**NCERT Solutions Class 10 Maths Chapter 2 Polynomials** – Here are all the NCERT solutions for Class 10 Maths Chapter 2. This solution contains questions, answers, images, explanations of the complete Chapter 2 titled Polynomials of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 2 Polynomials. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 2 Polynomials in one place.

## NCERT Solutions Class 10 Maths Chapter 2 Polynomials

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Class | 10 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 2 |

Chapter Name |
Polynomials |

### NCERT Solutions Class 10 Maths chapter 2 Polynomials

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### Question & Answer

Q.1:The graphs of y = p(x) are given in Figure 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 graph does not cut the x-axis at any point. (ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point. (iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points. (iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points. (v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points. (vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

Q.2: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) \( \ 6 x^{2}-3-7x \) (iv) \( \ 4 u^{2}+8 u \) (v) \( \ t^{2}-15\) (vi) \( \ 3 x^{2}-x-4\)

Ans :The value of \( \ x^{2}-2 x-8\) is zero if x +2 = 0 or x —4 0 \( \Rightarrow x=-2 \text { or } x=4\) Therefore, the zeroes of \( \ x^{2}-2 x-8\ are —2 and 4. Now ( \ =-2+4=2=\frac{-(-2)}{1}=\frac{-(\text { Cofficient of } x)}{\text { Cofficient of } x^{2}} \) \( \ =(-2) \times 4=-8=\frac{-8}{1}=\frac{\text { Constant term }}{\text { Cofficient of } x^{2}} \) (ii) \(4 s^{2}-4 s+1=(2 s-1)^{2}\) \(\begin{array}{l}{\text { The value of } 4 s^{2}-4 s+1 \text { is zero when } 2 s-1=0, \text { i.e., }^{s=\frac{1}{2}}} \\ {\text { Therefore, the zeroes of } 4 s^{2}-4 s+1 \text { are } \frac{1}{2} \text { and } \frac{1}{2}}\end{array}\) Sum of zeroes =\( \frac{1}{2}+\frac{1}{2}=1=\frac{-(-4)}{4}=\frac{-(\text { Coefficient of } s)}{\left(\text { Coefficient of } s^{2}\right)}\) Product of zeroes\(=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}=\frac{\text { Constant term }}{\text { Coefficient of } s^{2}}\) (iii) \(6 x^{2}-3-7 x=6 x^{2}-7 x-3=(3 x+1)(2 x-3)\) \(\begin{array}{l}{\text { The value of } 6 x^{2}-3-7 x \text { is zero when } 3 x+1=0 \text { or } 2 x-3=0, \text { i.e., }} \\ {x=\frac{-1}{3} \text { or } x=\frac{3}{2}} \\ {\text { Therefore, the zeroes of } 6 x^{2}-3-7 x \text { are } \frac{3}{3} \text { and } \frac{3}{2}}\end{array}\) \(\begin{array}{l}{\text { Sum of zeroes = } \frac{-1}{3}+\frac{3}{2}=\frac{7}{6}=\frac{-(-7)}{6}=\frac{-(\text { Coefficient of } x)}{6}} \\ {\text { Product of zeroes }=\frac{-1}{3} \times \frac{3}{2}=\frac{-1}{2}=\frac{-3}{6}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}}\end{array}\) (iv) \(\begin{array}{l}{4 u^{2}+8 u=4 u^{2}+8 u+0} \\ {=4 u(u+2)}\end{array}\) \(\begin{array}{l}{\text {The value of } 4 u^{2}+8 u \text { is zero when } 4 u=0 \text { or } u+2=0, \text { i.e., } u=0 \text { or }} \\ {u=-2} \\ {\text {Therefore, the zeroes of } 4 u^{2}+8 u \text { are } 0 \text { and }-2 \text { . }}\end{array}\) Sum of zeroes\(=^{0+(-2)=-2=\frac{-(8)}{4}}=\frac{-(\text { Coefficient of } u)}{\text { Coefficient of } u^{2}}\) Product of zeroes\( =^{0 \times(-2)}=0=\frac{0}{4}=\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}\) (v) \(\begin{array}{l}{t^{2}-15} \\ {=t^{2}-0.1-15} \\ {=(t-\sqrt{15})(t+\sqrt{15})}\end{array}\) \( \begin{array}{l}{\text { The value of } t^{2}-15 \text { is zero when } t-\sqrt{15}=0 \text { or } t+\sqrt{15}=0, \text { l.e., when }} \\ {t=\sqrt{15} \text { or } t=-\sqrt{15}} \\ {\text { Therefore, the zeroes of } t^{2}-15 \text { are } \sqrt{15} \text { and - } \sqrt{15} \text { . }}\end{array}\) Sum of zeroes =\(\sqrt{15}+(-\sqrt{15})=0=\frac{-0}{1}=\frac{-(\text { Coefficient of } t)}{\left(\text { Coefficient of } t^{2}\right)}\) Product of zeroes=\(\left( \begin{array}{l}{\sqrt{15} )(-\sqrt{15})=-15=\frac{-15}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}}\end{array}\right.\) \(\begin{array}{l}{\text { (vi) } 3 x^{2}-x-4} \\ {=(3 x-4)(x+1)} \\ {\text { The value of } 3 x^{2}-x-4 \text { is zero when } 3 x-4=0 \text { or } x+1=0, \text { i.e., }} \\ {\text { when } x=\frac{4}{3} \text { or } x=-1}\end{array}\) \(\begin{array}{l}{\text { Therefore, the zeroes of } 3 x^{2}-x-4 \text { are }^{\frac{4}{3}} \text { and }-1} \\ {\text { Sum of zeroes }=\frac{4}{3}+(-1)=\frac{1}{3}=\frac{-(-1)}{3}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}}\end{array}\) Product of zeroes\(=\frac{4}{3}(-1)=\frac{-4}{3}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

Q.3: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) \(4,1\)

Ans :(i) Let and ß are the zeroes of the polynomial \( a x^{2}+b x+c\), then we have \( \alpha+\beta=\frac{1}{4}=\frac{-b}{a}\) On comparing, \( \alpha \beta=-1=\frac{-4}{4}=\frac{C}{a}\) a=4,b=-1,and c=-4 (ii) Let and ß are the zeroes of the polynomial \( a x^{2}+b x+c\), then we have \( \alpha+\beta=\sqrt{2}=\frac{3 \sqrt{2}}{3}=\frac{-b}{a}\) On comparing, \( \alpha \beta=\frac{1}{3}=\frac{c}{a}\) \( a=3, b=-3 \sqrt{2} \text { and } c=1\) Hence, the required quadratic polynomial is \( x^{2}+0 . x+\sqrt{5}\) (iii) Let a and are the zeroes or the polynomial \( a x^{2}+b x+c\). then we have \( \alpha+\beta=0=\frac{0}{1}=\frac{-b}{a}\) on comparing \( \alpha \beta=\sqrt{5}=\frac{\sqrt{5}}{1}=\frac{c}{a}\) Hence, the required quadratic polynomial is : \( x^{2}+0 . x+\sqrt{5}\) (iv) Let a and are the zeroes Of the polynomial \( a x^{2}+b x+c\), then we have \( \alpha+\beta=1=\frac{1}{1}=\frac{-b}{a}\) \( \alpha \beta=1=\frac{1}{1}=\frac{c}{a}\) on comparing, \( a=1, \quad b=-1 \text { and } c=1\) Hence. the required quadratic polynomial is \( x^{2}-x+1\) (v) Let a and are the zeroes Of the polynomial \( a x^{2}+b x+c\), then we have \( \alpha+\beta=1=\frac{-1}{4}=\frac{-b}{a}\) \( \alpha \beta=1=\frac{1}{4}=\frac{c}{a}\) on comparing, a=4,b=1,and c=1 Hence. the required quadratic polynomial is \( 4 x^{2}+x+1\) (vi) Let a and are the zeroes Ofthe polynomial axa + bx + c, then we have \( \alpha+\beta=4=\frac{4}{1}=\frac{-b}{a}\) \( \alpha \beta=1=\frac{1}{1}=\frac{c}{a}\) on comparing , a=1, b=-4 and c=1 Hence, the required quadratic polynomial is \( x^{2}-4 x+1\)

Q.4:Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following : \( p(x)=x^{3}-3 x^{2}+5 x-3, \quad g(x)=x^{2}-2\) \( p(x)=x^{4}-3 x^{2}+4 x+5, \quad g(x)=x^{2}+1-x\) \( p(x)=x^{4}-5 x+6, \quad g(x)=2-x^{2}\)

Ans :(i) \( p(x)=x^{3}-3 x^{2}+5 x-3, g(x)=x^{2}-2\) (ii) \( p(x)=x^{4}-3 x^{2}+4 x+5=x^{4}+0 . x^{3}-3 x^{2}+4 x+5\) \( q(x)=x^{2}+1-x=x^{2}-x+1\) (iii) \( p(x)=x^{4}-5 x+6=x^{4}+0 x^{2}-5 x+6\) \( q(x)=2-x^{2}=-x^{2}+2\)

Q.5:Check whether the first polynomial is a factor of the second polynomial 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) \( x^{3}-3 x+1, x^{5}-4 x^{3}+x^{2}+3 x+1\)

Ans :(i)\( t^{2}-3,2 t^{4}+3 t^{3}-2 t^{2}-9 t-12\) \( t^{2}-3=t^{2}+0 . t-3\) Since the remainder is o, Hence,\( t^{2}-3\) is a factor of \( 2 t^{4}+3 t^{3}-2 t^{2}-9 t-12\) (ii) \( \quad x^{2}+3 x+1,3 x^{4}+5 x^{3}-7 x^{2}+2 x+2\) (iii) \(x^{3}-3 x+1, x^{5}-4 x^{3}+x^{2}+3 x+1\)

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**Exercise Covered under Maths Book Class 10 Maths Chapter 2 NCERT Solutions**

Here on this page we have covered following exercises from NCERT Maths Book for Class 10 Maths Chapter 2 Polynomials

- Exercise: 2.1 Class 10 Maths Solutions
- Exercise: 2.2 Class 10 Maths Solutions
- Exercise: 2.3 Class 10 Maths Solutions
- Exercise: 2.4 Class 10 Maths Solutions

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