NCERT Solutions Class 10 Maths Chapter 2
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NCERT Solutions for Class 10 Mathematics Chapter 2 Polynomials
Chapter 2 Class 10 Mathematics includes a detailed study of Polynomials. This chapter involves descriptions and practices of different equations and respective components. NCERT Class 10 Mathematics Chapter 2 textbook is one of the best study materials to understand the concept of Polynomials. The book has practice questions at the end of every chapter so that students can revise the concepts they have learned and prepare better for their exams.
Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 2 that have answers to all the questions given in the textbook. Whether students are looking for answers to questions or want to cross-check their answers, the NCERT Solutions by Extramarks are reliable learning material.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials –
Access NCERT Solutions for Class 10 Mathematics Chapter 2 – Polynomials
NCERT Solutions for Class 10 Mathematics Chapter 2 Polynomials – Free Download
The NCERT Solutions for Class 10 Mathematics Chapter 2 Polynomials can help students solve exercises quickly. In case students get stuck in any question while practicing, they can refer to the solutions, and get their doubts cleared without wasting any time.
NCERT Solutions for Class 10 Mathematics
In addition to NCERT Solutions for Class 10 Mathematics Chapter 2, Extramarks also offers NCERT Solutions for all chapters in Class 10 Mathematics.
- Chapter 1 – Real Numbers
- Chapter 2 – Polynomials
- Chapter 3 – Pair of Linear Equations in Two Variables
- Chapter 4 – Quadratic Equations
- Chapter 5 – Arithmetic Progressions
- Chapter 6 – Triangles
- Chapter 7 – Coordinate Geometry
- Chapter 8 – Introduction to Trigonometry
- Chapter 9 – Some Applications of Trigonometry
- Chapter 10 – Circles
- Chapter 11 – Constructions
- Chapter 12 – Areas Related to Circles
- Chapter 13 – Surface Areas and Volumes
- Chapter 14 – Statistics
- Chapter 15 – Probability
Polynomials: NCERT Solutions for Class 10 Mathematics Chapter 2 Summary
Chapter 2 includes the concepts of polynomials’ geometric and meanings of the terms relevant to it. The questions at the end of the NCERT Chapter 2 textbook will analyse the problem-solving skills of students. Therefore, Extramarks has ensured that every answer in solutions is solved in a step-by-step manner. The subject matter experts have kept the language simple and all the answers are highly accurate.
Benefits of Using NCERT Solutions for Class 10 Mathematics Polynomials
There are many benefits of using NCERT Solutions for Class 10 Mathematics Chapter 2 polynomials. Let’s take a look at a few of them:
- Concise Answers to Questions
NCERT Solutions for Class 10 Mathematics Chapter 2 are created by subject matter experts. You can rest assured that all the answers have been framed by keeping in mind the guidelines provided by CBSE. Hence students who practise according to NCERT Solutions Class 10 Mathematics Chapter 2 will perform very well in their examinations.
- Clear Concept of Chapter 2 Class 10 Mathematics
Just knowing the answer to the question does not mean you excel in that subject, but understanding the path to find the solution should be the main aim of the student. Concept clarity helps students solve any problem with ease. As discussed above, NCERT Solutions for Class 10 Mathematics Chapter 2 focus more on sorting out conceptual queries of the students, thus helping them strengthen their subject knowledge for future standards.
- Clearing Doubts
It is natural for students to have considerable doubts while practising the questions of Class 10 Mathematics Chapter 2. These doubts can be quickly resolved with NCERT Solutions for Class 10 Mathematics Chapter 2.
- Convenient Revision
NCERT Solutions for Class 10 Mathematics Chapter 2 make revision pretty convenient for students. They can refer to these solutions for last-minute preparations as well.
Related Questions
Q1. Write the following in the product form: a2b5.
Ans. a2b5 is – a×a×b×b×b×b×b
Q.1 The graphs of y = p(x) are given in figures below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Ans.
(i) The number of zeroes is 0 as the given graph does not intersect the x-axis.
(ii) The number of zeroes is 1 as the given graph intersects the x-axis at 1 point only.
(iii) The number of zeroes is 3 as the given graph intersects the x-axis at 3 points only.
(iv) The number of zeroes is 2 as the given graph intersects the x-axis at 2 points only.
(v) The number of zeroes is 4 as the given graph intersects the x-axis at 4 points only.
(vi) The number of zeroes is 3 as the given graph intersects the x-axis at 3 points only.
Q.2
Find the zeroes of the following quadratic polynomialsand verify the relationship between the zeroes andthe coefficients.(i)x2−2x−8(ii)4s2−4s+1(iii)6x2−3−7x(iv)4u2+8u(v)t2−15(vi)3x2−x−4
Ans.
(i)x2−2x−8=x2−4x+2x−8=x(x−4)+2(x−4)=(x+2)(x−4)So the value ofx2−2x−8is zero whenx=−2orx=4.Therefore, the zeroes ofx2−2x−8are−2and 4.Now,Sum of zeroes =−2+4=2=−(−2)1=−(Coefficient of x)Coefficient of x2Product of zeroes =−2×4=−81=Constant termCoefficient of s2(ii)4s2−4s+1=(2s−1)2So the value of4s2−4s+1is zero whenx=12.Therefore, the zeroes of4s2−4s+1are12and12.Now,Sum of zeroes =12+12=1=−(−4)4=−(Coefficient of s)Coefficient of s2Product of zeroes =12×12=14=Constant termCoefficient of s2(iii)6x2−3−7x=6x2−9x+2x−3=3x(2x−3)+1(2x−3)=(3x+1)(2x−3)So the value of6x2−3−7xis zero whenx=−13orx=32.Therefore, the zeroes of6x2−3−7xare−13and32.Now,Sum of zeroes =−13+32=76=−(−7)6=−(Coefficient of x)Coefficient of x2Product of zeroes =−13×32=−12=Constant termCoefficient of x2(iv)4u2+8u=4u(u+2)So the value of4u(u+2)is zero whenu=0orx=−2.Therefore, the zeroes of4u(u+2)are0and−2.Now,Sum of zeroes =0−2=−84=−(Coefficient of u)Coefficient of u2Product of zeroes =0×(−2)=0=04=Constant termCoefficient of u2(v)t2−15=t2−(√15)2=(t−√15)(t+√15)So the value oft2−15is zero whent=√15orx=−√15.Therefore, the zeroes oft2−15are√15and−√15.Now,Sum of zeroes =√15−√15=0=01=−(Coefficient of t)Coefficient of t2Product of zeroes =√15×(−√15)=−15=−151=Constant termCoefficient of t2(vi)3x2−x−4=3x2−4x+3x−4=x(3x−4)+1(3x−4)=(3x−4)(x+1)So the value of3x2−x−4is zero whenx=43orx=−1.Therefore, the zeroes of3x2−x−4are43and−1.Now,Sum of zeroes =43−1=13=−(−1)3=−(Coefficient of x)Coefficient of x2Product of zeroes =43×(−1)=−43=Constant termCoefficient of x2
Q.3
Find a quadratic polynomial each with the givennumbers as the sum and product of its zeroesrespectively.(i)14,−1(ii)√2,13(iii) 0,√5(iv) 1, 1 (v)−14,14(vi) 4, 1
Ans.
(i)14,−1Let the quadratic polynomial be ax2+bx+cand itszeroes beαandβ. Then, we haveα+β=14=−baandαβ=−1=caand thusa=4,b=−1,c=−4.So, the quadratic polynomial which fits the givenconditions is 4x2−x−4.(ii)√2,13Let the quadratic polynomial be ax2+bx+cand itszeroes beαandβ. Then, we haveα+β=√2=−baandαβ=13=caNow,α+β=√2=3√23=−baandαβ=13=caSo,a=3,b=−3√2,c=1.So, the quadratic polynomial which fits the givenconditions is 3x2−3√2x+1.(iii) 0,√5Let the quadratic polynomial be ax2+bx+cand itszeroes beαandβ. Then, we haveα+β=0=−baandαβ=√5=caSo,a=1,b=0,c=√5.So, the quadratic polynomial which fits the givenconditions is x2+√5.(iv) 1, 1Let the quadratic polynomial be ax2+bx+cand itszeroes beαandβ. Then, we haveα+β=1=−baandαβ=1=caSo,a=1,b=−1,c=1.So, the quadratic polynomial which fits the givenconditions is x2−x+1.(v)−14,14Let the quadratic polynomial be ax2+bx+cand itszeroes beαandβ. Then, we haveα+β=−14=−baandαβ=14=caSo,a=4,b=1,c=1.So, the quadratic polynomial which fits the givenconditions is 4x2+x+1.(vi) 4, 1Let the quadratic polynomial be ax2+bx+cand itszeroes beαandβ. Then, we haveα+β=4=−baandαβ=1=caSo,a=1,b=−4,c=1.So, the quadratic polynomial which fits the givenconditions is x2−4x+1.
Q.4
Divide the polynomialp(x)by the polynomialg(x)and find the quotient and remainder in each of thefollowing:(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
Ans.
(i)p(x)=x3−3x2+5x−3,g(x)=x2−2x2−2x−3x3−3x2+5x−3 x3−2x−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ −3x2+7x−3−3x2+6 +−7x−9Quotient=x−3Remainder=7x−9(ii)p(x)=x4−3x2+4x+5,g(x)=x2+1−xx2+1−xx2+x−3x4−3x2+4x+5 x4+x2−x3−−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x3−4x2+4x+5 x3−x2+x−+−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ −3x2+3x+5 −3x2+3x−3+−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯8Quotient=x2+x−3Remainder=8(iii)p(x)=x4−5x+6,g(x)=2−x22−x2−x2−2x4−5x+6 x4−2x2−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2x2−5x+6 2x2−4−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−5x+10Quotient=−x2−2Remainder=−5x+10
Q.5
Check whether the first polynomial is a facator ofthe second polymomial by dividing the secondpolynomial 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
Ans.
(i)t2−3,2t4+3t3−2t2−9t−12t2−32t2+3t+42t4+3t3−2t2−9t−12 2t4−6t2−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 3t3+4t2−9t−12 3t3−9t−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯4t2−124t2−12−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0Remainder = 0So,t2−3is a factor of 2t4+3t3−2t2−9t−12.(ii)x2+3x+1,3x4+5x3−7x2+2x+2x2+3x+13x2−4x+23x4+5x3−7x2+2x+2 3x4+9x3+3x2−−−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ −4x3−10x2+2x+2 −4x3−12x2−4x+++¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2x2+6x+2 2x2+6x+2−−−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0Remainder = 0So,x2+3x+1is a factor of 3x4+5x3−7x2+2x+2.(iii)x3−3x+1,x5−4x3+x2+3x+1x3−3x+1x2−1x5−4x3+x2+3x+1 x5−3x3+x2−+−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ −x3+3x+1 −x3+3x−1 +−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2Remainder = 2So,x3−3x+1is not a factor of x5−4x3+x2+3x+1.
Q.6
Obtain all other zeroes of 3x4+6x3−2x2−10x−5,if two of its zeroes are√53and−√53.
Ans.
p(x)=3x4+6x3−2x2−10x-5The two zeroes of p(x) are√53and-√53.Therefore,(x−√53)and(x+√53)are factors of p(x).Also,(x−√53)(x+√53)=x2−53and so x2−53is a factor of p(x).Now,3x2+6x+3x2−533x4+6x3−2x2−10x−53x4−5x2−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯6x3+3x2−10x−56x3−10x−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯3x2−53x2−5−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯03x4+6x3−2x2−10x−5=(x2−53)(3x2+6x+3)=3(x2−53)(x2+2x+1)=3(x2−53)(x+1)(x+1)Equating(x2−53)(x+1)(x+1) equal to zero, we getthe zeroes of the given polynomial.Hence, the zeroes of the given polynomial are√53,−√53,1and−1.
Q.7
On dividingx3−3x2+x+2by a polynomialg(x),the quotient and remainder werex−2and−2x+4,respectively. Findg(x).
Ans.
Dividend =x3−3x2+x+2Quotient =x−2Remainder =−2x+4Divisor = g(x) = ?We know that,Dividend=Divisor×Quotient+Remainder∴x3−3x2+x+2=g(x)×(x−2)+(−2x+4)⇒g(x)×(x−2)=x3−3x2+x+2+2x−4⇒g(x)=(x3−3x2+3x−2)(x−2)x−2x2−x+1x3−3x2+3x−2 x3−2x2−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−x2+3x−2−x2+2x+−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x−2x−2−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯0∴g(x)=x2−x+1
Q.8
Give examples of polynomialsp(x),g(x),q(x)andr(x)which satisfy the division algorithm and(i) degp(x)=degq(x)(ii) degq(x)=degr(x)(iii) degr(x)=0
Ans.
(i) degp(x)=degq(x)if divisor is constant thenDegree of quotient = degree of dividendLet dividendp(x)=2x2−2x+14anddivisor g(x)=2Then, wer havequotientq(x)=x2−x+7and remainderr(x)=0We find thatdegp(x)=degq(x)=2Let us check for division algorithm.dividend=divisor×quotient + remainderor2x2−2x+14=2(x2−x+7)+0=2x2−2x+14Thus, the division algorithm is satisfied.(ii) degq(x)=degr(x)Let us dividex3+xbyx2.Clearly we have,dividendp(x)=x3+xdivisor g(x)=x2quotientq(x)=xand remainderr(x)=xHere,degq(x)=degr(x)=1Let us check for division algorithm.dividend=divisor×quotient + remainderorx3+x=x2⋅x+x=x3+xThus, the division algorithm is satisfied.(iii) degr(x)=0Degree of remainder will be zero if the remainder is constant.Let us dividex2+1byx.Clearly we have,dividendp(x)=x2+1divisor g(x)=xquotientq(x)=xand remainderr(x)=1Here,degr(x)=0Let us check for division algorithm.dividend=divisor×quotient + remainderorx2+1=x⋅x+1=x2+1Thus, the division algorithm is satisfied.
Q.9
Verify that the numbers given alongside of the cubicpolynomials below are their zeroes. Also verify therelationship between the zeroes and the coefficientsin each case:(i) 2x3+x2−5x+2;12,1,−2(ii) x3−4x2+5x−2;2, 1, 1
Ans.
(i) 2x3+x2−5x+2;12,1,−2Letp(x)=2x3+x2−5x+2Thenp(12)=2(12)3+(12)2−5(12)+2=14+14−52+2=12+2−52=0p(1)=2(1)3+(1)2−5(1)+2=2+1−5+2=0p(−2)=2(−2)3+(−2)2−5(−2)+2=−16+4+10+2=0Therefore,12,1and−2are zeroes of the given poynomial.Comparing the given poynomial withax3+bx2+cx+d,we geta=2,b=1,c=−5,d=2Let the given roots areα,βandγ. Thenα=12,β=1,γ=−2Now,α+β+γ=12+1−2=−12=−baαβ+βγ+γα=12×1+1×(−2)+(−2)×12=−52=caαβγ=12×1×(−2)=−1=−22=−daThe relationship between roots and coefficients is verified.(ii) x3−4x2+5x−2;2, 1, 1Letp(x)=x3−4x2+5x−2Thenp(2)=(2)3−4(2)2+5(2)−2=8−16+10−2=0p(1)=(1)3−4(1)2+5(1)−2=1−4+5−2=0Therefore,2,1and1are zeroes of the given poynomial.Comparing the given poynomial withax3+bx2+cx+d,we geta=1,b=−4,c=5,d=−2Let the given roots areα,βandγ. Thenα=2,β=1,γ=1Now,α+β+γ=2+1+1=4=−−41=−baαβ+βγ+γα=2×1+1×1+1×2=5=51=caαβγ=2×1×1=2=−(−2)1=−daThe relationship between roots and coefficients is verified.
Q.10
Find a cubic polynomial with the sum of zeroes , sum of theproduct of its zeroes taken two at a time, and theproduct of its zeroes as 2,−7,−14 respectively.
Ans.
Let the polynomial beax3+bx2+cx+dand its zeroesbeα,βandγ.Given thatα+β+γ=2=−(−2)1=−baαβ+βγ+γα=−7=−71=caαβγ=−14=−141=−daIf we takea=1,b=−2,c=−7andd=14then therequired polynomial isx3−2x2−7x+14.
Q.11
If the zeroes of the polynomialx3−3x2+x+1are a−b, a,a+b, find a and b.
Ans.
The given polynomial isx3−3x2+x+1and its zeroesare a−b, a,a+b.Sum of the roots = (a−b)+a+(a+b)=3a=−Coefficient ofx2Coefficient ofx3=3⇒a=1Sum of the products of zero taken two at a time=a(a−b)+a(a+b)+(a+b)(a−b)=a2−ab+a2+ab+a2−b2=3a2−b2=Coefficient ofxCoefficient ofx3=1⇒3a2−b2=1⇒3.1−b2=1⇒b2=2⇒b=±√2Hence,a=1andb=√2or−√2
Q.12
If two zeroes of the polynomialx4−6x3−26x2+138x−35are2±√3,find other zeroes.
Ans.
Two zeroes of the polynomialx4−6x3−26x2+138x−35are2±√3.Therefore,[x−(2+√3)]and[x−(2−√3)]are factors of thegiven polynomial.Now,[x−(2+√3)][x−(2−√3)]=[(x−2)−√3)][(x−2)+√3)]=(x−2)2−(√3)2=x2−4x+4−3=x2−4x+1x2−4x+1is also a factor of the given polynomial.Therefore, we dividex4−6x3−26x2+138x−35byx2−4x+1to find other factors.x2−4x+1x4−6x3−26x2+138x−35x4−4x3+x2−+−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−2x3−27x2+138x−35−2x3+8x2−2x+−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−35x2+140x−35−35x2+140x−35+−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯0x4−6x3−26x2+138x−35=(x2−4x+1)(x2−2x−35)=[x−(2+√3)][x−(2−√3)](x2−7x+5x−35)=[x−(2+√3)][x−(2−√3)](x−7)(x+5)(x−7)and(x+5)arefactorsofx4−6x3−26x2+138x−35.Hence,7and−5arezeroesofx4−6x3−26x2+138x−35.
Q.13
If the polynomialx4−6x3+16x2−25x+10is dividedby another polynomialx2−2x+k,the remainder comesout to bex+a,findkanda.
Ans.
We know thatDividend = Divisor×Quotient + Remainderor Dividend−Remainder = Divisor×Quotientor(x4−6x3+16x2−25x+10)−(x+a)=(x2−2x+k)×Quotientorx4−6x3+16x2−26x+10−a=(x2−2x+k)×QuotientNow,x2−2x+kx2−4x+(8−k)x4−6x3+16x2−26x+10−a x4−2x3+kx2−+−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−4x3+(16−k)x2−26x+10−a−4x3+8x2−4kx+−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−(8−k)x2+(4k−26)x+10−a−(8−k)x2−(−2k+16)x+8k−k2+ + −+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(2k−10)x+k2−8k+10−aRemainder(2k−10)x+k2−8k+10−amust be zero.So,(2k−10)x+k2−8k+10−a=0⇒2k−10=0ork2−8k+10−a=0⇒k=5or52−8×5+10−a=0⇒k=5anda=−5
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NCERT Solutions for Class 10 Maths Related Chapters
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