NCERT Solutions Class 8 Maths Chapter 8 Algebraic Expressions and Identities
Chapter 8 of Class 8 Maths, "Algebraic Expressions and Identities", is an important topic that forms the foundation for understanding algebraic concepts in higher studies. This chapter introduces students to the principles of algebraic expressions, their operations, and numerous standard identities. Mastering these concepts is essential as they play a significant role in solving complex equations and mathematical problems encountered in advanced studies.
The NCERT Solutions for Class 8 Maths Chapter 8 provides detailed explanations and step-by-step solutions to textbook exercises, guiding students through the process of working with algebraic expressions and applying identities effectively. This blog aims to help students enhance their algebraic skills, boost their problem-solving abilities, and gain confidence in their math knowledge as they progress in their academic journey.
1.0Download Class 8 Maths Chapter 8 NCERT Solutions PDF Online
Downloading the NCERT Solutions of Class 8 Chapter 8 Algebraic Expressions and Identities PDF will help you understand the curriculum's key concepts. Below is the link to download the NCERT solutions for Class 8 Maths Chapter 8 PDF.
2.0Class 8 Maths Chapter 8: Breakdown of Exercises
3.0NCERT Questions with Solutions for Class 8 Maths Chapter 8 - Detailed Solutions
Exercise : 8.1
- Add the following :
(i) ab−bc,bc−ca,ca−ab
(ii) a−b+ab,b−c+bc,c−a+ac
(iii) 2p2q2−3pq+4,5+7pq−3p2q2
(iv) ℓ2+m2, m2+n2,n2+ℓ2,2ℓ m+2mn+2nℓ
Sol. (i) Writing the given expressions in separate rows with like terms one below the other, we have
ab−bc+bc−ca−ab+ca0+0+0
(ii) Writing the given expressions in separate rows with like terms one below the other, we have
or ab+bc+ac
(iii) Writing the given expressions in separate rows with like terms one below the other, we have
2p2q2−3pq+4−3p2q2+7pq+5−p2q2+4pq+9
(iv) Writing the given expressions in separate rows with like terms one below the other, we have
ℓ2+m2
+m2+n2
ℓ2+n2
2ℓ2+2m2+2n2+2ℓm+2mn+2nℓ+2ℓm+2mn+2nℓ
- (i) Subtract 4a−7ab+3b+12 from 12a −9ab+5b−3
(ii) Subtract 3xy+5yz−7zx from 5xy− 2yz−2zx+10xyz
(iii) Subtract 4p2q−3pq+5pq2−8p+7q -10 from 18−3p−11q+5pq−2q2 +5p2q.
Sol. Rearranging the terms of the given expressions, changing the sign of each term of the expression to be subtracted and adding the two expressions, we get,
(i) 12a−9ab+5 b−34a−7ab+3 b+12
(ii) Rearranging the terms of the given expressions, changing the sign of each term of the expression to be subtracted and adding the two expressions, we get,
(iii) Rearranging the terms of the given expressions, changing the sign of each term of the expression to be subtracted and adding the two expressions, we get.
5p2q−2pq2+5qp−11q−3p+18
4p2q+5q2−3pq+7q−8p−10
p2q−7q2+8pq−18q+5p+28−−+−++
- Identify the terms, their coefficients for each of the following expressions :
(i) 5xyz2−3zy
(ii) 1+x+x2
(iii) 4x2y2−4x2y2z2+z2
(iv) 3−pq+qr−rp
(v) 2x+2y−xy
(vi) 0.3a−0.6ab+0.5 b
Sol. (i) 5xyz2−3zy
In the expression 5x2z2−3zy, the terms are 5x′2z2 and −3zy.
Coefficient of xyz2 in the term 5x2z2 is 5 . Coefficient of zy in the term −3zy is -3 .
(ii) 1+x+x2
In the expression 1+x+x2, the terms are 1,x and x2.
Coefficient of x0 in the term 1 is 1 .
Coefficient of x in the term x is 1 .
Coefficient of x2 is the term x2 is 1 .
(iii) In the expression 4x2y2−4x2y2z2+z2, the terms are 4x2y2,−4x2y2z2 and z2.
Coefficient of x2y2 in the term 4x2y2 is 4 . Coefficient of x2y2z2 in the term −4x2y2z2 is -4 .
Coefficient of z2 in the term z2 is 1 .
(iv) 3 - pq + qr - rp
In the expression 3−pq+qr−rp, the terms are 3, -pq, qr and -rp
Coefficient of x0 in the term 3 is 3 .
Coefficient of pq in the term -pq is -1 . Coefficient of qr in the term qr is 1.
Coefficient of rp in the term -rp is -1 .
(v) In the expression 2x+2y−xy, the terms are 2x,2y and −xy
Coefficient of x in the term 2x is 21.
Coefficient of y in the term 2y is 21.
Coefficient of xy in the term -xy is -1 .
(vi) In the expression 0.3a−0.6ab+0.5b, the terms are 0.3a,−0.6ab and 0.5 b Coefficient of a in the term 0.3 a is 0.3 . Coefficient of ab in the term -0.6 ab is - 0.6.
Coefficient of b in the term 0.5b is 0.5 .
- Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x+y,1000,x+x2+x3+x4,7+y+5x,2y− 3y2,2y−3y2+4y3,5x−4y+3xy,4z−15z2, ab+bc+cd+da,pqr,p2q+q2,2p+2q.
Sol. The given polynomials are classified as under:
Monomials : 1000, pqr
Binomials : x+y,2y−3y2,4z−15z2,p2q+ pq2,2p+2q.
Trinomials: 7+y+5x,2y−3y2+4y3,5x −4y+3xy.
Polynomials that do not fit in any category :
x+x2+x3+x4,ab+bc+cd+da
Exercise : 8.2
- Find the product of the following pairs of monomials
(i) 4,7p
(ii) −4p,7p
(iii) −4p,7pq
(iv) 4p3,−3p
(v) 4p,0
Sol. (i) 4×7p=(4×7)×p=28p
(ii) −4p×7p=(−4×7)×(p×p)
=−28p1+1
=−28p2
(iii) −4p×7pq=(−4×7)×(p×p×q)
=−28p1+1q
=−28p2q
(iv) 4p3×(−3p)={4×(−3)}×(p3×p)
=−12×p3+1
=−12p4
(v) 4p×0=(4×0)×(p)
=0×p
=0
- Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively :
(p, q); (10m, 5n); (20x2,5y2); (4x, 3x2 ); (3mn, 4np)
Sol. We know that the area of a rectangle =ℓ×b, where ℓ= length and b= breadth .
Therefore, the areas of rectangles with pair of monomials (p, q); (10m, 5n); (20x2,5y2); (4x,3x2) and ( 3mn,4np ) as their lengths and breadths are given by p×q=pq
10 m×5n=(10×5)×(m×n)=50mn
20x2×5y2=(20×5)×(x2×y2)
=100x2y2
4x×3x2=(4×3)×(x×x2)
=12x3
and, 3mn×4np=(3×4)×(m×n×n×p) =12mn2p
- Complete the table of products :
Sol.
- Obtain the volume of rectangular boxes with the following length, breadth and height respectively :
(i) 5a,3a2,7a4
(ii) 2p,4q,8r
(iii) xy,2x2y,2xy2
(iv) a, 2b, 3c
Sol. (i) Required volume =5a×3a2×7a4
=(5×3×7)×(a×a2×a4)
=105a1+2+4=105a7
(ii) Required volume =2p×4q×8r
=(2×4×8)×(p×q×r)
=64pqr
(iii) Required volume =xy×2x2y×2xy2
=(1×2×2)×(xy×x2y×x2)
=(4)×(x1+2+1×y1+1+2)
=4x4y4
(iv) Required volume =a×2b×3c
=(1×2×3)×(a×b×c)
=6abc
- Obtain the product of
(i) xy,yz,zx
(ii) a,−a2,a3
(iii) 2,4y,8y2,16y3
(iv) a, 2b, 3c, 6abc
(v) m,−mn,mnp
Sol. (i) xy×yz×zx=x×x×y×y×z×z
=x1+1×y1+1×z1+1
=x2y2z2
(ii) a×(−a2)×a3
=[1×(−1)×1]×(a×a×a×a×a×a)
=(−1)×(a6)
=−a6
(iii) 2×(4y)×8y2×16y3
=(2×4×8×16)×(y×y2×y3)
=(1024)×(y1+2+3)
=1024y6
(iv) a×2 b×3c×6abc
=(2×3×6)×(a×b×c×abc)
=(36)×(a1+1×b1+1×c1+1)
=36a2b2c2
(v) m×−mn×mnp
=(1×−1×1)×(m×m×m×n×n×p)
=−1×m3×n2×p
=−m3n2p
Exercise : 8.3
- Carry out the multiplication of the expressions in each of the following pairs:
(i) 4p,q+r
(ii) ab, a - b
(iii) a+b,7a2b2
(iv) a2−9,4a
(v) pq+qr+rp,0
Sol. (i) 4p×(q+r)=4p×q+4p×r
=4pq+4pr
(ii) ab×(a−b)=ab×a−ab×(b) =a2b−ab2
(iii) (a+b)×(7a2b2)=7a2b2×a+7a2b2×b =7a3 b2+7a2 b3
(iv) (a2−9)×4a=a2×4a−9×4a =4a3−36a
(v) (pq+qr+rp)×0=0
- Complete the table:
Sol. (i) a(b+c+d)=a×b+a×c+a×d=ab +ac+ad
(ii) (x+y−5)×(5xy)
=5xy×x+5xy×y+5xy×(−5)
=5x2y+5xy2−25xy
(iii) 6p3−7p2+5p
(iv) 4p4q2−4p2q4
(v) a2bc+ab2c+abc2
- Find the product:
(i) (a2)×(2a22)×(4a26)
(ii) (32xy)×(10−9x2y2)
(iii) (−310pq3)×(56p3q)
(iv) x×x2×x3×x4
Sol. (i) (a2)×(2a22)×(4a26)=(1×2×4)×
(a2×a22×a26)=8a2+22+26=8a50
(ii) (32xy)×(10−9x2y2)
=(32×10−9)×(x×x2×y×y2)
=−53×x1+2×y1+2
=−53x3y3
(iii) (−310pq3)×(56p3q)
=(−310×56)×(p×q3×p3×q)
=(−4)×(p4×q4)=−4p4q4
(iv) x×x2×x3×x4=x1+2+3+4=x10
- (a) Simplify: 3x(4x−5)+3 and find its values for (i) x=3, (ii) x=21.
(b) Simplify: a(a2+a+1)+5 and find its value for
(i) a=0, (ii) a=1, (iii) a=−1.
Sol. (a) We have,
3x(4x−5)+3=3x×4x−3x×5+3 =12x2−15x+3
(i) When x=3, then
12(3)2−15(3)+3=66
(ii) When x=21, then
12(21)2−15×21+3=3−215+3=−23
(b) We have,
a(a2+a+1)+5
=a×a2+a×a+a×1+5
=a3+a2+a+5
(i) When a=0, then a3+a2+a+5
=03+02+0+5
=5
(ii) When a=1, then a3+a2+a+5
13+12+1+5
=1+1+1+5
=3+5=8
(iii) When a=−1, then a3+a2+a+5 =(−1)3+(−1)2+(−1)+5 =−1+1−1+5=4
- (a) Add: p(p−q),q(q−r) and r(r−p)
(b) Add: 2x(z−x−y) and 2y(z−y−x)
(c) Subtract: 3ℓ(ℓ−4m+5n) from
4ℓ(10n−3m+2ℓ)
(d) Subtract: 3a(a+b+c)−2 b(a−b+c) from 4c(−a+b+c)
Sol. (a) p(p−q)+q(q−r)+r(r−p)
=p×p−p×q+q×q−q×r+r×r −r×p
=p2−pq+q2−qr+r2−pr =p2+q2+r2−pq−qr−pr
(b) 2x(z−x−y)+2y(z−y−x)
=(2x×z)−(2x×x)−(2x×y)+(2y×z) −(2y×y)−(2y×x)
=2xz−2x2−2xy+2yz−2y2−2xy
=−2x2−2y2−4xy+2xz+2yz.
(c) 4ℓ(10n−3m+2ℓ)−3ℓ(ℓ−4m+5n) =(4ℓ×10n)−(4ℓ×3 m)+(4ℓ×2ℓ)+ (−3ℓ×ℓ)+(−3ℓ×−4 m)+(−3ℓ×5n) =40ℓn−12ℓ m+8ℓ2−3ℓ2+12ℓ m−15ℓn =5ℓ2+25ℓn
(d) 4c(−a+b+c)−{3a(a+b+c)−2 b(a−b+c)} ={4c×(−a)+4c×b+4c×c}−[{(3a ×a+3a×b+3a×c)}+{(−2b×a)+ (−2b×−b)+(−2b×c)}]
=−4ac+4bc+4c2−[3a2+3ab+3ac
−2ab+2b2−2bc]
=−4ac+4bc+4c2−3a2−3ab−3ac+
2ab−2 b2+2bc
=−3a2−2b2+4c2−7ac+6bc−ab
Exercise: 8.4
- Multiply the binomials:
(i) (2x+5) and (4x−3)
(ii) (y−8) and (3y−4)
(iii) (2.5ℓ−0.5 m) and (2.5ℓ+0.5 m)
(iv) (a+3 b) and (x+5)
(v) (2pq+3q2) and (3pq−2q2)
(vi) (43a2+3b2) and 4(a2−32b2)
Sol. (i) (2x+5) and (4x−3)
=2x(4x−3)+5(4x−3)
=8x2−6x+20x−15=8x2+14x−15
(ii) (y−8)×(3y−4)=y(3y−4)−8(3y−4)
=3y2−4y−24y+32
=3y2−28y+32
(iii) (2.5ℓ−0.5 m)×(2.5ℓ+0.5 m)
=2.5ℓ(2.5ℓ+0.5 m)−0.5 m(2.5ℓ+0.5 m)
=6.25ℓ2+1.25ℓ m−1.25ℓ m−0.25 m2
=6.25ℓ2−0.25 m2
(iv) (a+3b)×(x+5)
=a(x+5)+3b(x+5)
=ax+5a+3bx+15b
=ax+3bx+5a+15b
(v) (2pq+3q2)×(3pq−2q2)
=2pq(3pq−2q2)+3q2(3pq−2q2)
=6p2q2−4pq3+9pq3−6q4
=6p2q2+5pq3−6q4
(vi) (43a2+3b2) and 4(a2−32b2)
=(43a2+3b2)×(4a2−38b2)
=43a2(4a2−38 b2)+3 b2(4a2−38 b2)
=3a4−2a2b2+12a2b2−8b4
=3a4+10a2b2−8b4
- Find the product:
(i) (5−2x)(3+x)
(ii) (x+7y)(7x−y)
(iii) (a2+b)(a+b2)
(iv) (p2−q2)(2p+q)
Sol. (i) (5−2x)(3+x)=5(3+x)−2x(3+x) =15+5x−6x−2x2 =−2x2−x+15
(ii) (x+7y)(7x−y)
=x(7x−y)+7y(7x−y)=7x2−xy+49xy−7y2=7x2+48xy−7y2
(iii) (a2+b)(a+b2)
=a2(a+b2)+b(a+b2)=a3+a2b2+ab+b3
(iv) (p2−q2)(2p+q)=p2(2p+q)−q2(2p+q) =2p3+p2q−2pq2−q3
- Simplify :
(i) (x2−5)(x+5)+25
(ii) (a2+5)(b3+3)+5
(iii) (t+s2)(t2−s)
(iv) (a+b)(c−d)+(a−b)(c+d)+2(ac+bd)
(v) (x+y)(2x+y)+(x+2y)(x−y)
(vi) (x+y)(x2−xy+y2)
(vii) (1.5x−4y)(1.5x+4y+3)−4.5x+12y
(viii) (a+b+c)(a+b−c)
Sol. (i) (x2−5)(x+5)+25
=x2(x+5)−5(x+5)+25=x3+5x2−5x−25+25=x3+5x2−5x
(ii) (a2+5)(b3+3)+5
=a2(b3+3)+5(b3+3)+5
=a2b3+3a2+5b3+15+5
=a2b3+3a2+5b3+20
(iii) (t+s2)(t2−s)
=t(t2−s)+s2(t2−s)
=t3−ts+s2t2−s3
(iv) (a+b)(c−d)+(a−b)(c+d)+2(ac+bd)
=a(c−d)+b(c−d)+a(c+d)−b(c+d)
+2ac+2bd
=ac−ad+bc−bd+ac+ad−bc−bd
+2ac+2bd
=4ac
(v) (x+y)(2x+y)+(x+2y)(x−y)
=x(2x+y)+y(2x+y)+x(x−y)+2y(x−y)
=2x2+xy+2xy+y2+x2−xy+2xy−2y2
=(2+1)x2+(1+2−1+2)xy+(1−2)y2
=3x2+4xy−y2
(vi) (x+y)(x2−xy+y2)
=x(x2−xy+y2)+y(x2−xy+y2)
=x3−x2y+xy2+x2y−xy2+y3
=x3+y3
(vii) (1.5x−4y)(1.5x+4y+3)−4.5x+12y =1.5x(1.5x+4y+3)−4y(1.5x+4y+3)
−4.5x+12y
=1.5x×1.5x+1.5x×4y+1.5x×3
−4y×1.5x−4y×4y−4y×3−4.5x
+12y
=2.25x2+6xy+4.5x−6xy−16y2
−12y−4.5x+12y
=2.25x2+(6−6)xy+(4.5−4.5)x
−16y2+(−12+12)y
=2.25x2+(0)xy+(0)x−16y2+(0)y
=2.25x2+0+0−16y2+0
=2.25x2−16y2
(viii) (a+b+c)(a+b−c)
=a(a+b−c)+b(a+b−c)+c(a+b−c)
=a2+ab−ac+ab+b2−bc+ac+bc−c2
=a2+2ab+b2−c2