NCERT Solutions Class 10 Maths Chapter 2

Chapter 2 Class 10 Mathematics includes a detailed study of Polynomials. This chapter involves descriptions and practices of different equations and respective components. NCERT Class 10 Mathematics Chapter 2 textbook is one of the best study materials to understand the concept of Polynomials. The book has practice questions at the end of every chapter so that students can revise the concepts they have learned and prepare better for their exams.

Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 2 that have answers to all the questions given in the textbook. Whether students are looking for answers to questions or want to cross-check their answers, the NCERT Solutions by Extramarks are reliable learning material.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials –

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NCERT Solutions for Class 10 Mathematics Chapter 2 Polynomials – Free Download

The NCERT Solutions for Class 10 Mathematics Chapter 2 Polynomials can help students solve exercises quickly. In case students get stuck in any question while practicing, they can refer to the solutions, and get their doubts cleared without wasting any time.

NCERT Solutions for Class 10 Mathematics

In addition to NCERT Solutions for Class 10 Mathematics Chapter 2, Extramarks also offers NCERT Solutions for all chapters in Class 10 Mathematics.

Chapter 1 – Real Numbers
Chapter 2 – Polynomials
Chapter 3 – Pair of Linear Equations in Two Variables
Chapter 4 – Quadratic Equations
Chapter 5 – Arithmetic Progressions
Chapter 6 – Triangles
Chapter 7 – Coordinate Geometry
Chapter 8 – Introduction to Trigonometry
Chapter 9 – Some Applications of Trigonometry
Chapter 10 – Circles
Chapter 11 – Constructions
Chapter 12 – Areas Related to Circles
Chapter 13 – Surface Areas and Volumes
Chapter 14 – Statistics
Chapter 15 – Probability

Polynomials: NCERT Solutions for Class 10 Mathematics Chapter 2 Summary

Chapter 2 includes the concepts of polynomials’ geometric and meanings of the terms relevant to it. The questions at the end of the NCERT Chapter 2 textbook will analyse the problem-solving skills of students. Therefore, Extramarks has ensured that every answer in solutions is solved in a step-by-step manner. The subject matter experts have kept the language simple and all the answers are highly accurate.

Benefits of Using NCERT Solutions for Class 10 Mathematics Polynomials

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Concise Answers to Questions

NCERT Solutions for Class 10 Mathematics Chapter 2 are created by subject matter experts. You can rest assured that all the answers have been framed by keeping in mind the guidelines provided by CBSE. Hence students who practise according to NCERT Solutions Class 10 Mathematics Chapter 2 will perform very well in their examinations.

Clear Concept of Chapter 2 Class 10 Mathematics

Just knowing the answer to the question does not mean you excel in that subject, but understanding the path to find the solution should be the main aim of the student. Concept clarity helps students solve any problem with ease. As discussed above, NCERT Solutions for Class 10 Mathematics Chapter 2 focus more on sorting out conceptual queries of the students, thus helping them strengthen their subject knowledge for future standards.

Clearing Doubts

It is natural for students to have considerable doubts while practising the questions of Class 10 Mathematics Chapter 2. These doubts can be quickly resolved with NCERT Solutions for Class 10 Mathematics Chapter 2.

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Related Questions

Q1. Write the following in the product form: a2b5.

Ans. a2b5 is – a×a×b×b×b×b×b

Q.1 The graphs of y = p(x) are given in figures below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Ans.

(i) The number of zeroes is 0 as the given graph does not intersect the x-axis.
(ii) The number of zeroes is 1 as the given graph intersects the x-axis at 1 point only.
(iii) The number of zeroes is 3 as the given graph intersects the x-axis at 3 points only.
(iv) The number of zeroes is 2 as the given graph intersects the x-axis at 2 points only.
(v) The number of zeroes is 4 as the given graph intersects the x-axis at 4 points only.
(vi) The number of zeroes is 3 as the given graph intersects the x-axis at 3 points only.

Q.2

$\begin{array}{l} Find the zeroes of the following quadratic polynomials \\ and verify the relationship between the zeroes and \\ the coefficients. \\ (i) x^{2} - 2 x - 8 (ii) 4 s^{2} - 4 s + 1 (iii) 6 x^{2} - 3 - 7 x \\ (iv) 4 u^{2} + 8 u (v) t^{2} - 15 (vi) 3 x^{2} - x - 4 \end{array}$ $\begin{array}{l} Find the zeroes of the following quadratic polynomials \\ and verify the relationship between the zeroes and \\ the coefficients. \\ (i) x^{2} - 2 x - 8 (ii) 4 s^{2} - 4 s + 1 (iii) 6 x^{2} - 3 - 7 x \\ (iv) 4 u^{2} + 8 u (v) t^{2} - 15 (vi) 3 x^{2} - x - 4 \end{array}$

Ans.

$\begin{array}{l} (i) \\ x^{2} - 2 x - 8 = x^{2} - 4 x + 2 x - 8 \\ = x (x - 4) + 2 (x - 4) \\ = (x + 2) (x - 4) \\ So the value of x^{2} - 2 x - 8 is zero when x = - 2 or x = 4 . \\ Therefore, the zeroes of x^{2} - 2 x - 8 are - 2 and 4. \\ Now, \\ Sum of zeroes = - 2 + 4 = 2 = \frac{- (- 2)}{1} = \frac{- (Coefficient of x)}{{Coefficient of x}^{2}} \\ Product of zeroes = - 2 \times 4 = \frac{- 8}{1} = \frac{Constant term}{{Coefficient of s}^{2}} \\ (ii) \\ 4 s^{2} - 4 s + 1 = {(2 s - 1)}^{2} \\ So the value of 4 s^{2} - 4 s + 1 is zero when x = \frac{1}{2} . \\ Therefore, the zeroes of 4 s^{2} - 4 s + 1 are \frac{1}{2} and \frac{1}{2} . \\ Now, \\ Sum of zeroes = \frac{1}{2} + \frac{1}{2} = 1 = \frac{- (- 4)}{4} = \frac{- (Coefficient of s)}{{Coefficient of s}^{2}} \\ Product of zeroes = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{Constant term}{{Coefficient of s}^{2}} \\ (iii) \\ 6 x^{2} - 3 - 7 x = 6 x^{2} - 9 x + 2 x - 3 \\ = 3 x (2 x - 3) + 1 (2 x - 3) \\ = (3 x + 1) (2 x - 3) \\ So the value of 6 x^{2} - 3 - 7 x is zero when x = \frac{- 1}{3} or x = \frac{3}{2} . \\ Therefore, the zeroes of 6 x^{2} - 3 - 7 x are \frac{- 1}{3} and \frac{3}{2} . \\ Now, \\ Sum of zeroes = \frac{- 1}{3} + \frac{3}{2} = \frac{7}{6} = \frac{- (- 7)}{6} = \frac{- (Coefficient of x)}{{Coefficient of x}^{2}} \\ Product of zeroes = \frac{- 1}{3} \times \frac{3}{2} = \frac{- 1}{2} = \frac{Constant term}{{Coefficient of x}^{2}} \\ (iv) \\ 4 u^{2} + 8 u = 4 u (u + 2) \\ So the value of 4 u (u + 2) is zero when u = 0 or x = - 2 . \\ Therefore, the zeroes of 4 u (u + 2) are 0 and - 2 . \\ Now, \\ Sum of zeroes = 0 - 2 = \frac{- 8}{4} = \frac{- (Coefficient of u)}{{Coefficient of u}^{2}} \\ Product of zeroes = 0 \times (- 2) = 0 = \frac{0}{4} = \frac{Constant term}{{Coefficient of u}^{2}} \\ (v) \\ t^{2} - 15 = t^{2} - {(\sqrt{15})}^{2} = (t - \sqrt{15}) (t + \sqrt{15}) \\ So the value of t^{2} - 15 is zero when t = \sqrt{15} or x = - \sqrt{15} . \\ Therefore, the zeroes of t^{2} - 15 are \sqrt{15} and - \sqrt{15} . \\ Now, \\ Sum of zeroes = \sqrt{15} - \sqrt{15} = 0 = \frac{0}{1} = \frac{- (Coefficient of t)}{{Coefficient of t}^{2}} \\ Product of zeroes = \sqrt{15} \times (- \sqrt{15}) = - 15 = \frac{- 15}{1} = \frac{Constant term}{{Coefficient of t}^{2}} \\ (vi) \\ 3 x^{2} - x - 4 = 3 x^{2} - 4 x + 3 x - 4 \\ = x (3 x - 4) + 1 (3 x - 4) = (3 x - 4) (x + 1) \\ So the value of 3 x^{2} - x - 4 is zero when x = \frac{4}{3} or x = - 1 . \\ Therefore, the zeroes of 3 x^{2} - x - 4 are \frac{4}{3} and - 1 . \\ Now, \\ Sum of zeroes = \frac{4}{3} - 1 = \frac{1}{3} = \frac{- (- 1)}{3} = \frac{- (Coefficient of x)}{{Coefficient of x}^{2}} \\ Product of zeroes = \frac{4}{3} \times (- 1) = \frac{- 4}{3} = \frac{Constant term}{{Coefficient of x}^{2}} \end{array}$ $\begin{array}{l} (i) \\ x^{2} - 2 x - 8 = x^{2} - 4 x + 2 x - 8 \\ = x (x - 4) + 2 (x - 4) \\ = (x + 2) (x - 4) \\ So the value of x^{2} - 2 x - 8 is zero when x = - 2 or x = 4 . \\ Therefore, the zeroes of x^{2} - 2 x - 8 are - 2 and 4. \\ Now, \\ Sum of zeroes = - 2 + 4 = 2 = \frac{- (- 2)}{1} = \frac{- (Coefficient of x)}{{Coefficient of x}^{2}} \\ Product of zeroes = - 2 \times 4 = \frac{- 8}{1} = \frac{Constant term}{{Coefficient of s}^{2}} \\ (ii) \\ 4 s^{2} - 4 s + 1 = {(2 s - 1)}^{2} \\ So the value of 4 s^{2} - 4 s + 1 is zero when x = \frac{1}{2} . \\ Therefore, the zeroes of 4 s^{2} - 4 s + 1 are \frac{1}{2} and \frac{1}{2} . \\ Now, \\ Sum of zeroes = \frac{1}{2} + \frac{1}{2} = 1 = \frac{- (- 4)}{4} = \frac{- (Coefficient of s)}{{Coefficient of s}^{2}} \\ Product of zeroes = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{Constant term}{{Coefficient of s}^{2}} \\ (iii) \\ 6 x^{2} - 3 - 7 x = 6 x^{2} - 9 x + 2 x - 3 \\ = 3 x (2 x - 3) + 1 (2 x - 3) \\ = (3 x + 1) (2 x - 3) \\ So the value of 6 x^{2} - 3 - 7 x is zero when x = \frac{- 1}{3} or x = \frac{3}{2} . \\ Therefore, the zeroes of 6 x^{2} - 3 - 7 x are \frac{- 1}{3} and \frac{3}{2} . \\ Now, \\ Sum of zeroes = \frac{- 1}{3} + \frac{3}{2} = \frac{7}{6} = \frac{- (- 7)}{6} = \frac{- (Coefficient of x)}{{Coefficient of x}^{2}} \\ Product of zeroes = \frac{- 1}{3} \times \frac{3}{2} = \frac{- 1}{2} = \frac{Constant term}{{Coefficient of x}^{2}} \\ (iv) \\ 4 u^{2} + 8 u = 4 u (u + 2) \\ So the value of 4 u (u + 2) is zero when u = 0 or x = - 2 . \\ Therefore, the zeroes of 4 u (u + 2) are 0 and - 2 . \\ Now, \\ Sum of zeroes = 0 - 2 = \frac{- 8}{4} = \frac{- (Coefficient of u)}{{Coefficient of u}^{2}} \\ Product of zeroes = 0 \times (- 2) = 0 = \frac{0}{4} = \frac{Constant term}{{Coefficient of u}^{2}} \\ (v) \\ t^{2} - 15 = t^{2} - {(\sqrt{15})}^{2} = (t - \sqrt{15}) (t + \sqrt{15}) \\ So the value of t^{2} - 15 is zero when t = \sqrt{15} or x = - \sqrt{15} . \\ Therefore, the zeroes of t^{2} - 15 are \sqrt{15} and - \sqrt{15} . \\ Now, \\ Sum of zeroes = \sqrt{15} - \sqrt{15} = 0 = \frac{0}{1} = \frac{- (Coefficient of t)}{{Coefficient of t}^{2}} \\ Product of zeroes = \sqrt{15} \times (- \sqrt{15}) = - 15 = \frac{- 15}{1} = \frac{Constant term}{{Coefficient of t}^{2}} \\ (vi) \\ 3 x^{2} - x - 4 = 3 x^{2} - 4 x + 3 x - 4 \\ = x (3 x - 4) + 1 (3 x - 4) = (3 x - 4) (x + 1) \\ So the value of 3 x^{2} - x - 4 is zero when x = \frac{4}{3} or x = - 1 . \\ Therefore, the zeroes of 3 x^{2} - x - 4 are \frac{4}{3} and - 1 . \\ Now, \\ Sum of zeroes = \frac{4}{3} - 1 = \frac{1}{3} = \frac{- (- 1)}{3} = \frac{- (Coefficient of x)}{{Coefficient of x}^{2}} \\ Product of zeroes = \frac{4}{3} \times (- 1) = \frac{- 4}{3} = \frac{Constant term}{{Coefficient of x}^{2}} \end{array}$

Q.3

$\begin{array}{l} Find a quadratic polynomial each with the given \\ numbers as the sum and product of its zeroes \\ respectively . \\ (i) \frac{1}{4}, - 1 (ii) \sqrt{2}, \frac{1}{3} (iii) 0, \sqrt{5} \\ (iv) 1, 1 (v) \frac{- 1}{4}, \frac{1}{4} (vi) 4, 1 \end{array}$ $\begin{array}{l} Find a quadratic polynomial each with the given \\ numbers as the sum and product of its zeroes \\ respectively . \\ (i) \frac{1}{4}, - 1 (ii) \sqrt{2}, \frac{1}{3} (iii) 0, \sqrt{5} \\ (iv) 1, 1 (v) \frac{- 1}{4}, \frac{1}{4} (vi) 4, 1 \end{array}$

Ans.

$\begin{array}{l} (i) \frac{1}{4}, - 1 \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = \frac{1}{4} = \frac{- b}{a} and αβ = - 1 = \frac{c}{a} \\ and thus \\ a = 4, b = - 1, c = - 4 . \\ So, the quadratic polynomial which fits the given \\ {conditions is 4x}^{2} - x - 4 . \\ (ii) \sqrt{2}, \frac{1}{3} \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = \sqrt{2} = \frac{- b}{a} and αβ = \frac{1}{3} = \frac{c}{a} \\ Now, \\ α + β = \sqrt{2} = \frac{3 \sqrt{2}}{3} = \frac{- b}{a} \\ and αβ = \frac{1}{3} = \frac{c}{a} \\ So, a = 3, b = - 3 \sqrt{2}, c = 1 . \\ So, the quadratic polynomial which fits the given \\ {conditions is 3x}^{2} - 3 \sqrt{2} x + 1 . \\ (iii) 0, \sqrt{5} \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = 0 = \frac{- b}{a} and αβ = \sqrt{5} = \frac{c}{a} \\ So, a = 1, b = 0, c = \sqrt{5} . \\ So, the quadratic polynomial which fits the given \\ {conditions is x}^{2} + \sqrt{5} . \\ (iv) 1, 1 \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = 1 = \frac{- b}{a} and αβ = 1 = \frac{c}{a} \\ So, a = 1, b = - 1, c = 1 . \\ So, the quadratic polynomial which fits the given \\ {conditions is x}^{2} - x + 1 . \\ (v) \frac{- 1}{4}, \frac{1}{4} \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = \frac{- 1}{4} = \frac{- b}{a} and αβ = \frac{1}{4} = \frac{c}{a} \\ So, a = 4, b = 1, c = 1 . \\ So, the quadratic polynomial which fits the given \\ {conditions is 4x}^{2} + x + 1 . \\ (vi) 4, 1 \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = 4 = \frac{- b}{a} and αβ = 1 = \frac{c}{a} \\ So, a = 1, b = - 4, c = 1 . \\ So, the quadratic polynomial which fits the given \\ {conditions is x}^{2} - 4 x + 1 . \end{array}$ $\begin{array}{l} (i) \frac{1}{4}, - 1 \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = \frac{1}{4} = \frac{- b}{a} and αβ = - 1 = \frac{c}{a} \\ and thus \\ a = 4, b = - 1, c = - 4 . \\ So, the quadratic polynomial which fits the given \\ {conditions is 4x}^{2} - x - 4 . \\ (ii) \sqrt{2}, \frac{1}{3} \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = \sqrt{2} = \frac{- b}{a} and αβ = \frac{1}{3} = \frac{c}{a} \\ Now, \\ α + β = \sqrt{2} = \frac{3 \sqrt{2}}{3} = \frac{- b}{a} \\ and αβ = \frac{1}{3} = \frac{c}{a} \\ So, a = 3, b = - 3 \sqrt{2}, c = 1 . \\ So, the quadratic polynomial which fits the given \\ {conditions is 3x}^{2} - 3 \sqrt{2} x + 1 . \\ (iii) 0, \sqrt{5} \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = 0 = \frac{- b}{a} and αβ = \sqrt{5} = \frac{c}{a} \\ So, a = 1, b = 0, c = \sqrt{5} . \\ So, the quadratic polynomial which fits the given \\ {conditions is x}^{2} + \sqrt{5} . \\ (iv) 1, 1 \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = 1 = \frac{- b}{a} and αβ = 1 = \frac{c}{a} \\ So, a = 1, b = - 1, c = 1 . \\ So, the quadratic polynomial which fits the given \\ {conditions is x}^{2} - x + 1 . \\ (v) \frac{- 1}{4}, \frac{1}{4} \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = \frac{- 1}{4} = \frac{- b}{a} and αβ = \frac{1}{4} = \frac{c}{a} \\ So, a = 4, b = 1, c = 1 . \\ So, the quadratic polynomial which fits the given \\ {conditions is 4x}^{2} + x + 1 . \\ (vi) 4, 1 \\ {Let the quadratic polynomial be ax}^{2} + bx + c and its \\ zeroes be α and β . Then, we have \\ α + β = 4 = \frac{- b}{a} and αβ = 1 = \frac{c}{a} \\ So, a = 1, b = - 4, c = 1 . \\ So, the quadratic polynomial which fits the given \\ {conditions is x}^{2} - 4 x + 1 . \end{array}$

Q.4

$\begin{array}{l} Divide the polynomial p (x) by the polynomial g (x) \\ and find the quotient and remainder in each of the \\ following: \\ (i) p (x) = x^{3} - 3 x^{2} + 5 x - 3, g (x) = x^{2} - 2 \\ (ii) p (x) = x^{4} - 3 x^{2} + 4 x + 5, g (x) = x^{2} + 1 - x \\ (iii) p (x) = x^{4} - 5 x + 6, g (x) = 2 - x^{2} \end{array}$ $\begin{array}{l} Divide the polynomial p (x) by the polynomial g (x) \\ and find the quotient and remainder in each of the \\ following: \\ (i) p (x) = x^{3} - 3 x^{2} + 5 x - 3, g (x) = x^{2} - 2 \\ (ii) p (x) = x^{4} - 3 x^{2} + 4 x + 5, g (x) = x^{2} + 1 - x \\ (iii) p (x) = x^{4} - 5 x + 6, g (x) = 2 - x^{2} \end{array}$

Ans.

$\begin{array}{l} (i) p (x) = x^{3} - 3 x^{2} + 5 x - 3, g (x) = x^{2} - 2 \\ x^{2} - 2 \begin{matrix} x - 3 \\ x^{3} - 3 x^{2} + 5 x - 3 \end{matrix} \\ \underline{\begin{array}{l} x^{3} - 2 x \\ - + \end{array}} \\ - 3 x^{2} + 7 x - 3 \\ - 3 x^{2} + 6 \\ \frac{+ -}{7 x - 9} \\ Quotient = x - 3 \\ Remainder = 7 x - 9 \\ (ii) p (x) = x^{4} - 3 x^{2} + 4 x + 5, g (x) = x^{2} + 1 - x \\ x^{2} + 1 - x \begin{matrix} x^{2} + x - 3 \\ x^{4} - 3 x^{2} + 4 x + 5 \end{matrix} \\ \underline{\begin{array}{l} x^{4} + x^{2} - x^{3} \\ - - + \end{array}} \\ x^{3} - 4 x^{2} + 4 x + 5 \\ \underline{\begin{array}{l} x^{3} - x^{2} + x \\ - + - \end{array}} \\ - 3 x^{2} + 3 x + 5 \\ \underline{\begin{array}{l} - 3 x^{2} + 3 x - 3 \\ + - + \end{array}} \\ 8 \\ Quotient = x^{2} + x - 3 \\ Remainder = 8 \\ (iii) p (x) = x^{4} - 5 x + 6, g (x) = 2 - x^{2} \\ 2 - x^{2} \begin{matrix} - x^{2} - 2 \\ x^{4} - 5 x + 6 \end{matrix} \\ \underline{\begin{array}{l} x^{4} - 2 x^{2} \\ - + \end{array}} \\ 2 x^{2} - 5 x + 6 \\ \underline{\begin{array}{l} 2 x^{2} - 4 \\ - + \end{array}} \\ - 5 x + 10 \\ Quotient = - x^{2} - 2 \\ Remainder = - 5 x + 10 \end{array}$ $\begin{array}{l} (i) p (x) = x^{3} - 3 x^{2} + 5 x - 3, g (x) = x^{2} - 2 \\ x^{2} - 2 \begin{matrix} x - 3 \\ x^{3} - 3 x^{2} + 5 x - 3 \end{matrix} \\ \underline{\begin{array}{l} x^{3} - 2 x \\ - + \end{array}} \\ - 3 x^{2} + 7 x - 3 \\ - 3 x^{2} + 6 \\ \frac{+ -}{7 x - 9} \\ Quotient = x - 3 \\ Remainder = 7 x - 9 \\ (ii) p (x) = x^{4} - 3 x^{2} + 4 x + 5, g (x) = x^{2} + 1 - x \\ x^{2} + 1 - x \begin{matrix} x^{2} + x - 3 \\ x^{4} - 3 x^{2} + 4 x + 5 \end{matrix} \\ \underline{\begin{array}{l} x^{4} + x^{2} - x^{3} \\ - - + \end{array}} \\ x^{3} - 4 x^{2} + 4 x + 5 \\ \underline{\begin{array}{l} x^{3} - x^{2} + x \\ - + - \end{array}} \\ - 3 x^{2} + 3 x + 5 \\ \underline{\begin{array}{l} - 3 x^{2} + 3 x - 3 \\ + - + \end{array}} \\ 8 \\ Quotient = x^{2} + x - 3 \\ Remainder = 8 \\ (iii) p (x) = x^{4} - 5 x + 6, g (x) = 2 - x^{2} \\ 2 - x^{2} \begin{matrix} - x^{2} - 2 \\ x^{4} - 5 x + 6 \end{matrix} \\ \underline{\begin{array}{l} x^{4} - 2 x^{2} \\ - + \end{array}} \\ 2 x^{2} - 5 x + 6 \\ \underline{\begin{array}{l} 2 x^{2} - 4 \\ - + \end{array}} \\ - 5 x + 10 \\ Quotient = - x^{2} - 2 \\ Remainder = - 5 x + 10 \end{array}$

Q.5

$\begin{array}{l} Check whether the first polynomial is a facator of \\ the second polymomial by dividing the second \\ polynomial by the first polynomial: \\ (i) t^{2} - 3, 2 t^{4} + 3 t^{3} - 2 t^{2} - 9 t - 12 \\ (ii) x^{2} + 3 x + 1, 3 x^{4} + 5 x^{3} - 7 x^{2} + 2 x + 2 \\ (iii) x^{3} - 3 x + 1, x^{5} - 4 x^{3} + x^{2} + 3 x + 1 \end{array}$ $\begin{array}{l} Check whether the first polynomial is a facator of \\ the second polymomial by dividing the second \\ polynomial by the first polynomial: \\ (i) t^{2} - 3, 2 t^{4} + 3 t^{3} - 2 t^{2} - 9 t - 12 \\ (ii) x^{2} + 3 x + 1, 3 x^{4} + 5 x^{3} - 7 x^{2} + 2 x + 2 \\ (iii) x^{3} - 3 x + 1, x^{5} - 4 x^{3} + x^{2} + 3 x + 1 \end{array}$

Ans.

$\begin{array}{l} (i) t^{2} - 3, 2 t^{4} + 3 t^{3} - 2 t^{2} - 9 t - 12 \\ t^{2} - 3 \begin{matrix} 2 t^{2} + 3 t + 4 \\ 2 t^{4} + 3 t^{3} - 2 t^{2} - 9 t - 12 \end{matrix} \\ \underline{\begin{array}{l} 2 t^{4} - 6 t^{2} \\ - + \end{array}} \\ 3 t^{3} + 4 t^{2} - 9 t - 12 \\ \underline{\begin{array}{l} 3 t^{3} - 9 t \\ - + \end{array}} \\ 4 t^{2} - 12 \\ \underline{\begin{array}{l} 4 t^{2} - 12 \\ - + \end{array}} \\ 0 \\ Remainder = 0 \\ So, t^{2} - 3 is a factor of 2 t^{4} + 3 t^{3} - 2 t^{2} - 9 t - 12 . \\ (ii) x^{2} + 3 x + 1, 3 x^{4} + 5 x^{3} - 7 x^{2} + 2 x + 2 \\ x^{2} + 3 x + 1 \begin{matrix} 3 x^{2} - 4 x + 2 \\ 3 x^{4} + 5 x^{3} - 7 x^{2} + 2 x + 2 \end{matrix} \\ \underline{\begin{array}{l} 3 x^{4} + 9 x^{3} + 3 x^{2} \\ - - - \end{array}} \\ - 4 x^{3} - 10 x^{2} + 2 x + 2 \\ \underline{\begin{array}{l} - 4 x^{3} - 12 x^{2} - 4 x \\ + + + \end{array}} \\ 2 x^{2} + 6 x + 2 \\ \underline{\begin{array}{l} 2 x^{2} + 6 x + 2 \\ - - - \end{array}} \\ 0 \\ Remainder = 0 \\ So, x^{2} + 3 x + 1 is a factor of 3 x^{4} + 5 x^{3} - 7 x^{2} + 2 x + 2 . \\ (iii) x^{3} - 3 x + 1, x^{5} - 4 x^{3} + x^{2} + 3 x + 1 \\ x^{3} - 3 x + 1 \begin{matrix} x^{2} - 1 \\ x^{5} - 4 x^{3} + x^{2} + 3 x + 1 \end{matrix} \\ \underline{\begin{array}{l} x^{5} - 3 x^{3} + x^{2} \\ - + - \end{array}} \\ - x^{3} + 3 x + 1 \\ \underline{\begin{array}{l} - x^{3} + 3 x - 1 \\ + - + \end{array}} \\ 2 \\ Remainder = 2 \\ So, x^{3} - 3 x + 1 {is not a factor of x}^{5} - 4 x^{3} + x^{2} + 3 x + 1 . \end{array}$ $\begin{array}{l} (i) t^{2} - 3, 2 t^{4} + 3 t^{3} - 2 t^{2} - 9 t - 12 \\ t^{2} - 3 \begin{matrix} 2 t^{2} + 3 t + 4 \\ 2 t^{4} + 3 t^{3} - 2 t^{2} - 9 t - 12 \end{matrix} \\ \underline{\begin{array}{l} 2 t^{4} - 6 t^{2} \\ - + \end{array}} \\ 3 t^{3} + 4 t^{2} - 9 t - 12 \\ \underline{\begin{array}{l} 3 t^{3} - 9 t \\ - + \end{array}} \\ 4 t^{2} - 12 \\ \underline{\begin{array}{l} 4 t^{2} - 12 \\ - + \end{array}} \\ 0 \\ Remainder = 0 \\ So, t^{2} - 3 is a factor of 2 t^{4} + 3 t^{3} - 2 t^{2} - 9 t - 12 . \\ (ii) x^{2} + 3 x + 1, 3 x^{4} + 5 x^{3} - 7 x^{2} + 2 x + 2 \\ x^{2} + 3 x + 1 \begin{matrix} 3 x^{2} - 4 x + 2 \\ 3 x^{4} + 5 x^{3} - 7 x^{2} + 2 x + 2 \end{matrix} \\ \underline{\begin{array}{l} 3 x^{4} + 9 x^{3} + 3 x^{2} \\ - - - \end{array}} \\ - 4 x^{3} - 10 x^{2} + 2 x + 2 \\ \underline{\begin{array}{l} - 4 x^{3} - 12 x^{2} - 4 x \\ + + + \end{array}} \\ 2 x^{2} + 6 x + 2 \\ \underline{\begin{array}{l} 2 x^{2} + 6 x + 2 \\ - - - \end{array}} \\ 0 \\ Remainder = 0 \\ So, x^{2} + 3 x + 1 is a factor of 3 x^{4} + 5 x^{3} - 7 x^{2} + 2 x + 2 . \\ (iii) x^{3} - 3 x + 1, x^{5} - 4 x^{3} + x^{2} + 3 x + 1 \\ x^{3} - 3 x + 1 \begin{matrix} x^{2} - 1 \\ x^{5} - 4 x^{3} + x^{2} + 3 x + 1 \end{matrix} \\ \underline{\begin{array}{l} x^{5} - 3 x^{3} + x^{2} \\ - + - \end{array}} \\ - x^{3} + 3 x + 1 \\ \underline{\begin{array}{l} - x^{3} + 3 x - 1 \\ + - + \end{array}} \\ 2 \\ Remainder = 2 \\ So, x^{3} - 3 x + 1 {is not a factor of x}^{5} - 4 x^{3} + x^{2} + 3 x + 1 . \end{array}$

Q.6

$\begin{array}{l} Obtain all other zeroes of 3 x^{4} + 6 x^{3} - 2 x^{2} - 10 x - 5, \\ if two of its zeroes are \sqrt{\frac{5}{3}} and - \sqrt{\frac{5}{3}} . \end{array}$ $\begin{array}{l} Obtain all other zeroes of 3 x^{4} + 6 x^{3} - 2 x^{2} - 10 x - 5, \\ if two of its zeroes are \sqrt{\frac{5}{3}} and - \sqrt{\frac{5}{3}} . \end{array}$

Ans.

$\begin{array}{l} {p(x)=3x}^{4} {+6x}^{3} - {2x}^{2} - 10x-5 \\ The two zeroes of p(x) are \sqrt{\frac{5}{3}} and- \sqrt{\frac{5}{3}} . \\ Therefore, (x - \sqrt{\frac{5}{3}}) and (x+ \sqrt{\frac{5}{3}}) are factors of p(x). \\ Also, \\ (x - \sqrt{\frac{5}{3}}) (x+ \sqrt{\frac{5}{3}}) {= x}^{2} - \frac{5}{3} \\ {and so x}^{2} - \frac{5}{3} is a factor of p(x). \\ Now, \\ {3x}^{2} +6x+3 \\ x^{2} - \frac{5}{3} {3x}^{4} {+6x}^{3} - {2x}^{2} - 10x - 5 \\ \underline{\begin{array}{l} {3x}^{4} - {5x}^{2} \\ - + \end{array}} \\ {6x}^{3} {+3x}^{2} - 10x - 5 \\ \underline{\begin{array}{l} {6x}^{3} - 10x \\ - + \end{array}} \\ {3x}^{2} - 5 \\ \underline{\begin{array}{l} {3x}^{2} - 5 \\ - + \end{array}} \\ 0 \\ {3x}^{4} {+6x}^{3} - {2x}^{2} - 10x - 5= (x^{2} - \frac{5}{3}) ({3x}^{2} +6x+3) \\ =3 (x^{2} - \frac{5}{3}) (x^{2} +2x+1) \\ =3 (x^{2} - \frac{5}{3}) (x+1) (x+1) \\ Equating (x^{2} - \frac{5}{3}) (x+1) (x+1) equal to zero, we get \\ the zeroes of the given polynomial. \\ Hence, the zeroes of the given polynomial are \sqrt{\frac{5}{3}}, - \sqrt{\frac{5}{3}}, 1 \\ and - 1. \end{array}$ $\begin{array}{l} {p(x)=3x}^{4} {+6x}^{3} - {2x}^{2} - 10x-5 \\ The two zeroes of p(x) are \sqrt{\frac{5}{3}} and- \sqrt{\frac{5}{3}} . \\ Therefore, (x - \sqrt{\frac{5}{3}}) and (x+ \sqrt{\frac{5}{3}}) are factors of p(x). \\ Also, \\ (x - \sqrt{\frac{5}{3}}) (x+ \sqrt{\frac{5}{3}}) {= x}^{2} - \frac{5}{3} \\ {and so x}^{2} - \frac{5}{3} is a factor of p(x). \\ Now, \\ {3x}^{2} +6x+3 \\ x^{2} - \frac{5}{3} {3x}^{4} {+6x}^{3} - {2x}^{2} - 10x - 5 \\ \underline{\begin{array}{l} {3x}^{4} - {5x}^{2} \\ - + \end{array}} \\ {6x}^{3} {+3x}^{2} - 10x - 5 \\ \underline{\begin{array}{l} {6x}^{3} - 10x \\ - + \end{array}} \\ {3x}^{2} - 5 \\ \underline{\begin{array}{l} {3x}^{2} - 5 \\ - + \end{array}} \\ 0 \\ {3x}^{4} {+6x}^{3} - {2x}^{2} - 10x - 5= (x^{2} - \frac{5}{3}) ({3x}^{2} +6x+3) \\ =3 (x^{2} - \frac{5}{3}) (x^{2} +2x+1) \\ =3 (x^{2} - \frac{5}{3}) (x+1) (x+1) \\ Equating (x^{2} - \frac{5}{3}) (x+1) (x+1) equal to zero, we get \\ the zeroes of the given polynomial. \\ Hence, the zeroes of the given polynomial are \sqrt{\frac{5}{3}}, - \sqrt{\frac{5}{3}}, 1 \\ and - 1. \end{array}$

Q.7

$\begin{array}{l} On dividing x^{3} - 3 x^{2} + x + 2 by a polynomial g (x), \\ the quotient and remainder were x - 2 and - 2 x + 4, \\ respectively. Find g (x) . \end{array}$ $\begin{array}{l} On dividing x^{3} - 3 x^{2} + x + 2 by a polynomial g (x), \\ the quotient and remainder were x - 2 and - 2 x + 4, \\ respectively. Find g (x) . \end{array}$

Ans.

$\begin{array}{l} Dividend = x^{3} - 3 x^{2} + x + 2 \\ Quotient = x - 2 \\ Remainder = - 2 x + 4 \\ Divisor = g(x) = ? \\ We know that, \\ Dividend = Divisor \times Quotient + Remainder \\ ∴ x^{3} - 3 x^{2} + x + 2 = g(x) \times (x - 2) + (- 2 x + 4) \\ \Rightarrow g(x) \times (x - 2) = x^{3} - 3 x^{2} + x + 2 + 2 x - 4 \\ \Rightarrow g(x) = \frac{(x^{3} - 3 x^{2} + 3 x - 2)}{(x - 2)} \\ x - 2 \begin{matrix} x^{2} - x + 1 \\ x^{3} - 3 x^{2} + 3 x - 2 \end{matrix} \\ \underline{\begin{array}{l} x^{3} - 2 x^{2} \\ - + \end{array}} \\ - x^{2} + 3 x - 2 \\ \underline{\begin{array}{l} - x^{2} + 2 x \\ + - \end{array}} \\ x - 2 \\ \underline{\begin{array}{l} x - 2 \\ - + \end{array}} \\ 0 \\ ∴ g (x) = x^{2} - x + 1 \end{array}$ $\begin{array}{l} Dividend = x^{3} - 3 x^{2} + x + 2 \\ Quotient = x - 2 \\ Remainder = - 2 x + 4 \\ Divisor = g(x) = ? \\ We know that, \\ Dividend = Divisor \times Quotient + Remainder \\ ∴ x^{3} - 3 x^{2} + x + 2 = g(x) \times (x - 2) + (- 2 x + 4) \\ \Rightarrow g(x) \times (x - 2) = x^{3} - 3 x^{2} + x + 2 + 2 x - 4 \\ \Rightarrow g(x) = \frac{(x^{3} - 3 x^{2} + 3 x - 2)}{(x - 2)} \\ x - 2 \begin{matrix} x^{2} - x + 1 \\ x^{3} - 3 x^{2} + 3 x - 2 \end{matrix} \\ \underline{\begin{array}{l} x^{3} - 2 x^{2} \\ - + \end{array}} \\ - x^{2} + 3 x - 2 \\ \underline{\begin{array}{l} - x^{2} + 2 x \\ + - \end{array}} \\ x - 2 \\ \underline{\begin{array}{l} x - 2 \\ - + \end{array}} \\ 0 \\ ∴ g (x) = x^{2} - x + 1 \end{array}$

Q.8

$\begin{array}{l} Give examples of polynomials p (x), g (x), q (x) and r (x) \\ which satisfy the division algorithm and \\ (i) deg p (x) = deg q (x) (ii) deg q (x) = deg r (x) \\ (iii) deg r (x) = 0 \end{array}$ $\begin{array}{l} Give examples of polynomials p (x), g (x), q (x) and r (x) \\ which satisfy the division algorithm and \\ (i) deg p (x) = deg q (x) (ii) deg q (x) = deg r (x) \\ (iii) deg r (x) = 0 \end{array}$

Ans.

$\begin{array}{l} (i) deg p (x) = deg q (x) \\ if divisor is constant then \\ Degree of quotient = degree of dividend \\ Let dividend p (x) = 2 x^{2} - 2 x + 14 \\ and divisor g (x) = 2 \\ Then, wer have \\ quotient q (x) = x^{2} - x + 7 \\ and remainder r (x) = 0 \\ We find that \\ deg p (x) = deg q (x) = 2 \\ Let us check for division algorithm. \\ dividend = divisor \times quotient + remainder \\ or 2 x^{2} - 2 x + 14 = 2 (x^{2} - x + 7) + 0 = 2 x^{2} - 2 x + 14 \\ Thus, the division algorithm is satisfied. \\ (ii) deg q (x) = deg r (x) \\ Let us divide x^{3} + x by x^{2} . \\ Clearly we have, \\ dividend p (x) = x^{3} + x \\ divisor g (x) = x^{2} \\ quotient q (x) = x \\ and remainder r (x) = x \\ Here, \\ deg q (x) = deg r (x) = 1 \\ Let us check for division algorithm. \\ dividend = divisor \times quotient + remainder \\ or x^{3} + x = x^{2} \cdot x + x = x^{3} + x \\ Thus, the division algorithm is satisfied. \\ (iii) deg r (x) = 0 \\ Degree of remainder will be zero if the remainder is constant. \\ Let us divide x^{2} + 1 by x . \\ Clearly we have, \\ dividend p (x) = x^{2} + 1 \\ divisor g (x) = x \\ quotient q (x) = x \\ and remainder r (x) = 1 \\ Here, \\ deg r (x) = 0 \\ Let us check for division algorithm. \\ dividend = divisor \times quotient + remainder \\ or x^{2} + 1 = x \cdot x + 1 = x^{2} + 1 \\ Thus, the division algorithm is satisfied. \end{array}$ $\begin{array}{l} (i) deg p (x) = deg q (x) \\ if divisor is constant then \\ Degree of quotient = degree of dividend \\ Let dividend p (x) = 2 x^{2} - 2 x + 14 \\ and divisor g (x) = 2 \\ Then, wer have \\ quotient q (x) = x^{2} - x + 7 \\ and remainder r (x) = 0 \\ We find that \\ deg p (x) = deg q (x) = 2 \\ Let us check for division algorithm. \\ dividend = divisor \times quotient + remainder \\ or 2 x^{2} - 2 x + 14 = 2 (x^{2} - x + 7) + 0 = 2 x^{2} - 2 x + 14 \\ Thus, the division algorithm is satisfied. \\ (ii) deg q (x) = deg r (x) \\ Let us divide x^{3} + x by x^{2} . \\ Clearly we have, \\ dividend p (x) = x^{3} + x \\ divisor g (x) = x^{2} \\ quotient q (x) = x \\ and remainder r (x) = x \\ Here, \\ deg q (x) = deg r (x) = 1 \\ Let us check for division algorithm. \\ dividend = divisor \times quotient + remainder \\ or x^{3} + x = x^{2} \cdot x + x = x^{3} + x \\ Thus, the division algorithm is satisfied. \\ (iii) deg r (x) = 0 \\ Degree of remainder will be zero if the remainder is constant. \\ Let us divide x^{2} + 1 by x . \\ Clearly we have, \\ dividend p (x) = x^{2} + 1 \\ divisor g (x) = x \\ quotient q (x) = x \\ and remainder r (x) = 1 \\ Here, \\ deg r (x) = 0 \\ Let us check for division algorithm. \\ dividend = divisor \times quotient + remainder \\ or x^{2} + 1 = x \cdot x + 1 = x^{2} + 1 \\ Thus, the division algorithm is satisfied. \end{array}$

Q.9

$\begin{array}{l} Verify that the numbers given alongside of the cubic \\ polynomials below are their zeroes. Also verify the \\ relationship between the zeroes and the coefficients \\ in each case: \\ {(i) 2x}^{3} + x^{2} - 5 x + 2; \frac{1}{2}, 1, - 2 \\ {(ii) x}^{3} - 4 x^{2} + 5 x - 2; 2, 1, 1 \end{array}$ $\begin{array}{l} Verify that the numbers given alongside of the cubic \\ polynomials below are their zeroes. Also verify the \\ relationship between the zeroes and the coefficients \\ in each case: \\ {(i) 2x}^{3} + x^{2} - 5 x + 2; \frac{1}{2}, 1, - 2 \\ {(ii) x}^{3} - 4 x^{2} + 5 x - 2; 2, 1, 1 \end{array}$

Ans.

$\begin{array}{l} {(i) 2x}^{3} + x^{2} - 5 x + 2; \frac{1}{2}, 1, - 2 \\ Let p (x) = {2x}^{3} + x^{2} - 5 x + 2 \\ Then \\ p (\frac{1}{2}) = 2 {(\frac{1}{2})}^{3} + {(\frac{1}{2})}^{2} - 5 (\frac{1}{2}) + 2 \\ = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{2} + 2 - \frac{5}{2} = 0 \\ p (1) = 2 {(1)}^{3} + {(1)}^{2} - 5 (1) + 2 \\ = 2 + 1 - 5 + 2 = 0 \\ p (- 2) = 2 {(- 2)}^{3} + {(- 2)}^{2} - 5 (- 2) + 2 \\ = - 16 + 4 + 10 + 2 = 0 \\ Therefore, \frac{1}{2}, 1 and - 2 are zeroes of the given poynomial. \\ Comparing the given poynomial with {ax}^{3} + {bx}^{2} + cx + d, we get \\ a = 2, b = 1, c = - 5, d = 2 \\ Let the given roots are α, β and γ . Then \\ α = \frac{1}{2}, β = 1, γ = - 2 \\ Now, \\ α + β + γ = \frac{1}{2} + 1 - 2 = - \frac{1}{2} = \frac{- b}{a} \\ αβ + βγ + γα = \frac{1}{2} \times 1 + 1 \times (- 2) + (- 2) \times \frac{1}{2} = \frac{- 5}{2} = \frac{c}{a} \\ αβγ = \frac{1}{2} \times 1 \times (- 2) = - 1 = \frac{- 2}{2} = \frac{- d}{a} \\ The relationship between roots and coefficients is verified. \\ {(ii) x}^{3} - 4 x^{2} + 5 x - 2; 2, 1, 1 \\ Let p (x) = x^{3} - 4 x^{2} + 5 x - 2 \\ Then \\ p (2) = {(2)}^{3} - 4 {(2)}^{2} + 5 (2) - 2 \\ = 8 - 16 + 10 - 2 = 0 \\ p (1) = {(1)}^{3} - 4 {(1)}^{2} + 5 (1) - 2 \\ = 1 - 4 + 5 - 2 = 0 \\ Therefore, 2, 1 and 1 are zeroes of the given poynomial. \\ Comparing the given poynomial with {ax}^{3} + {bx}^{2} + cx + d, we get \\ a = 1, b = - 4, c = 5, d = - 2 \\ Let the given roots are α, β and γ . Then \\ α = 2, β = 1, γ = 1 \\ Now, \\ α + β + γ = 2 + 1 + 1 = 4 = - \frac{- 4}{1} = \frac{- b}{a} \\ αβ + βγ + γα = 2 \times 1 + 1 \times 1 + 1 \times 2 = 5 = \frac{5}{1} = \frac{c}{a} \\ αβγ = 2 \times 1 \times 1 = 2 = \frac{- (- 2)}{1} = \frac{- d}{a} \\ The relationship between roots and coefficients is verified. \end{array}$ $\begin{array}{l} {(i) 2x}^{3} + x^{2} - 5 x + 2; \frac{1}{2}, 1, - 2 \\ Let p (x) = {2x}^{3} + x^{2} - 5 x + 2 \\ Then \\ p (\frac{1}{2}) = 2 {(\frac{1}{2})}^{3} + {(\frac{1}{2})}^{2} - 5 (\frac{1}{2}) + 2 \\ = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{2} + 2 - \frac{5}{2} = 0 \\ p (1) = 2 {(1)}^{3} + {(1)}^{2} - 5 (1) + 2 \\ = 2 + 1 - 5 + 2 = 0 \\ p (- 2) = 2 {(- 2)}^{3} + {(- 2)}^{2} - 5 (- 2) + 2 \\ = - 16 + 4 + 10 + 2 = 0 \\ Therefore, \frac{1}{2}, 1 and - 2 are zeroes of the given poynomial. \\ Comparing the given poynomial with {ax}^{3} + {bx}^{2} + cx + d, we get \\ a = 2, b = 1, c = - 5, d = 2 \\ Let the given roots are α, β and γ . Then \\ α = \frac{1}{2}, β = 1, γ = - 2 \\ Now, \\ α + β + γ = \frac{1}{2} + 1 - 2 = - \frac{1}{2} = \frac{- b}{a} \\ αβ + βγ + γα = \frac{1}{2} \times 1 + 1 \times (- 2) + (- 2) \times \frac{1}{2} = \frac{- 5}{2} = \frac{c}{a} \\ αβγ = \frac{1}{2} \times 1 \times (- 2) = - 1 = \frac{- 2}{2} = \frac{- d}{a} \\ The relationship between roots and coefficients is verified. \\ {(ii) x}^{3} - 4 x^{2} + 5 x - 2; 2, 1, 1 \\ Let p (x) = x^{3} - 4 x^{2} + 5 x - 2 \\ Then \\ p (2) = {(2)}^{3} - 4 {(2)}^{2} + 5 (2) - 2 \\ = 8 - 16 + 10 - 2 = 0 \\ p (1) = {(1)}^{3} - 4 {(1)}^{2} + 5 (1) - 2 \\ = 1 - 4 + 5 - 2 = 0 \\ Therefore, 2, 1 and 1 are zeroes of the given poynomial. \\ Comparing the given poynomial with {ax}^{3} + {bx}^{2} + cx + d, we get \\ a = 1, b = - 4, c = 5, d = - 2 \\ Let the given roots are α, β and γ . Then \\ α = 2, β = 1, γ = 1 \\ Now, \\ α + β + γ = 2 + 1 + 1 = 4 = - \frac{- 4}{1} = \frac{- b}{a} \\ αβ + βγ + γα = 2 \times 1 + 1 \times 1 + 1 \times 2 = 5 = \frac{5}{1} = \frac{c}{a} \\ αβγ = 2 \times 1 \times 1 = 2 = \frac{- (- 2)}{1} = \frac{- d}{a} \\ The relationship between roots and coefficients is verified. \end{array}$

Q.10

$\begin{array}{l} Find a cubic polynomial with the sum of zeroes , sum of the \\ product of its zeroes taken two at a time, and the \\ product of its zeroes as 2, - 7, - 14 respectively. \end{array}$ $\begin{array}{l} Find a cubic polynomial with the sum of zeroes , sum of the \\ product of its zeroes taken two at a time, and the \\ product of its zeroes as 2, - 7, - 14 respectively. \end{array}$

Ans.

$\begin{array}{l} Let the polynomial be {ax}^{3} + {bx}^{2} + cx + d and its zeroes \\ be α, β and γ . \\ Given that \\ α + β + γ = 2 = \frac{- (- 2)}{1} = \frac{- b}{a} \\ αβ + βγ + γα = - 7 = \frac{- 7}{1} = \frac{c}{a} \\ αβγ = - 14 = \frac{- 14}{1} = \frac{- d}{a} \\ If we take a = 1, b = - 2, c = - 7 and d = 14 then the \\ required polynomial is x^{3} - 2 x^{2} - 7 x + 14 . \end{array}$ $\begin{array}{l} Let the polynomial be {ax}^{3} + {bx}^{2} + cx + d and its zeroes \\ be α, β and γ . \\ Given that \\ α + β + γ = 2 = \frac{- (- 2)}{1} = \frac{- b}{a} \\ αβ + βγ + γα = - 7 = \frac{- 7}{1} = \frac{c}{a} \\ αβγ = - 14 = \frac{- 14}{1} = \frac{- d}{a} \\ If we take a = 1, b = - 2, c = - 7 and d = 14 then the \\ required polynomial is x^{3} - 2 x^{2} - 7 x + 14 . \end{array}$

Q.11

$\begin{array}{l} If the zeroes of the polynomial x^{3} - 3 x^{2} + x + 1 \\ are a - b, a, a + b, find a and b. \end{array}$ $\begin{array}{l} If the zeroes of the polynomial x^{3} - 3 x^{2} + x + 1 \\ are a - b, a, a + b, find a and b. \end{array}$

Ans.

$\begin{array}{l} The given polynomial is x^{3} - 3 x^{2} + x + 1 and its zeroes \\ are a - b, a, a + b . \\ Sum of the roots = (a - b) + a + (a + b) \\ = 3 a = - \frac{Coefficient of x^{2}}{Coefficient of x^{3}} = 3 \\ \Rightarrow a = 1 \\ Sum of the products of zero taken two at a time \\ = a (a - b) + a (a + b) + (a + b) (a - b) \\ = a^{2} - ab + a^{2} + ab + a^{2} - b^{2} \\ = 3 a^{2} - b^{2} = \frac{Coefficient of x}{Coefficient of x^{3}} = 1 \\ \Rightarrow 3 a^{2} - b^{2} = 1 \\ \Rightarrow 3 .1 - b^{2} = 1 \\ \Rightarrow b^{2} = 2 \\ \Rightarrow b = \pm \sqrt{2} \\ Hence, a = 1 and b = \sqrt{2} or - \sqrt{2} \end{array}$ $\begin{array}{l} The given polynomial is x^{3} - 3 x^{2} + x + 1 and its zeroes \\ are a - b, a, a + b . \\ Sum of the roots = (a - b) + a + (a + b) \\ = 3 a = - \frac{Coefficient of x^{2}}{Coefficient of x^{3}} = 3 \\ \Rightarrow a = 1 \\ Sum of the products of zero taken two at a time \\ = a (a - b) + a (a + b) + (a + b) (a - b) \\ = a^{2} - ab + a^{2} + ab + a^{2} - b^{2} \\ = 3 a^{2} - b^{2} = \frac{Coefficient of x}{Coefficient of x^{3}} = 1 \\ \Rightarrow 3 a^{2} - b^{2} = 1 \\ \Rightarrow 3 .1 - b^{2} = 1 \\ \Rightarrow b^{2} = 2 \\ \Rightarrow b = \pm \sqrt{2} \\ Hence, a = 1 and b = \sqrt{2} or - \sqrt{2} \end{array}$

Q.12

$\begin{array}{l} If two zeroes of the polynomial x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 \\ are 2 \pm \sqrt{3}, find other zeroes . \end{array}$ $\begin{array}{l} If two zeroes of the polynomial x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 \\ are 2 \pm \sqrt{3}, find other zeroes . \end{array}$

Ans.

$\begin{array}{l} Two zeroes of the polynomial x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 \\ are 2 \pm \sqrt{3} . \\ Therefore, [x - (2 + \sqrt{3})] and [x - (2 - \sqrt{3})] are factors of the \\ given polynomial. \\ Now, \\ [x - (2 + \sqrt{3})] [x - (2 - \sqrt{3})] = [(x - 2) - \sqrt{3})] [(x - 2) + \sqrt{3})] \\ = {(x - 2)}^{2} - {(\sqrt{3})}^{2} \\ = x^{2} - 4 x + 4 - 3 \\ = x^{2} - 4 x + 1 \\ x^{2} - 4 x + 1 is also a factor of the given polynomial. \\ Therefore, we divide x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 \\ by x^{2} - 4 x + 1 to find other factors. \\ x^{2} - 4 x + 1 x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 \\ x^{4} - 4 x^{3} + x^{2} \\ - + - \\ \bar{\begin{array}{l} - 2 x^{3} - 27 x^{2} + 138 x - 35 \\ - 2 x^{3} + 8 x^{2} - 2 x \\ + - + \end{array}} \\ \bar{\begin{array}{l} - 35 x^{2} + 140 x - 35 \\ - 35 x^{2} + 140 x - 35 \\ + - + \end{array}} \\ \bar{0} \\ x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 \\ = (x^{2} - 4 x + 1) (x^{2} - 2 x - 35) \\ = [x - (2 + \sqrt{3})] [x - (2 - \sqrt{3})] (x^{2} - 7 x + 5 x - 35) \\ = [x - (2 + \sqrt{3})] [x - (2 - \sqrt{3})] (x - 7) (x + 5) \\ (x - 7) and (x + 5) are factors of x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 . \\ Hence, 7 and - 5 are zeroes of x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 . \end{array}$ $\begin{array}{l} Two zeroes of the polynomial x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 \\ are 2 \pm \sqrt{3} . \\ Therefore, [x - (2 + \sqrt{3})] and [x - (2 - \sqrt{3})] are factors of the \\ given polynomial. \\ Now, \\ [x - (2 + \sqrt{3})] [x - (2 - \sqrt{3})] = [(x - 2) - \sqrt{3})] [(x - 2) + \sqrt{3})] \\ = {(x - 2)}^{2} - {(\sqrt{3})}^{2} \\ = x^{2} - 4 x + 4 - 3 \\ = x^{2} - 4 x + 1 \\ x^{2} - 4 x + 1 is also a factor of the given polynomial. \\ Therefore, we divide x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 \\ by x^{2} - 4 x + 1 to find other factors. \\ x^{2} - 4 x + 1 x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 \\ x^{4} - 4 x^{3} + x^{2} \\ - + - \\ \bar{\begin{array}{l} - 2 x^{3} - 27 x^{2} + 138 x - 35 \\ - 2 x^{3} + 8 x^{2} - 2 x \\ + - + \end{array}} \\ \bar{\begin{array}{l} - 35 x^{2} + 140 x - 35 \\ - 35 x^{2} + 140 x - 35 \\ + - + \end{array}} \\ \bar{0} \\ x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 \\ = (x^{2} - 4 x + 1) (x^{2} - 2 x - 35) \\ = [x - (2 + \sqrt{3})] [x - (2 - \sqrt{3})] (x^{2} - 7 x + 5 x - 35) \\ = [x - (2 + \sqrt{3})] [x - (2 - \sqrt{3})] (x - 7) (x + 5) \\ (x - 7) and (x + 5) are factors of x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 . \\ Hence, 7 and - 5 are zeroes of x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 . \end{array}$

Q.13

$\begin{array}{l} If the polynomial x^{4} - 6 x^{3} + 16 x^{2} - 25 x + 10 is divided \\ by another polynomial x^{2} - 2 x + k, the remainder comes \\ out to be x + a, find k and a . \end{array}$ $\begin{array}{l} If the polynomial x^{4} - 6 x^{3} + 16 x^{2} - 25 x + 10 is divided \\ by another polynomial x^{2} - 2 x + k, the remainder comes \\ out to be x + a, find k and a . \end{array}$

Ans.

$\begin{array}{l} We know that \\ Dividend = Divisor \times Quotient + Remainder \\ or Dividend - Remainder = Divisor \times Quotient \\ or (x^{4} - 6 x^{3} + 16 x^{2} - 25 x + 10) - (x + a) = (x^{2} - 2 x + k) \times Quotient \\ or x^{4} - 6 x^{3} + 16 x^{2} - 26 x + 10 - a = (x^{2} - 2 x + k) \times Quotient \\ Now, \\ x^{2} - 2 x + k \begin{matrix} x^{2} - 4 x + (8 - k) \\ \begin{array}{l} x^{4} - 6 x^{3} + 16 x^{2} - 26 x + 10 - a \end{array} \end{matrix} \\ \underline{\begin{array}{l} x^{4} - 2 x^{3} + {kx}^{2} \\ - + - \end{array}} \\ \underline{\begin{array}{l} - 4 x^{3} + (16 - k) x^{2} - 26 x + 10 - a \\ - 4 x^{3} + 8 x^{2} - 4 kx \\ + - + \end{array}} \\ \underline{\begin{array}{l} - (8 - k) x^{2} + (4 k - 26) x + 10 - a \\ - (8 - k) x^{2} - (- 2 k + 16) x + 8 k - k^{2} \\ + + - + \end{array}} \\ (2 k - 10) x + k^{2} - 8 k + 10 - a \\ Remainder (2 k - 10) x + k^{2} - 8 k + 10 - a must be zero. \\ So, \\ (2 k - 10) x + k^{2} - 8 k + 10 - a = 0 \\ \Rightarrow 2 k - 10 = 0 or k^{2} - 8 k + 10 - a = 0 \\ \Rightarrow k = 5 or 5^{2} - 8 \times 5 + 10 - a = 0 \\ \Rightarrow k = 5 and a = - 5 \end{array}$ $\begin{array}{l} We know that \\ Dividend = Divisor \times Quotient + Remainder \\ or Dividend - Remainder = Divisor \times Quotient \\ or (x^{4} - 6 x^{3} + 16 x^{2} - 25 x + 10) - (x + a) = (x^{2} - 2 x + k) \times Quotient \\ or x^{4} - 6 x^{3} + 16 x^{2} - 26 x + 10 - a = (x^{2} - 2 x + k) \times Quotient \\ Now, \\ x^{2} - 2 x + k \begin{matrix} x^{2} - 4 x + (8 - k) \\ \begin{array}{l} x^{4} - 6 x^{3} + 16 x^{2} - 26 x + 10 - a \end{array} \end{matrix} \\ \underline{\begin{array}{l} x^{4} - 2 x^{3} + {kx}^{2} \\ - + - \end{array}} \\ \underline{\begin{array}{l} - 4 x^{3} + (16 - k) x^{2} - 26 x + 10 - a \\ - 4 x^{3} + 8 x^{2} - 4 kx \\ + - + \end{array}} \\ \underline{\begin{array}{l} - (8 - k) x^{2} + (4 k - 26) x + 10 - a \\ - (8 - k) x^{2} - (- 2 k + 16) x + 8 k - k^{2} \\ + + - + \end{array}} \\ (2 k - 10) x + k^{2} - 8 k + 10 - a \\ Remainder (2 k - 10) x + k^{2} - 8 k + 10 - a must be zero. \\ So, \\ (2 k - 10) x + k^{2} - 8 k + 10 - a = 0 \\ \Rightarrow 2 k - 10 = 0 or k^{2} - 8 k + 10 - a = 0 \\ \Rightarrow k = 5 or 5^{2} - 8 \times 5 + 10 - a = 0 \\ \Rightarrow k = 5 and a = - 5 \end{array}$

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