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Polynomials class 10 Maths NCERT Solutions Chapter is concerned with identifying the zeros of given polynomials. You will also learn about the division algorithm, which is quite significant. Students preparing for class 10 maths chapter 2 will be able to clear all their concepts on polynomials at the root level. Expert faculty of Toppr produced these solutions to assist students with their first term exam preparations. It covers all major concepts in detail, allowing students to understand the ideas better.

NCERT solutions for class 10 maths chapter 2 concentrates on the essential concepts such as the geometrical significance of polynomial zeros, the link between polynomial zeros and coefficients, and the polynomial division algorithm. All of these solutions are designed with the new CBSE pattern in mind so that students have a complete understanding of their tests.

Polynomial class 10 Questions and Answers are very useful for getting good grades in tests and properly preparing you with all of the important concepts. These NCERT Solutions are valuable tools that can assist you not only in covering the full syllabus but also in providing an in-depth analysis of the subjects. The Class 10 Maths NCERT Solutions Chapter 2 are available in pdf format below, and some of them are also included in the exercises.

Table of Content

Exercise 2.1

Question 1

The graphs of $y=p(x)$ are given in the figure, for some polynomials $p(x)$. Find the number of zeroes of $p(x)$, in each case.

Solution

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(i) The graph does not intersect the $x$-axis, so there are no zeros in $p(x)$.

(ii) The graph intersects the $x$-axis at one place, so here there is one zero in $p(x)$.

(iii) The graph intersects the $x$-axis at three places, so that there are three zeros in $p(x)$.

(iv) The graph intersects the $x$-axis at two places, so here there are two zeros in $p(x)$.

(v) The graph intersects the $x$-axis at four places, so there are four zeros in $p(x)$.

(vi) The graph intersects the $x$-axis at three places, so that there are three zeros in $p(x)$.

(ii) The graph intersects the $x$-axis at one place, so here there is one zero in $p(x)$.

(iii) The graph intersects the $x$-axis at three places, so that there are three zeros in $p(x)$.

(iv) The graph intersects the $x$-axis at two places, so here there are two zeros in $p(x)$.

(v) The graph intersects the $x$-axis at four places, so there are four zeros in $p(x)$.

(vi) The graph intersects the $x$-axis at three places, so that there are three zeros in $p(x)$.

Exercise 2.2

Question 1

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

(i) $x_{2}−2x−8$ (ii) $4s_{2}−4s+1$ (iii) $6x_{2}−3−7x$

(iv) $4u_{2}+8u$ (v) $t_{2}−15$ (vi) $3x_{2}−x−4$

(i) $x_{2}−2x−8$ (ii) $4s_{2}−4s+1$ (iii) $6x_{2}−3−7x$

(iv) $4u_{2}+8u$ (v) $t_{2}−15$ (vi) $3x_{2}−x−4$

Solution

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So, the value of $x_{2}−2x−8$ is zero when $x+2=0,x−4=0,$ i.e., when $x=−2$ or $x=4$.

Therefore, the zeros of $x_{2}−2x−8$ are -2 and 4.

Now,

$⇒$Sum of zeroes $=−2+4=2$=$−12 =−Coefficientofx_{2}Coefficientofx $

$⇒$Product of zeros $=(−2)×(4)=−8$ = $1−8 =Coefficientofx_{2}Constantterm $

Factorize the equation, we get$(2s−1)(2s−1)$

So, the value of $4s_{2}−4s+1$ is zero when $2s−1=0,2s−1=0$, i.e., when $s=21 $ or $s=21 $.

Therefore, the zeros of $4s_{2}−4s+1$ are $21 $ and $21 $.

Now,

So, the value of $4s_{2}−4s+1$ is zero when $2s−1=0,2s−1=0$, i.e., when $s=21 $ or $s=21 $.

Therefore, the zeros of $4s_{2}−4s+1$ are $21 $ and $21 $.

Now,

$⇒$Sum of zeroes = $21 +21 =1=−4−4 =−Coefficientofs_{2}Coefficientofs $

$⇒$Product of zeros = $21 ×21 =41 =41 =Coefficientofs_{2}Constantterm $

Factorize the equation, we get $(3x+1)(2x−3)$

So, the value of $6x_{2}−3−7x$ is zero when $3x+1=0,2x−3=0$, i.e., when $x=−31 $ or $x=23 $.

Therefore, the zeros of $6x_{2}−3−7x$ are $−31 $ and $23 $.

Now,

So, the value of $6x_{2}−3−7x$ is zero when $3x+1=0,2x−3=0$, i.e., when $x=−31 $ or $x=23 $.

Therefore, the zeros of $6x_{2}−3−7x$ are $−31 $ and $23 $.

Now,

$⇒$Sum of zeroes = $−31 +23 =67 =−6−7 =−Coefficientofx_{2}Coefficientofx $

$⇒$Product of zeros = $−31 ×23 =−21 =6−3 =2−1 =Coefficientofx_{2}Constantterm $

Factorize the equation, we get $4u(u+2)$

So, the value of $4u_{2}+8u$ is zero when $4u=0,u+2=0$, i.e., when $u=0$ or $u=−2$.

Therefore, the zeros of $4u_{2}+8u$ are $0$ and $−2.$

Now,

So, the value of $4u_{2}+8u$ is zero when $4u=0,u+2=0$, i.e., when $u=0$ or $u=−2$.

Therefore, the zeros of $4u_{2}+8u$ are $0$ and $−2.$

Now,

$⇒$Sum of zeroes = $0−2=−2=−48 =−2=−Coefficientofu_{2}Coefficientofu $

$⇒$Product of zeros = $−0x−2=0=40 =0=Coefficientofu_{2}Constantterm $

Factorize the equation, we get $t=±15 $

So, the value of $t_{2}−15$ is zero when $t+15 =0,t−15 =0$, i.e., when $t=15 $ or $t=−15 $.

Therefore, the zeros of $t_{2}−15$ are $±15 $.

Now,

So, the value of $t_{2}−15$ is zero when $t+15 =0,t−15 =0$, i.e., when $t=15 $ or $t=−15 $.

Therefore, the zeros of $t_{2}−15$ are $±15 $.

Now,

$⇒$Sum of zeroes = $15 −15 =0=−10 =0=−Coefficientoft_{2}Coefficientoft $

$⇒$Product of zeros = $15 ×−15 =−15=1−15 =Coefficientoft_{2}Constantterm $

Factorize the equation, we get$(x+1)(3x−4)$

So, the value of $3x_{2}−x−4$ is zero when x + 1 = 0, 3x - 4 = 0, i.e., when x = -1 or x = $34 $.

Therefore, the zeros of $3x_{2}−x−4$ are -1 and $34 $.

Now,

So, the value of $3x_{2}−x−4$ is zero when x + 1 = 0, 3x - 4 = 0, i.e., when x = -1 or x = $34 $.

Therefore, the zeros of $3x_{2}−x−4$ are -1 and $34 $.

Now,

$⇒$Sum of zeroes = $−1+34 =31 =−3−1 =−Coefficientofx_{2}Coefficientofx $

$⇒$Product of zeros = $−1×34 =−34 =3−4 =Coefficientofx_{2}Constantterm $

Question 2

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

(i) $41 $, $−1$ (ii) $2 ,31 $ (iii) $0,5 $

(iv) $1,1$ (v) $4−1 ,41 $ (vi) $4,1$

(i) $41 $, $−1$ (ii) $2 ,31 $ (iii) $0,5 $

(iv) $1,1$ (v) $4−1 ,41 $ (vi) $4,1$

Solution

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$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−41 x−1=0$

$4x_{2}−x−4=0$

Using the quadratic equation formula,

$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−2 x+31 =0$

Multiply by 3 to remove denominator,

$3x_{2}−32 x+1=0$

$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−2 x+31 =0$

Multiply by 3 to remove denominator,

$3x_{2}−32 x+1=0$

Using the quadratic equation formula,

$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−0x+5 =0$

$x_{2}+5 =0$

$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−0x+5 =0$

$x_{2}+5 =0$

Using the quadratic equation formula,

$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−1x+1=0$

$x_{2}−x+1=0$

$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−1x+1=0$

$x_{2}−x+1=0$

Using the quadratic equation formula,

$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−4−1 x+41 =0$

Multiply by $4$

$4x_{2}+x+1=0$

$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−4−1 x+41 =0$

Multiply by $4$

$4x_{2}+x+1=0$

Using the quadratic equation formula,

$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−4x+1=0$

$x_{2}−(Sumofroot)x+(Productofroot)=0$

Substitute the value in the formula, we get

$x_{2}−4x+1=0$

Exercise 2.3

Question 1

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder.

$p(x)=x_{3}−3x_{2}+5x−3$

$g(x)=x_{2}−2$

$p(x)=x_{3}−3x_{2}+5x−3$

$g(x)=x_{2}−2$

Solution

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$p(x)=x_{3}−3x_{2}+5x−3=x_{3}−3x_{2}−2x+6+7x−9$

$⇒p(x)=(x−3)(x_{2}−2)+(7x−9)$

$g(x)=x_{2}−2$

Dividing $p(x)$ by $g(x)$, we get

Quotient $q(x)=(x−3)$ and Remainder, $R(x)=(7x−9)$

Hence, option D.

Question 2

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

(i) $t_{2}−3,2t_{4}+3t_{3}−2t_{2}−9t−12$

(ii) $x_{2}+3x+1,3x_{4}+5x_{3}−7x_{2}+2x+2$

(iii) $x_{3}−3x+1,x_{5}−4x_{3}+x_{2}+3x+1$

(i) $t_{2}−3,2t_{4}+3t_{3}−2t_{2}−9t−12$

(ii) $x_{2}+3x+1,3x_{4}+5x_{3}−7x_{2}+2x+2$

(iii) $x_{3}−3x+1,x_{5}−4x_{3}+x_{2}+3x+1$

Solution

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Remainder is 0, hence $t_{2}−3$ is a factor of $2t_{4}+3t_{3}−2t_{2}−9t−12$

Question 3

Obtain all other zeroes of $3x_{4}+6x_{3}−2x_{2}−10x−5$, if two of its zeroes are $35 $ and $−35 $.

Solution

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Two zeroes are $35 $ and $−35 $

we get $x−35 =0$ and $x+35 =0$

So we can write it as, x = $35 $ and x = $−35 $

we get $x−35 =0$ and $x+35 =0$

Multiply both the factors we get,

$x_{2}−35 =0$

$x_{2}−35 =0$

Multiply by $3$ we get

$3x_{2}−5=0$ is the factor of $3x_{4}+6x_{3}−2x_{2}−10x−5$

$3x_{2}−5=0$ is the factor of $3x_{4}+6x_{3}−2x_{2}−10x−5$

Now divide, $3x_{4}+6x_{3}−2x_{2}−10x−5$ by $3x_{2}−5=0$ we get,

Quotient is $x_{2}+2x+1=0$

Quotient is $x_{2}+2x+1=0$

Compare the equation with quadratic formula,

$x_{2}−(Sumofroot)x+(Productofroot)=0$

$x_{2}−(Sumofroot)x+(Productofroot)=0$

$⇒$Sum of root $=2$

$⇒$Product of the root $=1$

So, we get

$⇒$$x_{2}+x+x+1=0$

$⇒$$x(x+1)+1(x+1)=0$

$⇒$$x+1=0,x+1=0$

$⇒$$x=−1,x=−1$

So, our zeroes are $−1,−1,$ $35 $ and $−35 $

Question 4

On dividing $x_{3}−3x_{2}+x+2$ by a polynomial $g(x)$, the quotient and remainder were $x−2$ and $−2x+4$, respectively. Find $g(x)$.

Solution

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Dividend $=p(x)=x_{3}−3x_{2}+x+2$

Quotient $=q(x)=x−2$

Remainder $=r(x)=2x+4$

By division algorithm, $p(x)=q(x)g(x)+r(x)$

$⇒g(x)=q(x)p(x)−r(x) $

$⇒g(x)=x−2x_{3}−3x_{2}+x+2+2x−4 $

$⇒g(x)=x−2x_{3}−3x_{2}+3x−2 $

So, $g(x)=x_{2}−x+1$

Question 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 $r(x)=0$

(i) deg $p(x)=$deg $q(x)$ (ii) deg $q(x)=$deg $r(x)$ (iii) deg $r(x)=0$

Solution

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Dividend = Divisor x quotient + Remainder

$p(x)=g(x)×q(x)+r(x)$

So here the degree of quotient will be equal to degree of dividend when the divisor is constant.

Let us assume the division of $4x_{2}$ by $2$.

Here, $p(x)=$$4x_{2}$

$g(x)=2$

$q(x)=$ $2x_{2}$ and $r(x)=0$

Here, $p(x)=$$4x_{2}$

$g(x)=2$

$q(x)=$ $2x_{2}$ and $r(x)=0$

Degree of $p(x)$ and $q(x)$ is the same i.e., $2$.

Checking for division algorithm,

$p(x)=g(x)×q(x)+r(x)$

$p(x)=g(x)×q(x)+r(x)$

$4x_{2}=2(2x_{2})$

Hence, the division algorithm is satisfied.

Let us assume the division of $x_{3}+x$ by $x_{2}$,

Here, p(x) = $x_{3}+x$, g(x) = $x_{2}$, q(x) = x and r(x) = x

Degree of q(x) and r(x) is the same i.e., 1.

Checking for division algorithm,

$p(x)=g(x)×q(x)+r(x)$

$x_{3}+x=x_{2}×x+x$

$x_{3}+x=x_{3}+x$

Hence, the division algorithm is satisfied.

$x_{3}+x=x_{3}+x$

Hence, the division algorithm is satisfied.

Degree of remainder will be 0 when remainder comes to a constant.

Let us assume the division of $x_{4}+1$ by $x_{3}$

Here, p(x) = $x_{4}+1$

g(x) = $x_{3}$

$q(x)=x$ and $r(x)=1$

Let us assume the division of $x_{4}+1$ by $x_{3}$

Here, p(x) = $x_{4}+1$

g(x) = $x_{3}$

$q(x)=x$ and $r(x)=1$

Degree of $r(x)$ is $0.$

Checking for division algorithm,

$p(x)=g(x)×q(x)+r(x)$

$x_{4}+1=x_{3}×x+1$

$x_{4}+1=x_{4}+1$

Hence, the division algorithm is satisfied.

Exercise 2.4

Question 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) $2x_{3}+x_{2}−5x+2;21 ,1,−2$

(i) $2x_{3}+x_{2}−5x+2;21 ,1,−2$

(ii) $x_{3}−4x_{2}+5x−2;2,1,1$

Solution

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Zeroes for this polynomial are $21 ,1,−2$

Substitute the $x=21 $ in equation (1)

$p(21 )=2(21 )_{3}+(21 )_{2}−5(21 )+2$

$=41 +41 +25 +2$

$=0$

Substitute the $x=1$ in equation (1)

$p(1)=2×1_{3}+1_{2}−5×1+2$

$=2+1−5+2=0$

$=2+1−5+2=0$

Substitute the $x=−2$ in equation (1)

$p(−2)=2(−2)_{3}+(−2)_{2}−5(−2)+2$

$=−16+4+10+2=0$

$=−16+4+10+2=0$

Therefore, $21 ,1,−2$ are the zeroes of the given polynomial.

Comparing the given polynomial with $ax_{3}+bx_{2}+cx+d$ we obtain,

$a=2,b=1,c=−5,d=2$

Let us assume $α=21 $, $β=1$, $γ=−2$

Sum of the roots = $α+β+γ=21 +1=2=2−1 =a−b $

Sum of the roots = $α+β+γ=21 +1=2=2−1 =a−b $

$αβ+βγ+αγ=21 +1(−2)+21 (−2)=2−5 =ac $

Product of the roots = $αβγ=21 ×x×(−2)=2−2 =ad $

Therefore, the relationship between the zeroes and coefficient are verified.

$p(x)=$$x_{3}−4x_{2}+5x−2$ .... (1)

Zeroes for this polynomial are $2,1,1$

Substitute $x=2$ in equation (1)

$p(2)=2_{3}−4×2_{2}+5×2−2$

$=8−16+10−2=0$

$p(2)=2_{3}−4×2_{2}+5×2−2$

$=8−16+10−2=0$

Substitute $x=1$ in equation (1)

$p(1)=x_{3}−4x_{2}+5x−2$

$=1_{3}−4(1)_{2}+5(1)−2$

$=1−4+5−2=0$

$=1_{3}−4(1)_{2}+5(1)−2$

$=1−4+5−2=0$

Therefore, $2,1,1$ are the zeroes of the given polynomial.

Comparing the given polynomial with $ax_{3}+bx_{2}+cx+d$ we obtain,

$a=1,b=−4,c=5,d=−2$

Let us assume $α=2$, $β=1$, $γ=1$

Let us assume $α=2$, $β=1$, $γ=1$

Sum of the roots = $α+β+γ=2+1+1=4=−1−4 a−b $

Multiplication of two zeroes taking two at a time=$αβ+βγ+αγ=(2)(1)+(1)(1)+(2)(1)=5=15 =ac $

Product of the roots = $αβγ=2×1×1=2=−1−2 =ad $

Therefore, the relationship between the zeroes and coefficient are verified.

Question 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 $2,7,14$ respectively.

Solution

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Practice more questions

A polynomial is an expression made up of variables, coefficients, and mathematical operations such as addition, subtraction, multiplication, division, and a non-negative integer exponent.

Polynomials are algebraic expressions with more than two terms. They can also be described as the sum of numerous phrases in which the same variable(s) has/have varying powers.

The degree of a polynomial in mathematics is the highest or greatest power of a variable in a polynomial equation. To determine the degree of any polynomial, just variables are evaluated; coefficients are ignored.

For example,

Equation -> x + 10 Degree -> 1

Equation -> 2x^{2} + x + 8 Degree -> 2

Equation -> 10x^{6} + 3x^{3} - 10 Degree -> 6

- Linear Polynomial - ax + b, where a, b are real numbers and a ≠ 0.
- Quadratic Polynomial - ax
^{2}+ bx + c, where a, b, c are real numbers & a ≠ 0. - Cubic Polynomial - ax
^{3}+ bx^{2}+ cx + d, where a, b, c, d are real numbers and a ≠ 0.

If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that

p(x) = g(x) × q(x) + r(x)

where r(x) = 0 or degree of r(x) < degree of g(x). This result is known as the Division Algorithm for polynomial

Related Chapters

- Chapter 1 : Real NumbersChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability

Q1. What can be the real-life applications of polynomials?

Answer: Professionals that need to perform difficult computations are the most likely to employ polynomials on a daily basis. Polynomials are used by civil engineers to design highways, buildings, and other constructions. Polynomials can be used to forecast traffic patterns and then use them to conduct traffic management measures.

Q2. How many exercises are in NCERT Solutions for Class 10 Maths Chapter 2?

Answer: The NCERT Solutions Class 10 Maths Chapter 2 Polynomials is made up of four exercises and 13 well-researched problems. These 13 sums are well-placed in relation to the theory's context. These issues come in a variety of styles, including equations, long answer type questions, word problems, proof-based questions, and many others. On Toppr, students can discover answers to all of the problems in these four exercises.

Q3. What are the Key Topics in NCERT Solutions Class 10 Maths Chapter 2?

Answer: The concept of a polynomials zero, degrees and coefficients, and the division algorithm of polynomials are all key subjects discussed in NCERT Solutions Class 10 Maths Chapter 2. Polynomials Chapter 2 is quite diversified and aids students in understanding concepts such as parabolas, graphs, and the geometric significance of polynomial zeros. Students will be able to answer polynomial problems after understanding these topics.

Related Chapters

- Chapter 1 : Real NumbersChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability