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      Class 10 Maths

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      • Class 10 Maths
      CoursesClass 10MathsClass 10 Maths
      • 01. Real Numbers
        9
        • Lecture1.1
          Real Numbers and Fundamental Theorem of Arithmetic 59 min
        • Lecture1.2
          Divisibility and Euclid’s Division Lemma 49 min
        • Lecture1.3
          Finding HCF and LCM Using Fundamental Theorem of Arithmetic 02 hour
        • Lecture1.4
          Miscellaneous Questions 31 min
        • Lecture1.5
          Chapter Notes – Real Numbers
        • Lecture1.6
          NCERT Solutions – Real Numbers Exercise 1.1 – 1.4
        • Lecture1.7
          Revision Notes Real Numbers
        • Lecture1.8
          R S Aggarwal Real Numbers
        • Lecture1.9
          R D Sharma Real Numbers
      • 02. Polynomials
        11
        • Lecture2.1
          Introduction to Polynomials and Its Zeroes 44 min
        • Lecture2.2
          Finding Zeroes of Quadratic Polynomial 01 hour
        • Lecture2.3
          Relationship b/w the Zeroes and coefficients of a Polynomial 01 hour
        • Lecture2.4
          Cubic Polynomial 35 min
        • Lecture2.5
          Division and Division Algorithm for Polynomials 01 hour
        • Lecture2.6
          Geometrical Meaning of Zeroes of Polynomial 15 min
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.4
        • Lecture2.9
          Revision Notes Polynomials
        • Lecture2.10
          R S Aggarwal Polynomials
        • Lecture2.11
          R D Sharma Polynomials
      • 03. Linear Equation
        9
        • Lecture3.1
          Solution of linear Equation in Two Variable by Graphical and Algebraic Methods 01 hour
        • Lecture3.2
          Some Basic Questions on Solving Pair of Linear Equations 01 hour
        • Lecture3.3
          Some Basics Problems on Numbers and Ages 40 min
        • Lecture3.4
          Problems on Money Matters, Time, Distance, Speed, Time and Work 01 hour
        • Lecture3.5
          Chapter Notes – Linear Equation
        • Lecture3.6
          NCERT Solutions – Linear Equation Exercise 3.1 – 3.7
        • Lecture3.7
          Revision Notes Linear Equation
        • Lecture3.8
          R S Aggarwal Linear Equation
        • Lecture3.9
          R D Sharma Linear Equation
      • 04. Quadratic Equation
        8
        • Lecture4.1
          Introduction and Finding The Roots of a Quadratic Equation 37 min
        • Lecture4.2
          Miscellaneous Questions 1 02 hour
        • Lecture4.3
          Miscellaneous Questions 2 02 hour
        • Lecture4.4
          Chapter Notes – Quadratic Equation
        • Lecture4.5
          NCERT Solutions – Quadratic Equation Exercise 4.1 – 4.4
        • Lecture4.6
          Revision Notes Quadratic Equation
        • Lecture4.7
          R S Aggarwal Quadratic Equation
        • Lecture4.8
          R D Sharma Quadratic Equation
      • 05. Arithmetic Progressions
        11
        • Lecture5.1
          Introduction to Arithmetic Progression 59 min
        • Lecture5.2
          Miscellaneous Questions Based on nth Term Formula of A.P. 55 min
        • Lecture5.3
          Middle term (s) of Finite A.P. and Arithmetic Mean 01 hour
        • Lecture5.4
          Selection of the Terms and Sum of nth Terms of A.P. 01 hour
        • Lecture5.5
          Miscellaneous Questions Based on Sum of nth Term of A.P. 01 hour
        • Lecture5.6
          Word Problems related to nth term and sum of nth terms of A.P. 01 hour
        • Lecture5.7
          Chapter Notes – Arithmetic Progressions
        • Lecture5.8
          NCERT Solutions – Arithmetic Progressions Exercise 5.1 – 5.4
        • Lecture5.9
          Revision Notes Arithmetic Progressions
        • Lecture5.10
          R S Aggarwal Arithmetic Progressions
        • Lecture5.11
          R D Sharma Arithmetic Progressions
      • 06. Some Applications of Trigonometry
        7
        • Lecture6.1
          Angle of Elevation and Depression; Problems involving Elevation and Double Elevation 01 hour
        • Lecture6.2
          Miscellaneous Problems Involving Double Elevation, Depression and Elevation 01 hour
        • Lecture6.3
          Miscellaneous Questions (Level-1, 2, 3, 4) 49 min
        • Lecture6.4
          Chapter Notes – Some Applications of Trigonometry
        • Lecture6.5
          NCERT Solutions – Some Applications of Trigonometry
        • Lecture6.6
          Revision Notes Some Applications of Trigonometry
        • Lecture6.7
          R D Sharma Some Applications of Trigonometry
      • 07. Coordinate Geometry
        17
        • Lecture7.1
          Introduction to Terms related to Coordinate Geometry 49 min
        • Lecture7.2
          Locating Coordinates of Point on the Axes 37 min
        • Lecture7.3
          Finding vertices of a Geometrical Fig. and Distance b/w Two Points 58 min
        • Lecture7.4
          Distance b/w Two Points Formula Based Examples 44 min
        • Lecture7.5
          Finding the type of Triangle and Quadrilateral 01 hour
        • Lecture7.6
          Find the Collinear or Non-collinear points and Missing Vertex of a Geometrical Figure 42 min
        • Lecture7.7
          Section Formula and Corollary 01 hour
        • Lecture7.8
          Finding the Section Ratio and Involving Equation of Line 53 min
        • Lecture7.9
          Proof Related to Mid-points and Centroid of a Triangle 59 min
        • Lecture7.10
          Finding Area of Triangle and Quadrilateral 53 min
        • Lecture7.11
          Collinearity 34 min
        • Lecture7.12
          Examples Based on If Areas are Given 34 min
        • Lecture7.13
          Chapter Notes – Coordinate Geometry
        • Lecture7.14
          NCERT Solutions – Coordinate Geometry Exercise 7.1 – 7.4
        • Lecture7.15
          Revision Notes Coordinate Geometry
        • Lecture7.16
          R S Aggarwal Coordinate Geometry
        • Lecture7.17
          R D Sharma Coordinate Geometry
      • 08. Triangles
        15
        • Lecture8.1
          Introduction, Similarity of Triangles, Properties of Triangles, Thales’s Theorem 52 min
        • Lecture8.2
          Questions Based on Thales’s Theorem, Converse Thales’s Theorem 01 hour
        • Lecture8.3
          Questions Based on Converse Thales’s Theorem, Questions Based On Mid-points and its Converse 59 min
        • Lecture8.4
          Questions Based on Internal Bisectors of an Angle, Similar Triangles, AAA criterion for Similarity 52 min
        • Lecture8.5
          SSS & SAS Criterion of Similarity, Question Based on Criteria for Similarity 02 hour
        • Lecture8.6
          Questions Based on Criteria for Similarity Conti. 01 hour
        • Lecture8.7
          Questions Based On Area of Similar Triangles 02 hour
        • Lecture8.8
          Questions Based On Area of Similar Triangles cont., Pythagoras Theorem Level-1 01 hour
        • Lecture8.9
          Pythagoras Theorem Level-1 cont. 01 hour
        • Lecture8.10
          Miscellaneous Questions 01 hour
        • Lecture8.11
          Chapter Notes – Triangles
        • Lecture8.12
          NCERT Solutions – Triangles Exercise 8.1 – 8.6
        • Lecture8.13
          Revision Notes Triangles
        • Lecture8.14
          R S Aggarwal Triangles
        • Lecture8.15
          R D Sharma Triangles
      • 09. Circles
        8
        • Lecture9.1
          Introduction to Circle and Tangent and Theorems-1, 2 01 hour
        • Lecture9.2
          Property of Chord of Circle and Theorem-3 52 min
        • Lecture9.3
          Theorem 4 and Angle in Semicircle 01 hour
        • Lecture9.4
          Chapter Notes – Circles
        • Lecture9.5
          NCERT Solutions – Circles Exercise
        • Lecture9.6
          Revision Notes Circles
        • Lecture9.7
          R S Aggarwal Circles
        • Lecture9.8
          R D Sharma Circles
      • 10. Areas Related to Circles
        10
        • Lecture10.1
          Area and Perimeter of Circle, Semicircle and Quarter circle 56 min
        • Lecture10.2
          Touching Circles and Example on Bending Wire, Wheel Rotation 58 min
        • Lecture10.3
          Sector of A Circle 51 min
        • Lecture10.4
          Segment of Circle and Area of Combination of Plane Figures 01 hour
        • Lecture10.5
          Questions Based on Area of Combination of Plane Figures 01 hour
        • Lecture10.6
          Chapter Notes – Areas Related to Circles
        • Lecture10.7
          NCERT Solutions – Areas Related to Circles
        • Lecture10.8
          Revision Notes Areas Related to Circles
        • Lecture10.9
          R S Aggarwal Areas Related to Circles
        • Lecture10.10
          R D Sharma Areas Related to Circles
      • 11. Introduction to Trigonometry
        7
        • Lecture11.1
          Introduction, Trigonometric Ratios, Sums when Sides of Triangle are given, Miscellaneous Questions 01 hour
        • Lecture11.2
          Trigonometric Ratios for Some Specific Angles, Trigonometric Ratio for 30 & 60, Value Table, Word Problems 48 min
        • Lecture11.3
          Complimentary angles in Inverse Multiplication, Miscellaneous Problems 01 hour
        • Lecture11.4
          Miscellaneous Questions 01 hour
        • Lecture11.5
          Chapter Notes – Introduction to Trigonometry
        • Lecture11.6
          NCERT Solutions – Introduction to Trigonometry
        • Lecture11.7
          Revision Notes Introduction to Trigonometry
      • 12. Surface Areas and Volumes
        9
        • Lecture12.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere, Type-1 : Surface Area and Volume of A Solid 01 hour
        • Lecture12.2
          Volume Related Questions, Based On Embarkment 48 min
        • Lecture12.3
          Based on the Rate of Flowing, Frustum of Cone (A) Surface Area (B) Volume 52 min
        • Lecture12.4
          Quantity, Percentage, Very Short Answer Types Questions 44 min
        • Lecture12.5
          Chapter Notes – Surface Areas and Volumes
        • Lecture12.6
          NCERT Solutions – Surface Areas and Volumes
        • Lecture12.7
          Revision Notes Surface Areas and Volumes
        • Lecture12.8
          R S Aggarwal Surface Areas and Volumes
        • Lecture12.9
          R D Sharma Surface Areas and Volumes
      • 13. Statistics
        12
        • Lecture13.1
          Introduction and Mean of Ungrouped Data 46 min
        • Lecture13.2
          Mean of Grouped Data 58 min
        • Lecture13.3
          Median of Ungrouped and grouped Data 01 hour
        • Lecture13.4
          Mode of Ungrouped and Grouped Data 49 min
        • Lecture13.5
          Mean, Median and Mode & Cumulative Frequency Curve-Less Than Type 53 min
        • Lecture13.6
          Cumulative Frequency Curve (Ogive)-More Than Type 52 min
        • Lecture13.7
          Miscellaneous Questions 38 min
        • Lecture13.8
          Chapter Notes – Statistics
        • Lecture13.9
          NCERT Solutions – Statistics
        • Lecture13.10
          Revision Notes Statistics
        • Lecture13.11
          R S Aggarwal Statistics
        • Lecture13.12
          R D Sharma Statistics
      • 14. Probability
        9
        • Lecture14.1
          Activities, Understanding term Probability, Experiments 54 min
        • Lecture14.2
          Calculations 43 min
        • Lecture14.3
          Event, Favorable Outcomes, Probability, Some important experiments- Tossing a coin, Rolling dice, Drawing a card 01 hour
        • Lecture14.4
          Complement of an event, Elementary Events, Equally Likely events, Some problems of Probability 58 min
        • Lecture14.5
          Chapter Notes – Probability
        • Lecture14.6
          NCERT Solutions – Probability
        • Lecture14.7
          Revision Notes Probability
        • Lecture14.8
          R S Aggarwal Probability
        • Lecture14.9
          R D Sharma Probability
      • 15. Construction
        7
        • Lecture15.1
          Introduction, Construction to Divide a Line Segment in Certain Ratio, Construction of Similar Triangles 44 min
        • Lecture15.2
          Construction of Different Angles, Construction of Angle Bisector 41 min
        • Lecture15.3
          Construction of Tangent of Circle at a Given Point 35 min
        • Lecture15.4
          Chapter Notes – Construction
        • Lecture15.5
          NCERT Solutions – Construction
        • Lecture15.6
          R S Aggarwal Construction
        • Lecture15.7
          R D Sharma Construction

        NCERT Solutions – Polynomials Exercise 2.1 – 2.4

        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.
                    (i)   1
                  (ii)  2
                 (iii)3
                 (iv)4
                (v)
        5 
                 (vi)   6

        Sol.
                (i) There are no zeroes as the graph does not intersect the x-axis.
                (ii) The number of zeroes is one as the graph intersects the x-axis at one point only.
                (iii) The number of zeroes is three as the graph intersects the x-axis at three points.
                (iv) The number of zeroes is two as the graph intersects the x-axis at two points.
                (v) The number of zeroes is four as the graph intersects the x-axis at four points.
                (vi) The number of zeroes is three as the graph intersects the x-axis at three points.

         

        Q.1       Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients :
                     (i) x2−2x−8                      
                     (ii) 4s2−4s+1
                     (iii) 6x2−3−7x                                         
                     (iv) 4u2+8u
                     (v) t2−15                                                    
                     (vi) 3x2−x−4

        Sol.       (i) We have, x2−2x−8 =x2+2x−4x−8
                                                           =x(x+2)−4(x+2)
                                                           =(x+2)(x−4)
        The value of x2−2x−8 is zero when the value of (x + 2) (x – 4) is zero, i.e.,
                          when x + 2 = 0 or x – 4 = 0 , i.e., when x = – 2 or x = 4.
        So, The zeroes of x2−2x−8 are – 2 and 4.
        Therefore , sum of the zeroes = (– 2) + 4 = 2
        =−CoefficientofxCoefficientofx2
        and product of zeroes = (– 2) (4) = – 8 =−81
        =ConstanttermCoefficientofx2

        (ii) We have, 4s2−4s+1 =4s2−2s−2s+1
                                                            =2s(2s−1)−1(2s−1)
                                                            =(2s−1)(2s−1)
                         The value of 4s2−4s+1 is zero when the value of  
                         (2s – 1) (2s – 1) is zero, i.e., when 2s – 1 = 0 or 2s – 1 = 0,
                         i.e., when s=12ors=12.
                         So, The zeroes of 4s2−4s+1are12and12
                         Therefore, sum of the zeroes =12+12=1
                                                                   =−CoefficientofsCoefficientofs2
                         and product of zeroes =(12)(12)=14
                                                         =ConstanttermCoefficientofs2
                   (iii) We have, 6x2−3−7x = 6x2−7x−3
                                                           =6x2−9x+2x−3
                                                           =3x(2x−3)+1(2x−3)
                                                           =(3x+1)(2x−3)
                         The value of 6x2−3−7x is zero when the value of (3x + 1) (2x – 3) is zero, i.e., when 3x + 1 = 0 or 2x – 3 = 0, i.e, when x=−13orx=32
                        So, The zeroes of 6x2−3−7xare−13and32
                        Therefore, sum of the zeroes =−13+32=76
                                                                   =−CoefficientofxCoefficientofx2
                         and product of zeroes =(−13)(32)=−12
                                                         =ConstanttermCoefficientofx2
                   (iv)  We have, 4u2+8u = 4u (u + 2) 
                          The value of 4u2+8u is zero when the value of 4u(u + 2) is zero, i.e., when u = 0 or u + 2 = 0, i.e., when u = 0 or u = – 2.
                          So, The zeroes of 4u2+8u and 0 and – 2 
                          Therefore, sum of the zeroes = 0 + (– 2) = – 2 
                                                                   =CoefficentofuCoefficientofu2
                          and , product of zeroes = (0) (–2) = 0 
                                                            =ConstanttermCoefficientofu2
                    (v) We have t2−15 =(t−15−−√)(t+15−−√)
                         The value of t2−15 is zero when the value of (t−15−−√)(t+15−−√) is zero,
                         i.e., when t−15−−√=0ort+15−−√ = 0 i.e., when t=15−−√ort=−15−−√
                         So, The zeroes of t2−15are15−−√and−15−−√
                         Therefore , sum of the zeroes = 15−−√+(−15−−√)=0
                         =−CoefficientoftCoefficientoft2
                         and, product of the zeroes = (15−−√)(−15−−√)=−15
                                                            =ConstanttermCoefficientoft2  
                   (vi) We have, 3x2−x−4 =  3x2+3x−4x−4
                                                          =3x(x+1)−4(x+1)
                                                          =(x+1)(3x−4)
                         The value of 3x2−x−4 is zero when the value of (x + 1) (3x – 4) is zero, i.e., when x + 1 = 0 or 3x – 4 = 0, i.e., when x = – 1 or  x=43.
                         So, The zeroes of 3x2−x−4are−1and43
                         Therefore , sum of the zeroes =−1+43=−3+43
                                                                      =13=CoefficientofxCoefficientofx2
                         and, product of the zeroes =(−1)(43)=−43
                                                                =ConstanttermCoefficientofx2


        Q.2       Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
                    (i) 14,−1                 
                    (ii) 2–√,13                      
                    (iii) 0,5–√
                    (iv) 1, 1
                    (v) −14,14                                                  
                    (vi) 4, 1
        Sol.      (i) Let the polynomial be ax2+bx+c, and its zeroes be αandβ. Then , 
                                                            α+β=14=−ba
                          and,                            αβ=−1=−44=ca
                         If a = 4,  then  b = – 1 and c = – 4.
                         Therefore, one quadratic polynomial which fits the given conditions is 4x2−x−4.

        (ii) Let the polynomial be ax2+bx+c, and its zeroes be αandβ. Then,
                                                         α+β=2–√=32√3=−ba
                        and                            αβ=13=ca
                        If a = 3, then b =32–√andc=1
                        So, One quadratic polynomial which fits the given conditions is 3x2−32–√x+1.

        (iii) Let the polynomial  be ax2+bx+c, and its zeroes be αandβ. Then,

        α+β=0=01=−ba
        and                           αβ=5–√=5√1=ca
                         If  a = 1, then b = 0 and c = 5–√
                         So, one quadratic polynomial which fits the given conditions is x2−0.x+5–√,i.e.,x2+5–√.

        (iv) Let the polynomial be ax2+bx+c and its zeroes be αandβ. Then,
                                                         α+β=1
                                                         =−(−1)1=−ba
        and                               αβ=1 =11=ca
                        If  a = 1, then b = – 1 and c = 1.
                        So, one quadratic polynomial which fits the given conditions is x2−x+1.

        (v) Let the polynomial be ax2+bx+c and its zeroes be αandβ. Then
                                                           α+β=−14=−ba 
                          and                               αβ=14=ca
                          If  a = 4 then b = – 1 and c = 1. 
                          So, one quadratic polynomial which fits the given conditions is 4x2−x+1.

        (vi) Let the polynomial be ax2+bx+c and its zeroes be αandβ. Then, 
                                                           α+β=4=−ba
        αβ=1=11=ca
                          If a = 1, then b = – 4 and c = 1
                          Therefore, one quadratic polynomial which fits the given conditions is x2−4x+1.

         

        Q.1       Divide the polynomial p(x) by the polynomial g (x) and find the quotient and remainder in each of the following : 
                    (i) p(x)=x3−3x2+5x−3,g(x)=x2−2
                    (ii) p(x)=x4−3x2+4x+5,g(x)=x2+1−x
                    (iii) p(x)=x4−5x+6,g(x)=2−x2
        Sol.      (i) We have,

        7

        Therefore, the quotient is x – 3 and the remainder is 7 x – 9

        (ii) Here, the dividend is already in the standard form and the divisor is also in the standard form.
        We have,
        8

        Therefore, the quotient is x2+x−3 and the remainder is 8.

        (iii) To carry out the division, we first write divisor in the standard form.
        So, divisor = −x2+2
        We have,
        22

        Therefore, the quotient is −x2+2
        and the remainder is – 5x + 10.


        Q.2         Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial : 
                      (i) t2−3,2t4+3t3−2t2−9t−12
                      (ii) x2+3x+1,3x4+5x3−7x2+2x+2
                      (iii) x3−3x+1,x5−4x3+x2+3x+1
        Sol.      (i) Let us divide 2t4+3t3−2t2−9t−12byt2−3.
                    We have,

        10

        Since the remainder is zero, therefore, t2−3 is a factor of 2t4+3t3−2t2−9t−12.

        (ii) Let us divide 3x4+5x3−7x2+2x+2byx2+3x+1.
        We get,
        18
        Since the remainder is zero, therefore x2+3x+1 is a factor of
        3x4+5x3−7x2+2x+2

        (iii) Let us divide x5−4x3+x2+3x+1 by x3−3x+1
        We get,
        19

        Here remainder is 2(≠ 0). Therefore,  x3−3x+1 is not a factor of
        x5−4x3+x2+3x+1.


        Q.3      Obtain all the zeroes of 3x4+6x3−2x2−10x−5 if two of its zeroes are 53−−√and−53−−√.
        Sol.      Since two zeroes are 53−−√and−53−−√, so (x−53−−√)and(x+53−−√) are the factors of the given polynomial.
                   Now, (x−53−−√)(x+53−−√)=x2−53 
                   ⇒ (3x2−5) is a factor of the given polynomial. 
                   Applying the division algorithm to the given polynomial and 3x2−5, we have

        18

        Therefore, 3x4+6x3−2x2−10x−5 = (3x2−5)(x2+2x+1)
        Now, x2+2x+1 =x2+x+x+1
                                            =x(x+1)+1(x+1)
                                           =(x+1)(x+1)
        So, its other zeroes are – 1 and – 1.
        Thus,  all the zeroes of the given fourth degree polynomial are
        53−−√,−53−−√,−1and−1.


        Q.4       On dividing x3−3x2+x+2 by a polynomial g(x) , the quotient and remainder were x – 2 and – 2x + 4 respectively. Find g(x).
        Sol.         Since on dividing x3−3x2+x+2 by a polynomial (x – 2) × g(x), the quotient and remainder were (x – 2) and (– 2x + 4) respectively, therefore,
        Therefore, Quotient × Divisor + Remainder = Dividend
        ⇒ (x−2)×g(x)+(−2x+4)  =x3−3x2+x−2
        ⇒ (x – 2) × g(x) =x3−3x2+x−2+2x−4
        ⇒ g(x)=x3−3x2+3x−2x−2                  … (1)
        Let us divide x3−3x2+3x−2byx−2. We get

        14

        Therefore, equation (1) gives g (x) =x2−x+1


        Q.5     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
        Sol.     There can be several examples for each of (i), (ii) and (iii). 
                   However, one example for each case may be taken as under : 
                (i) p(x)=2x2−2x+14,g(x)=2,
                     q(x)=x2−x+7,r(x)=0
               (ii)  p(x)=x3+x2+x+1,g(x)=x2−1,
                     q(x)=x+1,r(x)=2x+2
               (iii) p(x)=x3+2x2−x+2,
                     g(x)=x2−1,q(x)=x+2,r(x)=4

         

        Q.1        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) 2x3+x2−5x+2;12,1,−2 
                      (ii) x3−4x2+5x−2;2,1,1
        Sol.      (i) Comparing the given polynomial with ax3+bx2+cx+d, we get
                    a = 2 , b = 1, c = – 5 and d = 2.
        p(12)=2(12)3+(12)2−5(12)+2
        =14+14−52+2
        =1+1−10+84=04=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
        =2(−8)+4+10+2
        =−16+16=0
        Therefore, 12, 1 and – 2 are the zeroes of 2x3+x2−5x+2.
        So, α=12,β=1andγ=−2.
        Therefore, α+β+γ=12+1+(−2)=1+2−42
        =−12=−ba
        αβ+βγ+γα=(12)(1)+(1)(−2)+(−2)(12)
        =12−2−1
        =1−4−22=−52=ca
        and     αβγ=12×1×−2=1=−22=−dα

        (ii) Comparing the given polynomial with ax3+bx2+cx+d, we get
        a = 1, b = – 4, c = 5 and d = – 2.
        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 the zeros of  x3−4x2+5x−2
        Thus,     α=2,β=1andγ=1.
        Now α+β+γ=2+1+1=4=−(−4)1=−ba
        αβ+βγ+γα=(2)(1)+(1)(1)+(1)(2)
        =2+1+2=5=51=ca
        and           αβγ=(2)(1)(1)=2=−(−2)1=−da


        Q.2       Find a cubic polynomial with the sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively. 
        Sol.     Let zeroes be α,βandγ
        We have given α+β+γ=2
        αβ+βγ+γα=−7
        and αβγ=−14
        Cubic polynomial is given by (x−α)(x−β)(x−γ)=0
        ⇒x3−x2(α+β+γ)+x(αβ+βγ+γα)−αβγ=0
        ⇒x3−2x2−7x+14=0
        αβγ=−14=−141=−da
        If a = 1, then b = – 2, c = – 7 and d = 14
        So, one cubic polynomial which fits the given conditions will be x3−2x2−7x+14.


        Q.3       If the zeroes of the polynomial x3−3x2+x+1 are a – b, a, a + b , find a and b
        Sol.       Since (a – b) , a (a + b) are the zeroes of the polynomial x3−3x2+x+1,
                     Therefore, (a−b)+a+(a+b)=−(−3)1=3 
                      ⇒ 3a = 3 ⇒ a = 1 
                       (a – b) a + a(a + b) + (a + b) (a – b) =11=1
                      ⇒ a2−ab+a2+ab+a2−b2=1
                      ⇒ 3a2−b2=1
                      ⇒ 3(1)2−b2=1 [Since a = 1]
                      ⇒ 3−b2=1
                      ⇒ b2=2
                      ⇒ b=±2–√
                    Hence , a = 1 and b=±2–√.


        Q.4       If two zeroes of the polynomial x4−6x3−26x2+138x−35are2±3–√ , find other zeroes. 
        Sol.       Since 2±3–√, are two zeroes of the polynomial p(x)=x4−6x3−26x2+138x−35
                     Let x=2±3–√ ⇒ x−2=±3–√
                     Squaring we get x2−4x+4=3
                      ⇒ x2−4x+1=0
                     Let us divide p(x) by x2−4x+1 to obtain other zeroes.

        15

        Therefore, p(x)=x4−6x3−26x2+138x−35
        =(x2−4x+1)(x2−2x−35)
        =(x2−4x+1)(x2−7x+5x−35)
        =(x2−4x+1)[x(x−7)+5(x−7)]
        =(x2−4x+1)(x+5)(x−7)
        ⇒ (x + 5) and (x – 7) are other factors of p(x).
        Therefore,  – 5 and 7 are other zeroes of the given polynomial.


        Q.5       If the polynomial x4−6x3+16x2−25x+10 is divided by another polynomial x2−2x+k, the remainder comes out to be x + a find k and a. 
        Sol.       Let us divide x4−6x3+16x2−25x+10 by x2−2x+k

        19

        So, remainder = (2 k – 9) x – (8 – k) k + 10 
                    But the remainder is given as x + a. 
                    On comparing their coefficients , we have  2 k – 9 = 1 
                    ⇒ 2 k = 10 
                    ⇒ k = 5 
                    and                             – (8 – k)k + 10 = a 
                    ⇒ a = – (8 – 5)5 + 10  = – 3 × 5 + 10 = – 15 + 10 = – 5
                    Hence, k = 5 and a = – 5.

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