p(x) = x^2 – (-2\sqrt{3})x – 9 = x^2 + 2\sqrt{3}x – 9\]Factoring \(x^2 + 2\sqrt{3}x – 9 = 0\) (product \(= -9\), sum \(= 2\sqrt{3} \implies 3\sqrt{3}\) and \(-\sqrt{3}\)):
\[x^2 + 3\sqrt{3}x – \sqrt{3}x – 9 = x(x + 3\sqrt{3}) – \sqrt{3}(x + 3\sqrt{3}) = (x – \sqrt{3})(x + 3\sqrt{3}) = 0\]Zeroes: \(x = \sqrt{3}\) and \(x = -3\sqrt{3}\).
The polynomial is:
\[p(x) = x^2 – \left(-\frac{3}{2\sqrt{5}}\right)x – \frac{1}{2} = x^2 + \frac{3\sqrt{5}}{10}x – \frac{1}{2} = \frac{1}{10}(10x^2 + 3\sqrt{5}x – 5)\]Factoring \(10x^2 + 3\sqrt{5}x – 5 = 0\) (product \(= -50\), sum \(= 3\sqrt{5} \implies 5\sqrt{5}\) and \(-2\sqrt{5}\)):
\[\sqrt{5}x(2\sqrt{5}x + 5) – 1(2\sqrt{5}x + 5) = (\sqrt{5}x – 1)(2\sqrt{5}x + 5) = 0\]Setting each factor to zero:
\[\sqrt{5}x – 1 = 0 \implies x = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\] \[2\sqrt{5}x + 5 = 0 \implies x = -\frac{5}{2\sqrt{5}} = -\frac{\sqrt{5}}{2}\]Zeroes: \(x = \frac{\sqrt{5}}{5}\) and \(x = -\frac{\sqrt{5}}{2}\).
