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NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials

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NCERT Class 10 Maths Chapter 2: Complete Resource for Polynomials

Mathematics is a crucial subject for Class 10 students. The syllabus is designed in such a way that the students can gather knowledge and develop a conceptual foundation to carry on to the advanced classes. Class 10 Maths Chapter 2 will need special assistance from the NCERT Solutions for Class 10 Maths Chapter 2 prepared by the top mentors of Vedantu. Download the Class 10 Maths Chapter 2 Solutions PDF file for free and access it offline for your convenience.


Chapter 2 Maths Class 10 is based on polynomials. The different types of equations and their components have been described in this NCERT Maths Class 10 Chapter 2. You can easily learn the new concepts and solve the exercise questions by using the NCERT Solutions Class 10 Maths Chapter 2 and complete this chapter. Refer to the solutions while practising the problems and resolve your doubts in no time. NCERT Solutions for all subjects and classes are available on Vedantu. Science students who are looking for NCERT Solutions for Class 10 Science will also find the solutions curated by our Master Teachers really helpful.


Class:

NCERT Solutions for Class 10

Subject:

Class 10 Maths

Chapter Name:

Chapter 2 - Polynomials

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 1 Polynomials

Chapter

Dropped Topics


Polynomials


Page Number 33 - 37

2.4 Division algorithm for polynomials

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Exercises under NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT Solutions for Class 10 Maths Chapter 2, "Polynomials," is based on the concept of polynomials and their applications. The chapter consists of the following exercises:

Exercise 2.1: This exercise discusses the concept of polynomials and their terms.

Exercise 2.2: This exercise covers the addition and subtraction of polynomials.

Exercise 2.3: This exercise explains the multiplication of polynomials and the use of identities such as (a+b)², (a-b)², and (a+b)(a-b).

Exercise 2.4: This exercise deals with the factorization of quadratic polynomials.


Access NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

Exercise - 2.1

1. The graphs of \[\text{y=p(x)}\] are given in following figure, for some Polynomials \[\text{p(x)}\]. Find the number of zeroes of \[\text{p(x)}\], in each case.


Straight line parallel to x-axis


Ans: The graph does not intersect the \[\text{x-axis}\] at any point. Therefore, it does not have any zeroes.


Graph intersect x-axis at one point


Ans: The graph intersects at the \[\text{x-axis}\] at only \[\text{1}\]point. Therefore, the number of zeroes is \[\text{1}\].


Graph intersect x-axis at three points


Ans: The graph intersects at the \[\text{x-axis}\] at \[\text{3}\] points. Therefore, the number of zeroes is \[\text{3}\].


Graph intersect x-axis at twice


Ans: The graph intersects at the \[\text{x-axis}\] at \[\text{2}\] points. Therefore, the number of zeroes is \[\text{2}\]. 


Graph intersect x-axis at one point and touch at two points


Ans: The graph intersects at the \[\text{x-axis}\] at \[\text{4}\] points. Therefore, the number of zeroes is \[\text{4}\].


Graph intersect x-axis at four points


Ans: The graph intersects at the \[\text{x-axis}\] at \[\text{3}\] points. Therefore, the number of zeroes is \[\text{3}\]. 

 

 Exercise - 2.2

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

(i) \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\]        

Ans:  Given: \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\].

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(x}-\text{4)(x+2)}\]

The value of \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\] is zero. 

when \[\text{x}-\text{4=0}\] or \[\text{x+2=0}\]. i.e., \[\text{x = 4}\] or \[\text{x = }-\text{2}\]

Therefore, the zeroes of \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\] are \[\text{4}\] and \[-2\].

Now, Sum of zeroes\[\text{=4}-\text{2= 2 =}-\dfrac{\text{2}}{\text{1}}\text{=}-\dfrac{\text{Coefficient of x}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\] $ \therefore Sum\,of\,zeroes=-\dfrac{Coefficient\,of\,x}{Coefficient\,of\,{{x}^{2}}} $ 

Product of zeroes \[\text{=4 }\times\text{ (}-\text{2)=}-\text{8=}\dfrac{\text{(}-\text{8)}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore \text{Product of zeroes}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ .

(ii) \[\mathbf{4{{s}^{2}}-4s+1}\]

Ans: Given: \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow {{\text{(2s}-\text{1)}}^{\text{2}}}\]

The value of \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\] is zero. 

when \[\text{2s}-\text{1=0}\], \[\text{2s}-\text{1=0}\]. i.e., \[\text{s =}\dfrac{\text{1}}{\text{2}}\] and \[\text{s =}\dfrac{\text{1}}{\text{2}}\]

Therefore, the zeroes of \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\] are \[\dfrac{\text{1}}{\text{2}}\] and \[\dfrac{\text{1}}{\text{2}}\].

Now, Sum of zeroes\[\text{=}\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{2}}\text{=1=}\dfrac{\text{(}-\text{4)}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

\[\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

Product of zeroes\[=\dfrac{\text{1}}{\text{2}}\text{ }\times\text{ }\dfrac{\text{1}}{\text{2}}=\dfrac{\text{1}}{\text{4}}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

 $ \therefore Product of zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}} $ .

(iii) \[\mathbf{6{{x}^{2}}-3-7x}\]

Ans: Given: \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\]

\[\Rightarrow \text{6}{{\text{x}}^{\text{2}}}-\text{7x}-\text{3}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(3x+1)(2x}-\text{3)}\]

The value of  \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\] is zero. 

when \[\text{3x+1=0}\] or \[\text{2x}-\text{3=0}\]. i.e., \[\text{x =}\dfrac{-\text{1}}{\text{3}}\] or \[\text{x =}\dfrac{\text{3}}{\text{2}}\].

Therefore, the zeroes of  \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\] are \[\dfrac{\text{-1}}{\text{3}}\] and \[\dfrac{\text{3}}{\text{2}}\].

Now, Sum of zeroes\[\text{=}\dfrac{-\text{1}}{\text{3}}\text{+}\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{\text{7}}{\text{6}}\text{=}\dfrac{-\text{(}-\text{7)}}{\text{6}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore Sum of zeroes\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

Product of zeroes\[\text{=}\dfrac{-\text{1}}{\text{3}}\text{ }\times\text{ }\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{-\text{3}}{\text{6}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

(iv) \[\mathbf{4{{u}^{2}}+8u}\]                          

Ans: Given: \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\]

\[\Rightarrow \text{4}{{\text{u}}^{\text{2}}}\text{+8u+0}\] 

\[\Rightarrow \text{4u(u+2)}\]

The value of \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\] is zero. 

when \[\text{4u=0}\] or \[\text{u+2=0}\]. i.e., \[\text{u = 0}\] or \[\text{u =}-\text{2}\]

Therefore, the zeroes of \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\] are \[\text{0}\] and \[\text{-2}\].

Now, Sum of zeroes\[\text{=0+(}-\text{2)=}-\text{2=}\dfrac{-\text{8}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}\]

 $ \therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}} $ 

Product of zeroes\[\text{=0 }\times\text{ (}-\text{2)= 0 =}\dfrac{\text{0}}{\text{4}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}\]

 $ \therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}} $ 

(v) \[\mathbf{{{t}^{2}}-15}\]

Ans: Given: \[{{\text{t}}^{\text{2}}}-\text{15}\]

\[\Rightarrow {{\text{t}}^{\text{2}}}-\text{0t}-\text{15}\] 

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(t}-\sqrt{\text{15}}\text{)(t +}\sqrt{\text{15}}\text{)}\] 

The value of \[{{\text{t}}^{\text{2}}}-\text{15}\] is zero. 

when \[\text{t}-\sqrt{\text{15}}\text{=0}\] or \[\text{t+}\sqrt{\text{15}}\text{=0}\], i.e., \[\text{t=}\sqrt{\text{15}}\] or \[\text{t=}-\sqrt{\text{15}}\]

Therefore, the zeroes of \[{{\text{t}}^{\text{2}}}-\text{15}\] are \[\sqrt{\text{15}}\] and \[-\sqrt{\text{15}}\].

Now, Sum of zeroes\[\text{=}\sqrt{\text{15}}\text{+(}-\sqrt{\text{15}}\text{)=0=}\dfrac{-\text{0}}{\text{1}}\text{=}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}\]

 $ \therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}} $ 

Product of zeroes\[\text{=}\left( \sqrt{\text{15}} \right)\times \left( -\sqrt{\text{15}} \right)\text{=}-\text{15=}\dfrac{-\text{15}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}\]

 $ \therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}} $ .

(vi) \[\mathbf{3{{x}^{2}}-x-4}\]

Ans:  Given: \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\]

Now factorize the given polynomial to get the roots.

 $ \Rightarrow \left( 3x-4 \right)(x+1) $ 

The value of \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\] is zero.

when \[\text{3x}-\text{4=0}\] or \[\text{x+1=0}\], i.e., \[\text{x=}\dfrac{\text{4}}{\text{3}}\] or \[\text{x=}-\text{1}\]

Therefore, the zeroes of \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\] are \[\dfrac{\text{4}}{\text{3}}\] and \[\text{-1}\].

Now, Sum of zeroes\[\text{=}\dfrac{\text{4}}{\text{3}}\text{+(}-\text{1)=}\dfrac{\text{1}}{\text{3}}\text{=}\dfrac{-\text{(}-\text{1)}}{\text{3}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

Product of zeroes\[\text{=}\dfrac{\text{4}}{\text{3}}\times \text{(}-\text{1)=}\dfrac{-\text{4}}{\text{3}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ .


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

(i)  \[\mathbf{\dfrac{1}{4},-1}\]  

Ans: Given: \[\dfrac{\text{1}}{\text{4}}\text{,-1}\]

Let the zeroes of polynomial be \[\text{ }\alpha\text{ }\] and \[\text{ }\beta\text{ }\].

Then, 

\[\text{ }\alpha\text{ + }\beta\text{  =}\dfrac{\text{1}}{\text{4}}\]

\[\text{ }\alpha\text{  }\beta\text{ =}-\text{1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

 $ \Rightarrow {{x}^{2}}-\dfrac{1}{4}x-1 $ 

 $ \Rightarrow 4{{x}^{2}}-x-4 $ 

Therefore, the quadratic polynomial is \[\text{4}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\].

(ii) \[\mathbf{\sqrt{2},\dfrac{1}{3}}\]      

Ans: Given: \[\sqrt{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}\]

Let the zeroes of polynomial be \[\text{ }\alpha \text{ }\] and \[\text{ }\beta\text{ }\].

Then, \[\text{ }\alpha\text{ + }\beta\text{  =}\sqrt{\text{2}}\]

\[\text{ }\alpha\text{  }\beta\text{ =}\dfrac{\text{1}}{\text{3}}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

\[\Rightarrow {{\text{x}}^{\text{2}}}-\sqrt{2}\text{x+}\dfrac{1}{3}\]

\[\Rightarrow 3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}\]

Therefore, the quadratic polynomial is \[3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}\].

(iii) \[\mathbf{0,\sqrt{5}}\]  

(here, root is missing)

Ans: Given: \[\text{0,}\sqrt{\text{5}}\]

Let the zeroes of polynomial be \[\text{ } \alpha \text{ }\] and \[\text{ } \beta \text{ }\].

Then,

\[\text{ } \alpha \text{ + } \beta \text{  = 0}\]

\[\text{ } \alpha \text{  } \beta \text{ =}\sqrt{\text{5}}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

\[\Rightarrow {{\text{x}}^{\text{2}}}-0\text{x+}\sqrt{5}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{5}\]

Therefore, the quadratic polynomial is \[{{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{5}}\].

(iv) \[\mathbf{1,1}\]     

Ans: Given: \[\text{1,1}\]

Let the zeroes of polynomial be \[\text{ } \alpha \text{ }\] and \[\text{ } \beta \text{ }\].

Then,

\[\text{ } \alpha \text{ + } \beta \text{  =1}\]

\[\text{ } \alpha \text{  } \beta \text{  =1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

\[\Rightarrow {{\text{x}}^{\text{2}}}-\text{1x+1}\]

Therefore, the quadratic polynomial is \[{{\text{x}}^{\text{2}}}-\text{x+1}\].

(v) \[\mathbf{-\dfrac{1}{4},\dfrac{1}{4}}\]      

Ans: Given: \[-\dfrac{\text{1}}{\text{4}}\text{,}\dfrac{\text{1}}{\text{4}}\]

Let the zeroes of polynomial be \[\text{ } \alpha \text{ }\] and \[\text{ } \beta \text{ }\].

Then,

\[\text{ } \alpha \text{ + } \beta\text{ =}-\dfrac{\text{1}}{\text{4}}\]

\[\text{ } \alpha \text{  } \beta \text{  =}\dfrac{\text{1}}{\text{4}}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

 $ \Rightarrow {{\text{x}}^{\text{2}}}-\left( -\dfrac{1}{4} \right)\text{x+}\dfrac{1}{4} $ 

 $ \Rightarrow \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1} $ 

Therefore, the quadratic polynomial is  $ \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1} $ .

(vi) \[\mathbf{4,1}\]

Ans: Given: \[\text{4,1}\]

Let the zeroes of polynomial be \[\text{ } \alpha \text{ }\] and \[\text{ }\beta \text{ }\].

Then,

\[\text{ } \alpha \text{ + } \beta \text{  = 4}\]

\[\text{ } \alpha \text{  } \beta \text{  =1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

 $ \Rightarrow {{\text{x}}^{\text{2}}}-\text{4x+1} $ 

Therefore, the quadratic polynomial is  $ {{\text{x}}^{\text{2}}}-\text{4x+1} $ .


Exercise - 2.3

1. Divide the polynomial \[\text{p(x)}\] by the polynomial \[\text{g(x)}\] and find the quotient and remainder in each of the following:

(i) \[\mathbf{p(x) = {{x}^{3}}-3{{x}^{2}}+5x-3, g(x)={{x}^{2}}-2}\]

Ans: Given: \[\text{p(x)=}{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+5x}-\text{3}\] and \[\text{g(x)=}{{\text{x}}^{\text{2}}}-\text{2}\] 

Then, divide the polynomial \[\text{p(x)}\] by \[\text{g(x)}\].

\[\begin{align}  & {{\text{x}}^{\text{2}}}-\text{2}\overset{\text{x}-\text{3}}{\overline{\left){\begin{align}  & {{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+5x}-\text{3} \\  & {{\text{x}}^{\text{3}}}\text{        }-\text{2x} \\  & \underline{-\text{          +         }} \\ \end{align}}\right.}} \\  & \text{               }-\text{3}{{\text{x}}^{\text{2}}}\text{+7x}-\text{3} \\  & \text{               }-\text{3}{{\text{x}}^{\text{2}}}\text{       + 6} \\  & \text{                 }\underline{\text{+            }-\text{  }} \\  & \text{                   }\underline{\text{        7x}-\text{9 }} \\  & \text{               } \\ \end{align}\]

Therefore, Quotient\[\text{= x}-\text{3}\]and Remainder\[\text{= 7x}-\text{9}\].

(ii) \[\mathbf{p(x) = {{x}^{4}}-3{{x}^{2}}+ 4x + 5, g(x) = {{x}^{2}}+1-x}\]

Ans: Given: \[\text{p(x) = }{{\text{x}}^{\text{4}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ 4x + 5}\Rightarrow {{\text{x}}^{\text{4}}}\text{+0}{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ 4x + 5}\]

\[\text{g(x) = }{{\text{x}}^{\text{2}}}\text{+1}-\text{x}\Rightarrow {{\text{x}}^{\text{2}}}-\text{x+1}\] 

Then, divide the polynomial \[\text{p(x)}\] by \[\text{g(x)}\].

\[\begin{align}  & {{\text{x}}^{\text{2}}}-\text{x+1}\overset{{{\text{x}}^{\text{2}}}\text{+x}-\text{3}}{\overline{\left){\begin{align}  & {{\text{x}}^{\text{4}}}\text{ + 0}{{\text{x}}^{3}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ 4x + 5} \\  & {{\text{x}}^{\text{4}}}  -{{\text{x}}^{\text{3}}}\text{  + }{{\text{x}}^{\text{2}}} \\ \end{align}}\right.}} \\  & \text{               }\underline{-\text{    +     }-\text{                   }} \\  & \text{                       }{{\text{x}}^{\text{3}}} -\text{4}{{\text{x}}^{\text{2}}}\text{+ 4x +5} \\  & \text{                       }{{\text{x}}^{\text{3}}} -{{\text{x}}^{\text{2}}}\text{  +x} \\  & \text{               }\underline{\text{     }-\text{   +      }-\text{            }} \\  & \text{                          }-\text{3}{{\text{x}}^{\text{2}}}\text{+ 3x + 5} \\  & \text{               }\underline{\begin{align}  & \text{           }-\text{3}{{\text{x}}^{\text{2}}}\text{+ 3x}-\text{3} \\  & \text{            +      }-\text{     +   } \\ \end{align}} \\  & \text{                              }\underline{\text{                8 }} \\ \end{align}\]

Therefore, Quotient\[\text{= }{{\text{x}}^{\text{2}}}\text{+ x}-\text{3}\] and Remainder\[\text{=8}\].

(iii)  \[\mathbf{p(x)={{x}^{4}}-5x+6, g(x)=2-{{x}^{2}}}\]

Ans: Given: \[\text{p(x) = }{{\text{x}}^{\text{4}}}-\text{5x+6}\Rightarrow {{\text{x}}^{\text{4}}}\text{+0}{{\text{x}}^{3}}\text{+0}{{\text{x}}^{\text{2}}}-\text{5x+6}\]

\[\text{g(x) = 2}-{{\text{x}}^{\text{2}}}\Rightarrow \text{ }-{{\text{x}}^{\text{2}}}\text{+2}\] 

Then, divide the polynomial \[\text{p(x)}\] by \[\text{g(x)}\].

\[\begin{align}  & -{{\text{x}}^{\text{2}}}\text{+2}\overset{-{{\text{x}}^{\text{2}}}-\text{2}}{\overline{\left){{{\text{x}}^{\text{4}}}\text{+0}{{\text{x}}^{3}}\text{+ 0}{{\text{x}}^{\text{2}}}-\text{5x+6}}\right.}} \\  & \text{            }{{\text{x}}^{\text{4}}}\text{        }-\text{2}{{\text{x}}^{\text{2}}} \\  & \text{            }\underline{-\text{          +                     }} \\  & \text{                            2}{{\text{x}}^{\text{2}}}-\text{5x + 6} \\  & \text{                            2}{{\text{x}}^{\text{2}}}\text{       }-\text{4} \\  & \text{            }\underline{\text{              }-\text{            +   }} \\  & \text{            }\underline{\text{                    }-\text{5x + 10}} \\ \end{align}\]

Therefore, Quotient\[\text{=}-{{\text{x}}^{\text{2}}}-\text{2}\] and Remainder\[\text{=}-\text{5x+10}\]


2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i)  \[\mathbf{{{t}^{2}}-3, 2{{t}^{4}}+3{{t}^{3}}-2{{t}^{2}}-9t-12}\]

Ans: Given: \[{{\text{t}}^{\text{2}}}-\text{3, 2}{{\text{t}}^{\text{4}}}\text{+3}{{\text{t}}^{\text{3}}}-\text{2}{{\text{t}}^{\text{2}}}-\text{9t}-\text{12}\]

Let us take first polynomial is  $ {{\text{t}}^{\text{2}}}-\text{3}\Rightarrow \text{ }{{\text{t}}^{\text{2}}}\text{+0t}-\text{3} $ .

And second polynomial is \[\text{2}{{\text{t}}^{\text{4}}}\text{+3}{{\text{t}}^{\text{3}}}-\text{2}{{\text{t}}^{\text{2}}}-\text{9t}-\text{12}\].

Now, divide the second polynomial by first polynomial.

\[{{\text{t}}^{\text{2}}}\text{+ 0}{{\text{t}}^{\text{2}}}-\text{3}\overset{\text{2}{{\text{t}}^{\text{2}}}\text{+ 3t + 4}}{\overline{\left){\begin{align}  & \text{2}{{\text{t}}^{\text{4}}}\text{+3}{{\text{t}}^{\text{3}}}-\text{2}{{\text{t}}^{\text{2}}}-\text{9t}-\text{12} \\  & \text{2}{{\text{t}}^{\text{4}}}\text{+0}{{\text{t}}^{\text{3}}}-\text{6}{{\text{t}}^{\text{2}}} \\  & \underline{-\text{   }-\text{    +                    }} \\  & \text{       2}{{\text{t}}^{\text{3}}}\text{+ 4}{{\text{t}}^{\text{2}}}-\text{9t}-\text{12} \\  & \text{       3}{{\text{t}}^{\text{3}}}\text{+ 0}{{\text{t}}^{\text{2}}}-\text{9t} \\  & \underline{\text{      }-\text{  }-\text{     +              }} \\  & \text{               4}{{\text{t}}^{\text{2}}}\text{+0t}-\text{12} \\  & \underline{\text{               4}{{\text{t}}^{\text{2}}}\text{+0t}-\text{12    }} \\  & \underline{\text{                      0               }} \\ \end{align}}\right.}}\]

Since the remainder is \[\text{0}\]

Therefore, \[{{\text{t}}^{\text{2}}}-\text{3}\] is a factor of \[\text{2}{{\text{t}}^{\text{4}}}\text{+3}{{\text{t}}^{\text{3}}}-\text{2}{{\text{t}}^{\text{2}}}-\text{9t}-\text{12}\].

(ii) \[\mathbf{{{x}^{2}}+3x+1, 3{{x}^{4}}+5{{x}^{3}}-7{{x}^{2}}+2x+2}\]

Ans: Given: \[{{\text{x}}^{\text{2}}}\text{+ 3x + 1, 3}{{\text{x}}^{\text{4}}}\text{+5}{{\text{x}}^{\text{3}}}-\text{7}{{\text{x}}^{\text{2}}}\text{+ 2x+ 2}\]

Let us take first polynomial is \[{{\text{x}}^{\text{2}}}\text{+ 3x + 1}\].

And second polynomial is \[\text{3}{{\text{x}}^{\text{4}}}\text{+5}{{\text{x}}^{\text{3}}}-\text{7}{{\text{x}}^{\text{2}}}\text{+ 2x+ 2}\]

Now, divide the second polynomial by first polynomial.

\[{{\text{x}}^{\text{2}}}\text{+ 3x + 1}\overset{\text{3}{{\text{x}}^{\text{2}}}-\text{ 4x + 2}}{\overline{\left){\begin{align}  & \text{3}{{\text{x}}^{\text{4}}}\text{+ 5}{{\text{x}}^{\text{3}}}-\text{7}{{\text{x}}^{\text{2}}}\text{+ 2x + 2} \\  & \text{3}{{\text{x}}^{\text{4}}}\text{+ 9}{{\text{x}}^{\text{3}}}\text{+ 3}{{\text{x}}^{\text{2}}} \\  & \underline{-\text{    }-\text{     }-\text{                    }} \\  & \text{     }-\text{4}{{\text{x}}^{\text{3}}}-\text{10}{{\text{x}}^{\text{2}}}\text{+ 2x + 2} \\  & \text{     }-\text{4}{{\text{x}}^{\text{3}}}-\text{12}{{\text{x}}^{\text{2}}}-\text{4x} \\  & \underline{\text{      +       +         +         }} \\  & \text{                   2}{{\text{x}}^{\text{2}}}\text{+ 6x + 2} \\  & \text{                   2}{{\text{x}}^{\text{2}}}\text{+ 6x + 2} \\  & \underline{\text{                  }-\text{  }-\text{     }-\text{  }} \\  & \underline{\text{                      0               }} \\ \end{align}}\right.}}\]

Since the remainder is \[\text{0}\]

Therefore, \[{{\text{x}}^{\text{2}}}\text{+ 3x + 1}\] is a factor of \[\text{3}{{\text{x}}^{\text{4}}}\text{+ 5}{{\text{x}}^{\text{3}}}-\text{7}{{\text{x}}^{\text{2}}}\text{+ 2x + 2}\].

(iii) \[\mathbf{{{x}^{2}}-3x+1, {{x}^{5}}-4{{x}^{3}}+{{x}^{2}}+3x+1}\]

Ans: Given: \[{{\text{x}}^{\text{2}}}-\text{3x + 1, }{{\text{x}}^{\text{5}}}-\text{4}{{\text{x}}^{\text{3}}}\text{+ }{{\text{x}}^{\text{2}}}\text{+ 3x+1}\].

Let us take first polynomial is \[{{\text{x}}^{\text{2}}}-\text{3x + 1}\].

And second polynomial is \[{{\text{x}}^{\text{5}}}-\text{4}{{\text{x}}^{\text{3}}}\text{+ }{{\text{x}}^{\text{2}}}\text{+ 3x+1}\].

Now, divide the second polynomial by first polynomial.

\[{{\text{x}}^{\text{2}}}-\text{3x + 1}\overset{{{\text{x}}^{\text{2}}}-\text{1}}{\overline{\left){\begin{align}  & {{\text{x}}^{\text{5}}}-\text{4}{{\text{x}}^{\text{3}}}\text{+ }{{\text{x}}^{\text{2}}}\text{+ 3x+1} \\  & {{\text{x}}^{\text{5}}}-\text{3}{{\text{x}}^{\text{3}}}\text{+ }{{\text{x}}^{\text{2}}} \\  & \underline{-\text{  +      }-\text{                }} \\  & \text{    }-{{\text{x}}^{\text{3}}}\text{         + 3x +1} \\  & \text{    }-{{\text{x}}^{\text{3}}}\text{         + 3x}-\text{1} \\  & \underline{\text{      +           }-\text{      + }} \\  & \text{ }\underline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,} \end{align}}\right.}}\]

Since the remainder\[\ne \text{0}\].

Therefore, \[{{\text{x}}^{\text{2}}}-\text{3x + 1}\] is not a factor of \[{{\text{x}}^{\text{5}}}-\text{4}{{\text{x}}^{\text{3}}}\text{+ }{{\text{x}}^{\text{2}}}\text{+ 3x +1}\].

3.Obtain all other zeroes of \[\text{3}{{\text{x}}^{\text{4}}}\text{+6}{{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}}-\text{10x}-\text{5}\], if two of its zeroes are \[\sqrt{\dfrac{\text{5}}{\text{3}}}\] and \[-\sqrt{\dfrac{\text{5}}{\text{3}}}\]

Ans: Let us assume, \[\text{p(x)=3}{{\text{x}}^{\text{4}}}\text{+6}{{\text{x}}^{\text{3}}}\text{-2}{{\text{x}}^{\text{2}}}\text{-10x-5}\]

Then, given two zeroes are \[\sqrt{\dfrac{\text{5}}{\text{3}}}\] and \[-\sqrt{\dfrac{\text{5}}{\text{3}}}\].

\[\therefore \left( \text{x-}\sqrt{\dfrac{\text{5}}{\text{3}}} \right)\left( \text{x+}\sqrt{\dfrac{\text{5}}{\text{3}}} \right)\text{=}\left( {{\text{x}}^{\text{2}}}-\dfrac{\text{5}}{\text{3}} \right)\] is a factor of  \[\text{3}{{\text{x}}^{\text{4}}}\text{+6}{{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}}-\text{10x}-\text{5}\]

Now, divide the given polynomial by \[{{\text{x}}^{\text{2}}}-\dfrac{\text{5}}{\text{3}}\]

\[{{\text{x}}^{\text{2}}}\text{+ 0x}-\dfrac{\text{5}}{\text{3}}\overset{\text{3}{{\text{x}}^{\text{2}}}\text{+ 6x + 3}}{\overline{\left){\begin{align}  & \text{3}{{\text{x}}^{\text{4}}}\text{+6}{{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}}-\text{10x}-\text{5} \\  & \text{3}{{\text{x}}^{\text{4}}}\text{+0}{{\text{x}}^{\text{3}}}-\text{5}{{\text{x}}^{\text{2}}} \\  & \underline{-\text{   }-\text{      +                     }} \\  & \text{        6}{{\text{x}}^{\text{3}}}\text{+ 3}{{\text{x}}^{\text{2}}}-\text{10x}-\text{5} \\  & \text{        6}{{\text{x}}^{\text{3}}}\text{+ 0}{{\text{x}}^{\text{2}}}-\text{10x} \\  & \underline{       -\text{     }-\text{     +                }} \\  & \text{                  3}{{\text{x}}^{\text{2}}}\text{+ 0x}-\text{5} \\  & \text{                  3}{{\text{x}}^{\text{2}}}\text{+ 0x}-\text{5} \\  & \underline{\text{                }-\text{    }-\text{     +   }} \\  & \underline{\text{                       0             }} \\  & \text{        } \\ \end{align}}\right.}}\]  

\[\therefore \text{3}{{\text{x}}^{\text{4}}}\text{+ 6}{{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}}-\text{10x}-\text{5=}\left( {{\text{x}}^{\text{2}}}-\dfrac{\text{5}}{\text{3}} \right)\left( \text{3}{{\text{x}}^{\text{2}}}\text{+6x+3} \right)\]

\[\Rightarrow \text{3}\left( {{\text{x}}^{\text{2}}}-\dfrac{\text{5}}{\text{3}} \right)\left( {{\text{x}}^{\text{2}}}\text{+ 2x + 1} \right)\] 

\[\Rightarrow \left( \text{3}{{\text{x}}^{\text{2}}}-\text{5} \right)\left( {{\text{x}}^{\text{2}}}\text{+ 2x +1} \right)\]

Now, factorize the polynomial \[{{\text{x}}^{\text{2}}}\text{+ 2x +1}\]

\[\Rightarrow {{\left( \text{x+1} \right)}^{\text{2}}}\]

Hence, its zero is given by \[\text{x+1=0}\]

\[\Rightarrow \text{x=}-\text{1}\]

As it has the term \[{{\left( \text{x+1} \right)}^{\text{2}}}\], then, there will be \[\text{2}\] zeroes at \[\text{x =}-\text{1}\]. 

Therefore, the zeroes of the given polynomial are \[\sqrt{\dfrac{\text{5}}{\text{3}}}\text{,}-\sqrt{\dfrac{\text{5}}{\text{3}}}\text{,}-\text{1}\] and \[-\text{1}\].


4. On dividing \[{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x + 2}\] by a polynomial \[\text{g(x)}\], the quotient and remainder were \[\text{x}-\text{2}\] and \[-\text{2x + 4}\], respectively. Find \[\text{g(x)}\].

Ans: Let us take the dividend as \[\text{p(x) }\]. Then, \[\text{p(x) = }{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x + 2}\]  

And the divisor is \[\text{g(x)}\]. Then, find the value of \[\text{g(x)}\].

Quotient\[\text{=}\left( \text{x}-\text{2} \right)\]

Remainder\[\text{=(}-\text{2x+ 4)}\]

\[\text{Dividend = Divisor  }\times \text{  Quotient + Remainder}\]

\[{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x + 2 = g(x) } \times \text{ (x}-\text{2)+(}-\text{2x+4)}\]

\[{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x + 2 + 2x}-\text{4 = g(x)}\times \text{(x}-\text{2)}\] 

\[{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ 3x}-\text{2 = g(x)}\times \text{(x}-\text{2)}\] 

Hence, \[\text{g(x)}\] is the quotient when we divide \[\left( {{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+3x}-\text{2} \right)\] by \[\left( \text{x}-\text{2} \right)\].

\[\text{x}-\text{2}\overset{{{\text{x}}^{\text{2}}}-\text{x+1}}{\overline{\left){\begin{align}  & {{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ 3x }-\text{ 2} \\  & {{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}} \\  & \underline{-\text{   +                }} \\  & \text{    }-{{\text{x}}^{\text{2}}}\text{+ 3x}-\text{2} \\  & \text{    }-{{\text{x}}^{\text{2}}}\text{+ 2x} \\  & \underline{\text{     +    }-\text{         }} \\  & \underline{\begin{align}  & \text{               x}-2 \\  & \text{               x}-2 \\  & \underline{\text{             }-\text{  +   }} \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\ \end{align}} \\ \end{align}}\right.}}\]

\[\therefore \text{g(x)=}\left( {{\text{x}}^{\text{2}}}-\text{x+1} \right)\]


5. Give examples of polynomial \[p(x), g(x), q(x)\] and \[r(x)\], which satisfy the division algorithm and

(i) \[\mathbf{deg p(x)=deg q(x)}\]

Ans: According to the division algorithm, if \[\text{p(x)}\] and \[\text{g(x)}\] are two polynomials with 

\[\text{g(x)}\ne \text{0}\], then we can find polynomials \[\text{q(x)}\] and \[\text{r(x)}\] such that

\[\text{p(x) = g(x)  }\times \text{  q(x) + r(x)}\],

Where \[\text{r(x)=0}\] or degree of \[\text{r(x)}\] degree of \[\text{g(x)}\]

Degree of a polynomial is the highest power of the variable in the polynomial.

Given: \[\text{deg p(x) = deg q(x)}\]

The degree of the quotient will be equal to the degree of dividend when the divisor is constant (i.e., when any polynomial is divided by a constant).

Let us assume the division of \[\text{6}{{\text{x}}^{\text{2}}}\text{+ 2x + 2}\] by \[\text{2}\].

Here, \[\text{p(x) = 6}{{\text{x}}^{\text{2}}}\text{+ 2x + 2}\]

\[\text{g(x) = 2}\]

\[\text{q(x) = 3}{{\text{x}}^{\text{2}}}\text{+ x + 1}\] and \[\text{r(x) = 0}\]

Degree of \[\text{p(x)}\]and \[\text{q(x)}\] is the same i.e., \[\text{2}\].

Checking for division algorithm,

\[\text{p(x) = g(x)  }\times\text{  q(x) + r(x)}\]

\[\text{6}{{\text{x}}^{\text{2}}}\text{+ 2x + 2 = 2}\left( \text{3}{{\text{x}}^{\text{2}}}\text{+ x + 1} \right)\] 

\[\text{6}{{\text{x}}^{\text{2}}}\text{+ 2x + 2}\,\text{= 6}{{\text{x}}^{\text{2}}}\text{+ 2x + 2}\] 

Therefore, the division algorithm is satisfied. 

(ii) \[\mathbf{deg q(x)=deg r(x)}\]

Ans: Given: \[\text{deg q(x) = deg r(x)}\]

Let us assume the division of \[{{\text{x}}^{\text{3}}}\text{+x}\] by \[{{\text{x}}^{\text{2}}}\]

Here, \[\text{p(x) = }{{\text{x}}^{\text{3}}}\text{+ x}\]

\[\text{g(x) = }{{\text{x}}^{\text{2}}}\] 

\[\text{q(x) = x}\] and \[\text{r(x) = x}\]

Clearly, the degree of \[\text{q(x)}\] and \[\text{r(x)}\] is the same, i.e.,

Checking for division algorithm,

\[\text{p(x) = g(x)  }\times\text{  q(x) + r(x)}\]

\[{{\text{x}}^{\text{3}}}\text{+ x =}\left( {{\text{x}}^{\text{2}}} \right)\text{ }\times\text{  x + x}\] 

\[{{\text{x}}^{\text{3}}}\text{+ x = }{{\text{x}}^{\text{3}}}\text{+ x}\] 

Therefore, the division algorithm is satisfied.

(iii) $ \mathbf{deg r\left( x \right)=0 }$ 

Ans: Given: \[\text{deg r(x) = 0}\]

Degree of remainder will be \[\text{0}\] when remainder comes to a constant.

Let us assume the division of \[{{\text{x}}^{\text{3}}}\text{+1}\] by \[{{\text{x}}^{\text{2}}}\].

Here, \[\text{p(x) = }{{\text{x}}^{\text{3}}}\text{+1}\]

\[\text{g(x) = }{{\text{x}}^{\text{2}}}\] 

\[\text{q(x) = x}\] and \[\text{r(x) =1}\]

Clearly, the degree of \[\text{r(x)}\] is \[\text{0}\]

Checking for division algorithm,

\[\text{p(x) = g(x)  }\times\text{  q(x) + r(x)}\]

\[{{\text{x}}^{\text{3}}}\text{+1=}\left( {{\text{x}}^{\text{2}}} \right)\text{ }\times\text{  x +1}\] 

\[{{\text{x}}^{\text{3}}}\text{+1= }{{\text{x}}^{\text{3}}}\text{+1}\] 

Therefore, the division algorithm is satisfied. 


Exercise - 2.4

1. Verify that the numbers given alongside the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) \[\mathbf{2{{x}^{3}}+ {{x}^{2}}-5x + 2;  \dfrac{1}{2},1,-2}\]

Ans: Let us assume \[\text{p(x) = 2}{{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{2}}}-\text{5x + 2}\]

And zeroes for this polynomial are \[\dfrac{\text{1}}{\text{2}}\text{,1,}-\text{2}\]. Then,

\[\text{p}\left( \dfrac{\text{1}}{\text{2}} \right)\text{ = 2}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{3}}}\text{+}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}-\text{5}\left( \dfrac{\text{1}}{\text{2}} \right)\text{+ 2}\]  

\[\text{=}\dfrac{\text{1}}{\text{4}}\text{+}\dfrac{\text{1}}{\text{4}}-\dfrac{\text{5}}{\text{2}}\text{+2}\]

\[\text{= 0}\]

\[\text{p(1) = 2 }\times \text{ }{{\text{1}}^{\text{3}}}\text{+}{{\text{1}}^{2}}-\text{5 }\times\text{ 1+2}\]

\[\text{= 0}\] 

\[\text{p(}-\text{2)=2(}-\text{2}{{\text{)}}^{\text{3}}}\text{+}{{\left( -\text{2} \right)}^{\text{2}}}-\text{5}\left( -\text{2} \right)\text{+2}\]

\[\text{=}-\text{16 + 4 +10 + 2}\]

\[\text{= 0}\] 

Therefore, \[\dfrac{\text{1}}{\text{2}}\text{,1}\] and \[\text{-2}\] are the zeroes of the given polynomial.

Comparing the given polynomial with \[\text{a}{{\text{x}}^{\text{3}}}\text{+b}{{\text{x}}^{\text{2}}}\text{+cx+d}\] to get, \[\text{a = 2, b = 1, c =}-\text{5,}\] and \[\text{d = 2}\].

Let us take \[\text{ } \alpha\text{  =}\dfrac{\text{1}}{\text{2}}\text{,  }\beta\text{  =1,}\] and \[\text{  }\gamma\text{  =}-\text{2}\].

Then,

\[\text{ } \alpha \text{  +  }\beta\text{  +  } \gamma\text{  =}\dfrac{\text{1}}{\text{2}}\text{+1+}\left( -\text{2} \right)\text{=}-\dfrac{\text{1}}{\text{2}}\text{=}\dfrac{-\text{b}}{\text{a}}\]

\[\therefore \text{ } \alpha \text{  +  } \beta \text{  +  } \gamma\text{  =}\dfrac{-\text{b}}{\text{a}}\]  

\[\text{ } \alpha \text{  } \beta \text{  +  } \beta \text{  } \gamma \text{  +  } \alpha \text{  } \gamma \text{  = }\dfrac{\text{1}}{\text{2}}\text{ } \times \text{ 1+1}\left( -\text{2} \right)\text{+}\dfrac{\text{1}}{\text{2}}\left( -\text{2} \right)\text{ = }\dfrac{-\text{5}}{\text{2}}\text{ = }\dfrac{\text{c}}{\text{a}}\]

\[\therefore \text{ } \alpha \text{  }\beta\text{  +  }\beta \text{  }\gamma\text{  +  } \alpha\text{  } \gamma \text{  =}\dfrac{\text{c}}{\text{a}}\]

\[\text{ }\alpha\text{  }\beta\text{  }\gamma \text{  =}\dfrac{\text{1}}{\text{2}}\text{ } \times\text{ 1 }\times \text{ }\left( -\text{2} \right)\text{ = }\dfrac{-\text{1}}{\text{1}}\text{ = }\dfrac{-\left( \text{2} \right)}{\text{2}}\text{ = }\dfrac{-\text{d}}{\text{a}}\]  

\[\therefore \text{ } \alpha\text{  } \beta \text{  } \gamma \text{  = }\dfrac{-\text{d}}{\text{a}}\]

Therefore, the relationship between the zeroes and the coefficients is verified.

(ii) \[\mathbf{{{x}^{3}}-4{{x}^{2}}+5x-2; 2,1,1}\]

Ans: Let us assume \[\text{p(x) = }{{\text{x}}^{\text{3}}}-\text{4}{{\text{x}}^{\text{2}}}\text{+ 5x}-\text{2}\]

And zeroes for this polynomial are \[\text{2,1,1}\]. Then,

\[\text{p(2) = }{{\text{2}}^{\text{3}}}-\text{4}\left( {{\text{2}}^{\text{2}}} \right)\text{+5}\left( \text{2} \right)-\text{2}\]

\[\text{= 8}-\text{16+10}-\text{2}\]

\[\text{= 0}\]        

 \[\text{p(1) = }{{\text{1}}^{\text{3}}}-\text{4}\left( {{\text{1}}^{\text{2}}} \right)\text{+5}\left( \text{1} \right)-\text{2}\]

\[\text{=1}-\text{4+5}-\text{2}\]

\[\text{= 0}\]   

Therefore, \[\text{2,1,}\] and \[\text{1}\] are the zeroes of the given polynomial.

Comparing the given polynomial with \[\text{a}{{\text{x}}^{\text{3}}}\text{+b}{{\text{x}}^{\text{2}}}\text{+cx+d}\] to get,\[\text{a = 1, b =}-\text{4, c = 5,}\]and \[\text{d =}-\text{2}\].

Let us take \[\text{ }\alpha\text{  = 2,  }\beta \text{  =1,}\] and \[\text{  } \gamma \text{  =1}\].

Then,

\[\text{ } \alpha \text{  +  } \beta \text{  +  } \gamma \text{  = 2+1+1= 4 =}\dfrac{-\text{(}-\text{4)}}{\text{1}}=\dfrac{-\text{b}}{\text{a}}\]

\[\therefore \text{ } \alpha \text{  +  } \beta \text{  +  } \gamma \text{  =}\dfrac{-\text{b}}{\text{a}}\]

\[\text{ } \alpha \text{  } \beta \text{  +  } \beta \text{  } \gamma \text{  +  } \alpha \text{  } \gamma \text{  =(2)(1)+(1)(1)+(2)(1)}\]

\[\text{= 2+1+2 = 5 =}\dfrac{\left( \text{5} \right)}{\text{1}}\text{=}\dfrac{\text{c}}{\text{a}}\]

\[\therefore \text{ } \alpha \text{  } \beta \text{  +  } \beta \text{  } \gamma \text{  +  } \alpha \text{  } \gamma \text{  =}\dfrac{\text{c}}{\text{a}}\]

\[\text{ } \alpha \text{  } \beta \text{  } \gamma \text{  = 2 } \times \text{ 1 } \times \text{ 1= 2 =}\dfrac{-\left( -\text{2} \right)}{\text{1}}\text{=}\dfrac{-\text{d}}{\text{a}}\]

\[\therefore \text{ } \alpha \text{  } \beta \text{  }\gamma\text{  = }\dfrac{-\text{d}}{\text{a}}\]

Therefore, the relationship between the zeroes and the coefficients is verified.


2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as \[\text{2,}-\text{7,}-\text{14}\] respectively.

Ans: Let the polynomial be \[\text{a}{{\text{x}}^{\text{3}}}\text{+ b}{{\text{x}}^{\text{2}}}\text{+ cx + d}\] and the zeroes be \[\text{ }\alpha\text{ ,  }\beta\text{ ,}\] and \[\text{ }\gamma\text{ }\].

Then given that,

\[\text{ }\alpha\text{ + }\beta\text{ + }\gamma\text{  =}\dfrac{\text{2}}{\text{1}}\text{=}\dfrac{-\text{b}}{\text{a}}\]

\[\text{ }\alpha\text{  }\beta\text{ + }\beta\text{  }\gamma\text{ + }\alpha\text{  }\gamma\text{  =}\dfrac{-\text{7}}{\text{1}}\text{=}\dfrac{\text{c}}{\text{a}}\]

\[\text{ }\alpha\text{  }\beta\text{  }\gamma\text{  =}\dfrac{-\text{14}}{\text{1}}\text{=}\dfrac{-\text{d}}{\text{a}}\]

If \[\text{a=1}\], then \[\text{b =}-\text{2, c =}-\text{7, d =14}\]

Therefore, the polynomial is \[{{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}}-\text{7x +14}\].


3. If the zeroes of polynomial \[{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x +1}\] are \[\text{a}-\text{b, a, a+b}\], find \[\text{a}\] and \[\text{b}\].

Ans: Let us assume \[\text{p(x) = }{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x +1}\].

And the zeroes are \[\text{a-b, a, a+b}\].

Let us assume  $ \text{ } \alpha\text{  = a-b,  }\beta\text{  = a} $  and  $ \text{ }\gamma\text{  = a+b} $.

Comparing the given polynomial with \[\text{p}{{\text{x}}^{\text{3}}}\text{+ q}{{\text{x}}^{\text{2}}}\text{+ rx + t}\] to get, 

\[\text{p = 1, q =}-\text{3, r =1,}\] and \[\text{t =1}\].

Then,

\[\text{ }\alpha \text{ + }\beta\text{ + }\gamma\text{  = a}-\text{b+ a + a+b}\]

\[\Rightarrow \dfrac{-\text{q}}{\text{p}}\text{=3a}\]

\[\Rightarrow \dfrac{-\left( -\text{3} \right)}{\text{1}}\text{=3a}\]

\[\Rightarrow \text{3=3a}\]

\[\therefore \text{a =1}\]

Then, the zeroes are \[\text{1-b,1+b}\].

Now,

\[\text{ } \alpha \text{  } \beta \text{  } \gamma \text{  = 1}\left( \text{1-b} \right)\left( \text{1+b} \right)\]

\[\Rightarrow \dfrac{-\text{t}}{\text{p}}\text{=1}-{{\text{b}}^{\text{2}}}\]

\[\Rightarrow \dfrac{-\text{1}}{\text{1}}\text{=1}-{{\text{b}}^{\text{2}}}\] 

\[\Rightarrow \text{1}-{{\text{b}}^{\text{2}}}\text{=}-\text{1}\]

\[\Rightarrow {{\text{b}}^{\text{2}}}=2\]

\[\therefore \text{b =  }\!\!\pm\!\!\text{ }\sqrt{\text{2}}\]

Therefore, \[\text{a=1}\] and \[\text{b =}\pm \sqrt{\text{2}}\]. 


4. If two zeroes of the polynomial \[{{\text{x}}^{\text{4}}}-\text{6}{{\text{x}}^{\text{3}}}-\text{26}{{\text{x}}^{\text{2}}}\text{+138x}-\text{35}\] are \[\text{2 }\!\!\pm\!\!\text{ }\sqrt{\text{3}}\], find other zeroes.

Ans: Given that \[\text{2+}\sqrt{\text{3}}\] and \[\text{2}-\sqrt{\text{3}}\] are zeroes of the given polynomial.

Therefore, \[\left( \text{x}-\text{2}-\sqrt{\text{3}} \right)\left( \text{x}-\text{2+}\sqrt{\text{3}} \right)\text{= }{{\text{x}}^{\text{2}}}\text{+4}-\text{4x}-\text{3}\] 

\[\text{= }{{\text{x}}^{\text{2}}}-\text{4x+1}\] 

Hence, \[{{\text{x}}^{\text{2}}}-\text{4x+1}\] is a factor of the given polynomial.

For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing the polynomial \[{{\text{x}}^{\text{4}}}-\text{6}{{\text{x}}^{\text{3}}}-\text{26}{{\text{x}}^{\text{2}}}\text{+138x}-\text{35}\] by \[{{\text{x}}^{\text{2}}}-\text{4x+1}\].

\[{{\text{x}}^{\text{2}}}-\text{4x+1}\overset{{{\text{x}}^{\text{2}}}-\text{2x}-\text{35}}{\overline{\left){\begin{align}  & {{\text{x}}^{\text{4}}}-\text{6}{{\text{x}}^{\text{3}}}-\text{26}{{\text{x}}^{\text{2}}}\text{+138x}-\text{35} \\  & {{\text{x}}^{\text{4}}}-\text{4}{{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{2}}} \\  & \underline{-\text{  +      }-\text{                          }} \\  & \text{    }-\text{2}{{\text{x}}^{\text{3}}}-\text{27}{{\text{x}}^{\text{2}}}\text{+138x}-\text{35} \\  & \text{    }-\text{2}{{\text{x}}^{\text{3}}}\text{+ 8}{{\text{x}}^{\text{2}}}   -\text{2x} \\  & \underline{\text{     +      }-\text{        +                 }} \\  & \text{              }-\text{35}{{\text{x}}^{\text{2}}}\text{+140x}-\text{35} \\  & \text{              }-\text{35}{{\text{x}}^{\text{2}}}\text{+140x}-\text{35} \\  & \underline{\text{               +       }-\text{         +   }} \\  & \underline{\text{                       0                   }} \\ \end{align}}\right.}}\]

Clearly, \[{{\text{x}}^{\text{4}}}-\text{6}{{\text{x}}^{\text{3}}}-\text{26}{{\text{x}}^{\text{2}}}\text{+138x}-\text{35=}\left( {{\text{x}}^{\text{2}}}-\text{4x+1} \right)\left( {{\text{x}}^{\text{2}}}-\text{2x}-\text{35} \right)\]

Then, \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{35}\] is also a factor of the given polynomial.

And,\[{{\text{x}}^{\text{2}}}-\text{2x}-\text{35=}\left( \text{x}-\text{7} \right)\left( \text{x+5} \right)\]

Therefore, the value of the polynomial is also zero when \[\text{x-7=0}\] Or \[\text{x+5=0}\]

Hence, \[\text{x=7}\] or \[\text{-5}\]

Therefore, \[\text{7}\] and \[\text{-5}\] are also zeroes of this polynomial.


5. If the polynomial \[{{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-25x-10}\] is divided by another polynomial \[{{\text{x}}^{\text{2}}}\text{-2x+k}\], the remainder comes out to be \[\text{x+a}\], find \[\text{k}\] and \[\text{a}\].

Ans:Given:\[{{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-25x-10}\] and \[{{\text{x}}^{\text{2}}}\text{-2x+k}\].

Then, the remainder is \[\text{x+a}\]

By division algorithm,

\[\text{Dividend = Divisor }\!\!\times\!\!\text{ Quotient+Remainder}\]  

\[\text{Dividend-Remainder= Divisor }\!\!\times\!\!\text{ Quotient}\]  

\[{{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-25x-10-x-a}\Rightarrow {{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-26x+10-a}\] will be perfectly divisible by \[{{\text{x}}^{\text{2}}}\text{-2x+k}\].

Let us divide \[{{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-26x+10-a}\] by \[{{\text{x}}^{\text{2}}}\text{-2x+k}\]

\[{{\text{x}}^{\text{2}}}\text{-2x+k}\overset{{{\text{x}}^{\text{2}}}\text{-4x+}\left( \text{8-k} \right)}{\overline{\left){\begin{align}  & {{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-26x+10-a} \\  & {{\text{x}}^{\text{4}}}\text{-2}{{\text{x}}^{\text{3}}}\text{+k}{{\text{x}}^{\text{2}}} \\  & \underline{\text{-  +     -                              }} \\  & \text{    -4}{{\text{x}}^{\text{3}}}\text{+}\left( \text{16-k} \right){{\text{x}}^{\text{2}}}\text{-26x} \\  & \text{    -4}{{\text{x}}^{\text{3}}}\text{+          8}{{\text{x}}^{\text{2}}}\text{-4kx} \\  & \underline{\text{   +       -               +                }} \\  & \left( \text{8-k} \right){{\text{x}}^{\text{2}}}\text{-}\left( \text{26-4k} \right)\text{x+10 - a} \\  & \left( \text{8-k} \right){{\text{x}}^{\text{2}}}\text{-}\left( \text{16-2k} \right)\text{x+}\left( \text{8k-}{{\text{k}}^{\text{2}}} \right) \\  & \underline{\text{-              +                -            }} \\  & \underline{\left( \text{-10+2k} \right)\text{x+}\left( \text{10-a-8k+}{{\text{k}}^{\text{2}}} \right)\text{ }}\text{     }  \end{align}}\right.}}\]

Hence, the reminder \[\left( \text{-10+2k} \right)\text{x+}\left( \text{10-a-8k+}{{\text{k}}^{\text{2}}} \right)\] will be \[\text{0}\].

Then, \[\left( \text{-10+2k} \right)\text{=0}\] and \[\left( \text{10-a-8k+}{{\text{k}}^{\text{2}}} \right)\text{=0}\]

For \[\left( \text{-10+2k} \right)\text{=0}\]

\[\text{2k=10}\]

\[\therefore \text{k=5}\]

For \[\left( \text{10-a-8k+}{{\text{k}}^{\text{2}}} \right)\text{=0}\]

\[\text{10-a-8 }\!\!\times\!\!\text{ 5+25=0}\]

\[\text{10-a-40+25=0}\]

\[\text{-5-a=0}\]

\[\therefore \text{a=-5}\]

Hence, \[\text{k=5}\] and \[\text{a=-5}\].\[\]


NCERT Solutions for Class 10 Maths Chapter 2 Polynomials - Free PDF Download

You can opt for Chapter 2 - Polynomials NCERT Solutions for Class 10 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.


Important Topics under NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Chapter 2 Polynomials is an important chapter in the mathematics syllabus for Class 10. This chapter on Polynomials includes a number of topics, and in order to internalize the ideas properly, students are advised to go through the important topics under Polynomials, thoroughly. We have provided the following list of important topics covered in this chapter for a better understanding of the concept of Polynomials.

  • Definition

  • Degree of a polynomial

  • Types of polynomial

  1. Constant polynomial

  2. Linear polynomial

  3. Quadratic polynomial

  4. Cubic polynomial

  • Value of a polynomial

  • Zero of a polynomial

  • Graph of a polynomial

  • Meaning of the zeroes of a quadratic polynomial

Importance of Polynomials

Polynomials are expressions that have more than two algebraic terms. They can also be defined as the sum of several terms where the same variable/variables has/have different powers.

They are important because they have applications in most mathematical expressions. They are used to represent appropriate relations between different variables or numbers. We encourage students to learn from this chapter to be able to solve tricky problems easily in exams.


NCERT Solutions for Class 10 Maths Chapter 2 Exercises

Chapter 2 Polynomials all  Exercises in PDF Format

Exercise 2.1

1 Questions & Solutions

Exercise 2.2

2 Questions & Solutions

Exercise 2.3

5 Questions & Solutions

Exercise 2.4

5 Questions & Solutions

Polynomials: NCERT Solutions for Class 10 Maths Chapter 2 Summary

Class 10 is an important level for your academic excellence. This is the first time you will experience CBSE board exams held nationwide. Your performance will be determined by the depth of your knowledge and your answering skills. To ace the exam, you can refer to the Maths Ch 2 Class 10 NCERT Solutions prepared by the top mentors. By using the Ch 2 Maths Class 10 NCERT Solutions, you can prepare the chapter better.


The exercises in the Class 10 Maths Chapter 2 have questions related to the new concepts you will learn. These questions in Chapter 2 Maths Class 10 have been designed to deliver a platform to judge your problem-solving skills. You will need the best Maths Class 10 Chapter 2 Solutions for understanding the various approaches to solve the problems and prepare the chapter well.


If you study Chapter 2 Class 10 Maths, you will discover concepts related to the geometrical meanings of polynomials. The graphical representation of the different polynomial equations in Polynomial Class 10 NCERT Solutions will help you understand the concepts well. If you need further help you can proceed to the simplest explanation given in the NCERT Solutions for Class 10th Maths Chapter 2. These simple answers to the exercise questions have been formatted by the experienced teachers. They know where students generally feel confused. They aim to solve the general queries that arise in the minds of Class 10 students in the NCERT Solution Class 10 Maths Chapter 2 in such a way that the complex concepts become easily comprehensible.


Benefits of Using NCERT Solutions for Class 10 Maths Polynomials

The Polynomials Class 10 Solutions have been designed to deliver the following benefits to the students.

Answers to Exercise Questions

All the answers provided in the NCERT Solutions of Class 10 Maths Chapter 2 are framed by the expert teachers of Vedantu. You can rely on the quality of the answers in Class 10 Maths Ch 2 Solutions. Maths NCERT Solutions Class 10 Chapter 2 is formulated following the CBSE guidelines.

Understanding the Concepts of Chapter 2 Polynomials Class 10

As mentioned earlier, new concepts will be taught in Chapter 2 Polynomials. These concepts will then have to be used to solve the problems in the exercises. You can easily grasp the concepts by using the NCERT Solutions Class 10 Maths Ch 2. The simplification and utilization of the concepts in the answers will help you solve the problems in the future.

Developing a Strategy

There is no easiest way to develop a strategy other than using Class 10 Ch 2 Maths NCERT Solutions to practice solving the exercise problems. Learn efficient approaches and develop a strong strategy to answer questions and save time during an exam.

Doubt Clarification

The doubts arising during studying Maths Class 10 Chapter 2 can be resolved using the simplest NCERT Class 10 Maths Chapter 2 Solutions prepared by the teachers. You can now resolve your doubts on your own and complete preparing the chapter efficiently.

Quick Revision

Revise Class 10th Maths Chapter 2 before an exam by referring to the Class 10 Maths Chapter 2 Solutions and save time. Use your time to complete other chapters and their NCERT solutions too.


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FAQs on NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials

1. How can You Practice Polynomials Questions for Class 10?

Ans: To practice Polynomials questions for Class 10, solve a variety of exercises from textbooks or practice guides. Understand the concepts, formulas, and methods of polynomial manipulation. Attempt sample papers or previous year's question papers to familiarize yourself with different types of polynomial problems and enhance problem-solving skills.

2. Why use Class 10 Maths Chapter 2 NCERT Solutions?

Ans: Using Class 10 Maths Chapter 2 NCERT solutions is beneficial as they provide accurate and comprehensive answers to all the questions in the textbook. These solutions are prepared by subject-matter experts and are aligned with the latest CBSE curriculum. By using NCERT solutions, students can clarify their doubts, understand the step-by-step solution approach, and improve their problem-solving skills. It helps in building a strong foundation in mathematics and ensures thorough preparation for exams.

3. What are the real-life applications of Class 10th Mathematics Chapter 2?

Ans: Class 10 Mathematics Chapter 2, "Polynomials," has several real-life applications. Some include:

  • Finance: Polynomials are used in financial calculations, such as compound interest, loan amortization and investment analysis.

  • Engineering: Polynomial equations are utilized in various engineering fields, including electrical circuit analysis, control systems, and structural design.

  • Data Analysis: Polynomial regression is used to fit curves to data points, helping in trend analysis and predicting future values.

  • Physics: Polynomial equations are applied in physics for modeling motion, calculating forces, and understanding phenomena like projectile motion.

  • Computer Graphics: Polynomials play a crucial role in computer graphics, enabling the creation of smooth curves and surfaces in animations and 3D rendering.

4. Which questions and examples are important in Class 10 Maths Chapter 2?

Ans: Students can find a list of the most important questions and examples from Chapter 2 of Class 10 Maths on Vedantu (vedantu.com). Subject experts at Vedantu prepare the important questions, revision notes, and other study materials for the students. Hence, to find out the most important questions that may be asked from the Polynomials chapter in Class 10 Maths, students can refer to the important questions and revision notes on Vedantu (vedantu.com).

5. What is a polynomial in Class 10 NCERT?

Ans: Polynomial is an algebraic expression that has variables and coefficients in it. The meaning of a polynomial is “many terms”. An example of a polynomial is $4x^2 + 3x + 7$. They are usually a sum or difference of exponents and variables. For an expression to be a polynomial, it should not contain the square root, negative powers, or fractional powers on the variables. It should also not have variables in the denominator of any fractions.