Complete NCERT solutions for class 8 Maths Chapter 9 Algebraic Expressions

Solution for Exercise 9.1

1.Identify the terms, their coefficients for each of the following expressions :
(i) 5xyz23zy

(ii) 1+x+x2

(iii) 4x2y24x2y2z2+z2

(iv) 3pq+qr?rp

(v) x2+y2?xy

(vi)0.3a0.6ab+0.5b


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i) We have the expression 5xyz23zy, the terms are 5xyz2 and ?3zy.
Coefficient of xyz2 in the term 5xyz2 is 5.

Coefficient of zy in the term 3yz is 3.

(ii)We have the expression 1+x+x2, the terms are 1, x and x2.

Coefficient of the term 1 is 1.

Coefficient of x in the term x is 1.

Coefficient of x2 in the term x2 is 1.

(iii) We have the expression 4xy4xyz+z , 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) We have the expression 3pq+qrrp, the terms are 3, – pq, qr and – rp.

Coefficient of 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) We have the expression x2+y2?xy, the terms are x2,72 and xy.

Coefficient of x in the term x2 is 12

Coefficient of y in the term y2 is 12

Coefficient of xy in the term xy is -1.

(vi) In the expression 0.3a?0.6ab+0.5b, the terms are 0.3a,006aband 0.5b.

Coefficient of a in the term 0.3a is 0.3.

Coefficient of ab in the term – 0.6ab is – 0.6.

Coefficient of b in the term 0.5b is 0.5.

2.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,2y3y2,2y3y2+4y3,5x4y+3xy,4z15z2,ab+bc+cd+da,pqr,p2q+pq2,2p+2q.


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

The calssification of the given expressions are:

Monomials : 1000, pqr

Binomials : x+y,2y3y2,4z?15z2,p2q+pq2,2p+2q.

Trinomials : 7+y+5x,2y3y2+4y3,5x4y+3x.

And the polynomials which don't have any category:

x+x2+x3+x4,ab+be+cd+da.

3.Add the following :
(i) abbe,beca,eaab

(ii)ab+ab,b?c+be,ca+ac

(iii) 2p2q23pq+4,5+7pq3p2q2

(iv)l2+m2,m2+n2,n2+l2,2lm+2mn+2nl


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i) We have the expressions as:abbc,bcca,caab
Adding them :

(abbc)+(bcca)+(caab)

?abbc+bcca+caab

?(abab)+(bcbc)+(caca) [Placing the like terms together]

?0+0+0

?0

(ii) We have the expressions as:ab+ab,bc+bc,ca+ac

Adding these we have:

(ab+ab)+(bc+bc)+(ca+ac)

?ab+ab+bc+bc+ca+ac

?(aa)+(bb)+(cc)+ab+bc+ac[Placing all the like terms together]

?0+0+0+ab+bc+ac

?ab+bc+ac

(iii)We have the expressions as:2p2q23pq+4,5+7pq3p2q2

Adding these we get: 2p2q23pq+4+5+7pq3p2q2

?(2p2q23p2q2)+(3pq+7pq)+(4+5)[Placing like terms together]

??p2q2+4pq+9

(iv)We have the expressions as:l2+m2,m2+n2,n2+l2,2lm+2mn+2nl
Adding these we get: (l2+m2)+(m2+n2)+(n2+l2)+(2lm+2mn+2nl)

?(l2+l2)+(m2+m2)+(n2+n2)+(2lm+2mn+2nl)[Placing like terms together]

?(2l2)+(2m2)+(2n2)+(2lm+2mn+2nl

?2(l2+m2+n2+lm+mn+nl)

4.(a) Subtract 4a7ab+3b+12 from 12a9ab+5b3
(b) Subtract 3xy+5yz7zx from 5xy2yz2zx+10xyz

(c) Subtract 4p2q3pq+5pq28p+7q10 from 183p11q+5pq2pq2+5p2q


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(a)We have the expressions to subtract:4a7ab+3b+12 from 12a9ab+5b3
Subtracting these we get: (12a9ab+5b3)?(4a7ab+3b+12)

(12a9ab+5b3)+(?4a+7ab?3b?12)[Taking (-) inside the second bracket]

?(12a?4a)+(9ab+7ab)+(+5b?3b)+(3?12)[Placing like terms together]

?(8a)+(?2ab)+(2b)+(?15)

?(8a?2ab+2b?15)

(b)We have the expressions to subtract:3xy+5yz7zx from 5xy2yz2zx+10xyz
Subtracting these we get: (5xy2yz2zx+10xyz)?(3xy+5yz7zx)

(5xy2yz2zx+10xyz)+(?3xy?5yz+7zx)[Taking (-) inside the second bracket]

?(5xy?3xy)+(2yz?5yz)+(2zx+7zx)+(+10xyz)[Placing like terms together]

?(2xy)+(?7yz)+(5zx)+(10xyz)

?(2xy?7yz+5zx+10xyz)

(c)We have the expressions to subtract:4p2q3pq+5pq28p+7q10 from 183p11q+5pq2pq2+5p2q
Subtracting these we get: (183p11q+5pq2pq2+5p2q)?(4p2q3pq+5pq28p+7q10)

?(183p11q+5pq2pq2+5p2q)+(?4p2q+3pq?5pq2+8p?7q+10)[Taking (-) inside the second bracket]

?(+18+10)+(?3p+8p)+(?11q?7q)+(5pq+3pq)+(?2pq2?5pq2)+(+5p2q?4p2q)[Placing like terms together]

?(+28)+(+5p)+(?18q)+(8pq)+(?7pq2)+(+p2q)

?(+28+5p?18q+8pq?7pq2+p2q)

4.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


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i)Given that, length = 5a, breadth = 3a2, height = 7a4

?Volume of the box =l×b×h=5a×3a2×7a4=105a7cu. units

(ii)Given that, length = 2p, breadth = 4q, height = 8r

?Volume of the box =l×b×h=2p×4q×8r=64pqr cu. units

(iii)Given that, length = xy, breadth =2x2y, height = 2xy^2\)

Volume of the box =l×b×h=xy×2x2y×2xy2=(1×2×2)×xy×x2y×xy2=4x4y4cu. units

(iv)Given that, length = a, breadth = 2b, height = 3c

?Volume of the box = l×b×h=a×2b×3c=(1×2×3)abc=6abc cu. units

Solution for Exercise 9.2

1.Find the product of the following pairs of monomials :
(i) 4, 7p

(ii) -4p, 7p

(iii) – 4p, 7pq

(iv) 4p3 , – 3p

(v) 4p, 0


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i)Multiplying the given expressions: 4x×7p=(4×7)×p=28p

(ii) Multiplying the given expressions:4p×7p=(?4×7)×(p×p)

=?28p(1+1)=28p2

(iii)Multiplying the given expressions: ?4p×7pq=(?4×7)×(p×p×q)

=?28p(1+1)q=28p2q

(iv)Multiplying the given expressions: 4p3×3p=(4×3)×(p3×p)

=12p(3+1)==?12p4

(v) Multiplying the given expressions:4p×0=(4×0)×p=0×p=0

2.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)


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i) Given that: Length = p units and breadth = q units

Area of the rectangle = length×breadth=p×q=pq sq units

(ii)Given that: Length = 10 m units, breadth = 5n units

?Area of the rectangle =length×breadth=10m×5n=(10×5)×m×n=50mnsq units

(iii) Given that: Length = 20x2 units, breadth = 5y2 units

?Area of the rectangle = length × breadth = 20x2 × 5y2 = (20 × 5) × x2 × y2 = 100x2y2 sq units

(iv)Given that: Length = 4x units, breadth = 3x2 units

?Area of the rectangle = length×breadth=4x×3x2=(4×3)×x×x2=12x3 sq units

(v) Given that: Length = 3mn units, breadth = 4np units

?Area of the rectangle = length×breadth=3mn×4np=(3×4)×mn×np=12mn2p sq units

3.Complete the table of products :


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

Completed table is as under :

5. 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


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i) Product=xy×yz×zx=x2y2z2

(ii) Product=a×(?a2)×a3=(?a)6

(iii)Product=2×4y×8y2×16y3=(2×4×8×16)×y×y2×y3=1024y6

(iv)Product=a×2b×3c×6abc=(1×2×3×6)×a×b×c×abc=36a2b2c2

(v)Product=m×(?mn)×mnp=[1×(?1)×1]m×mn×mnp=?m3n2p

Solution for Exercise 9.3

1.Carry out the multiplication of the expressions in each of the following pairs:
(i) 4p,q+r

(ii) ab,ab

(iii) a+b,7a2b2

(iv) a29,4a

(v) pq+qr+rp,0


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

Multiplication of the expressions are as follows:
(i) 4p×(q+r)=(4p×q)+(4p×r)=4pq+4pr

(ii) ab,ab=ab×(ab)=(ab×a)(ab×b)=a2bab2

(iii) (a+b)×7a2b2=(a×7a2b2)+(b×7a2b2)=7a3b2+7a2b3

(iv) (a29)×4a=(a2×4a)(9×4a)=4a336a

(v) (pq+qr+rp)×0=0

2. Complete the table.


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

Products of the given expressions in the table are done as follows:
.(i) a×(b+c+d)=(a×b)+(a×c)+(a×d)=ab+ac+ad

(ii) (x+y5)(5xy)=(x×5xy)+(y×5xy)(5×5xy)=5x2y+5xy225xy

(iii) p×(6p27p+5)=(p×6p2)(p×7p)+(p×5)=6p37p2+5p

(iv) 4p2q2×(p2q2)=4p2q2×p24p2q2×q2=4p4q24p2q4

(v) (a+b+c)×(abc)=(a×abc)+(b×abc)+(c×abc)=a2bc+ab2c+abc2

Therefore, the completed table is as follows:


3.Find the products
(i)(a2)×(2a22)×(4a26)

(ii)(23xy)×(?910x2y2)

(iii)(?103pq3)×(65p3q)

(iv)x×x2×x3×x4


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i)(a2)×(2a22)×(4a26)

=(1×2×4)×a2+22+26=8a50

(ii)(23xy)×(?910x2y2)

=(23×(?910))×x(1+2)y(1+2)

(iii)(?103pq3)×(65p3q)

=(?103×65)×p(1+3)q(1+3)

=?4p4q4

(iv)x×x2×x3×x4

\(= x^{(1+2+3+4)}=x^{10\)

4.(a) Simplify: 3x(4x5)+3 and find its values for (i) x = 3 (ii) x = 12.
(b) Simplify: a(a2+a+1)+5 and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(a) Given the expression: 3x(4x5)+3=4x×3x5×3x+3=12x215x+3
(i) So for x = 3, we have

12×(3)215×3+3=12×945+3=10842=66

(ii)For x=12, we have:

?12(12)2?15(12)+3

?12×(14)?152+3

?3?152+3

?6?15+62=12?152=?32

(b) We have a(a2+a+1)+5

=(a2×a)+(a×a)+(1×a)+5

=a3+a2+a+5

(i) For a = 0, we have:

=(0)3+(0)2+(0)+5=5

(ii) For a = 1, we have:

=(1)3+(1)2+(1)+5=1+1+1+5=8

(iii) For a = -1, we have:

=(?1)3+(?1)2+(?1)+5=?1+11+5=4

5.(a) Add:p(pq),q(qr)andr(rp)
(b) Add:2x(zxy)and2y(zyx)

(c) Subtract:3l(l4m+5n)from4l(10n3m+2l)

(d) Subtract:3a(a+b+c)2b(ab+c)from4c(?a+b+c)


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(a) p(pq)+q(qr)+r(rp)

=(p×p)(p×q)+(q×q)(q×r)+(r×r)(r×p)[Opening the brackets]

= p2pq+q2qr+r2rp

=p2+q2+r2pqqrrp

(b) 2x(zxy)+2y(zyx)

= (2x×z)(2x×x)(2x×y)+(2y×z)(2y×y)(2y×x)[Opening the brackets]

= 2xz2x22xy+2yz2y22xy

=?2x22y2+2xz+2yz4xy

=?2x22y24xy+2yz+2xz

(c) 4l(10n3m+2l)3l(l4m+5n)

=(4l×10n)(4l×3m)+(4l×2l)(3l×l)(3l×?4m)(3l××5n)[Opening the brackets]

=40ln12lm+8l23l2+12lm15ln

=(40ln15ln)+(?12lm+12lm)+(8l23l2)

=25ln+0+5l2

=25ln+5l2

=5l2+25ln

(d)[4c(?a+b+c)][3a(a+b+c)2b(ab+c)]

=(?4ac+4bc+4c2)(3a2+3ab+3ac2ab+2b22bc)[Opening the brackets]

=?4ac+4bc+4c23a23ab3ac+2ab2b2+2bc

=?3a22b2+4c2ab+6bc7ac

Solution for Exercise 9.4

1.Multiply the binomials:
(i) (2x+5)and (4x3)

(ii) (y8) and (3y4)

(iii) (2.5l0.5m) and (2.5l+0.5m)

(iv) (a+3b) and (x+5)

(v) (2pq+3q2) and (3pq2q2)

(vi) (34a2+3b2) and 4(a223b2)


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i)Multiplying the given expressions as:(2x+5)×(4x3)

=2x×(4x3)+5×(4x3)

=(2x×4x)(3×2x)+(5×4x)(5×3)

=8x26x+20x15

=8x2+14x15

(ii)Multiplying the given expressions as: (y8)×(3y4)

=y×(3y4)8×(3y4)

=(y×3y)(y×4)(8×3y)+(?8×?4)

=3y24y24y+32

=3y228y+32

(iii)Multiplying the given expressions as: (2.5l0.5m)×(2.5l+0.5m)

=(2.5l×2.5l)+(2.5l×0.5m)(0.5m×2.5l)(0.5m×0.5m)

=6.25l2+1.25ml1.25ml0.25m2

=6.25l2+00.25m2

=6.25l20.25m2

(iv)Multiplying the given expressions as: (a+3b)×(x+5)

=a×(x+5)+36×(x+5)

=(a×x)+(a×5)+(36×x)+(36×5)

=ax+5a+3bx+15b

(v) Multiplying the given expressions as:(2pq+3q2)×(3pq2q2)

=2pq×(3pq2q2)+3q2(3pq2q2)

=(2pq×3pq)(2pq×2q2)+(3q2×3pq)(3q2×2q2)

=6p2q24pq3+9pq36q4

=6p2q2+5pq36q4

(vi)Multiplying the given expressions as:(34a2+3b2)×4(a223b2)

=(34a2+3b2)×(4a2?83b2)

=34a2×(4a2?83b2)+3b2×(4a2?83b2)

=(34a2×4b2)?(34a2×83b2)+(3b2×4a2)?(3b2×83b2)

=3a4?2a2b2+12a2b2?8b4

=3a4+10a2b2?8b4

2.Find the product:
(i) (52x)(3+x)

(ii)(x+7y)(7xy)

(iii) (a2+b)(a+b2)

(iv)(p2q2)(2p+q)


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i)Product of the expression:(52x)(3+x)

=5(3+x)2x(3+x)

=(5×3)+(5×x)(2x×3)(2x×x)

=15+5x6x2x2

(ii)Product of the expression: (x+7y)(7xy)

=x(7xy)+7y(7xy)

=(x×7x)(x×y)+(7y×7x)(7y×y)

=7x2xy+49xy7y2

=7x2+48xy7y2

(iii)Product of the expression: (a2+b)(a+b2)

=a2(a+b2)+b(a+b2)

=(a2×a)+(a2×b2)+(b×a)+(b×b2)

=a3+a2b2+ab+b3

(iv)Product of the expression:(p2q2)(2p+q)

=p2(2p+q)q2(2p+q)

=(p2×2p)+(p2×q)(q2×2p)(q2×q)

=2p3+p2q2pq2q3

3.Simplify: (i) (x25)(x+5)+25

(ii) (a2+5)(b3+3)+5

(iii) (t+s2)(t2s)

(iv)(a+b)(cd)+(ab)(c+d)+2(ac+bd)

(v)(x+y)(2x+y)+(x+2y)(xy)

(vi) (x+y)(x2xy+y2)

(vii)(1.5x4y)(1.5x+4y+3)4.5x+12y

(viii)(a+b+c)(a+bc)


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i)Simplifying the expression, we have: (x25)(x+5)+25

=x2×(x+5)+5×(x+5)+25

=x3+5x25x25+25

=x3+5x25x+0

=x3+5x25x

(ii)Simplifying the expression, we have: (a2+5)(b3+3)+5

=a2×(b3+3)+5×(b3+3)+5

=a2b3+3a2+5b3+15+5

=a2b3+3a2+5b3+20

(iii)Simplifying the expression, we have:(t+s2)(t2s)

=t×(t2s)+s2×(t2s)

=t3st+s2t2s3

=t3+s2t2sts3

(iv) Simplifying the expression, we have:(a+b)(cd)+(ab)(c+d)+2(ac+bd)

=a×(cd)+b×(cd)+a×(c+d)b×(c+d)+2ac+2bd

=acad+bcbd+ac+adbcbd+2ac+2bd

=ac+ac+2ac+bcbcad+adbdbd+2bd

=4ac+0+0+0

=4ac

(v) Simplifying the expression, we have:(x+y)(2x+y)+(x+2y)(xy)

=x×(2x+y)+y×(2x+y)+x×(xy)+2y×(xy)

=2x2+xy+2xy+y2+x2xy+2xy2y2

=2x2+x2+xy+2xyxy+2xy+y22y2

=3x2+4xyy2

(vi)Simplifying the expression, we have:(x+y)(x2xy+y2)

=x×(x2xy+y2)+y×(x2xy+y2)

=x3x2y+x2y+xy2xy2+y3

=x30+0+y3

=x3+y3

(vii)Simplifying the expression, we have:(1.5x4y)(1.5x+4y+3)4.5x.+12y

=1.5x×(1.5x+4y+3)4y×(1.5x+4y+3)4.5x+12y

=2.25x2+6xy+4.5x6xy16y212y4.5x+12y

=2.25x2+6xy6xy+4.5x4.5x+12y12y16y2

=2.25x2+0+0+016y62

=2.25x216y2

(viii)Simplifying the expression, we have: (a+b+c)(a+bc)

=a(a+bc)+b(a+bc)+c(a+bc)

=a2+abac+ab+b2bc+ac+bcc2

=a2+ab+abbc+bcac+ac+b2c2

=a2+2ab+b2c2+0+0

=a2+2ab+b2c2

Solution for Exercise 9.5

1.Use a suitable identity to get each of the following products:
(i) (x+3)(x+3)

(ii) (2y+5)(2y+5)

(iii) (2a7)(2a7)

(iv)(3a12)(3a12)

(v)(1.1m0.4)(1.1m+0.4)

(vi)(a2+b2)(?a2+b2)

(vii) (6x7)(6x+7)

(viii) (?a+c)(?a+c)

(ix)(x2+3y4)(x2+3y4)

(x) (7a9b)(7a9b)


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i)Given: (x+3)(x+3)
=(x+3)2

=(x)2+2×x×3+(3)2[?(a+b)2=a2+2ab+b2]

=x2+6x+9

(ii)Given: (2y+5)(2y+5)
=(2y+5)2

=(2y)2+2×2y×5+(5)2[?(a+b)2=a2+2ab+b2]

=4y2+20y+25

(iii)Given: (2a?7)(2a?7)
=(2a?7)2

=(2a)2?2×(2a)×7+(7)2[?(a?b)2=a2?2ab+b2]

=4a2?28a+49

(iv)Given: (3a?12)(3a?12)
=(3a?12)2

=(3a)2?2×(3a)×(12)+(12)2[?(a?b)2=a2?2ab+b2]

=9a2?3a+14

(v)Given: (1.1m?0.4)(1.1m+0.4)
=((1.1m)2?(0.4)2[?(a?b)(a+b)=a2?b2]

=1.21a2?0.16

(vi)Given: (a2+b2)(?a2+b2)
=(b2+a2)(b2?a2)

=(b2)2?(a2)2[?(a?b)(a+b)=a2?b2]

=(b4?a4)

(vii)Given: (6x?7)(6x+7)
=(6x)2?(7)2[?(a?b)(a+b)=a2?b2]

=36x2?49

(viii)Given: (?a+c)(?a+c)
=[(?a)+c]2

=(?a)2+2×(?a)×c+(c)2[?(a+b)2=a2+2ab+b2]

=a2?2ac+c2

(ix)Given:(x2+3y4)(x2+3y4)

=(x2+3y4)2

=(x2)2+2×(x2)×(3y4)+(3y4)2[?(a+b)2=a2+2ab+b2]

=x24+34xy+9y216

(x)Given: (7a?9b)(7a?9b)
=(7a?9b)2

=(7a)2?2×(7a)×(9b)+(9b)2[?(a?b)2=a2?2ab+b2]

=49a2?126ab+81b2

2.Use the identity (x+a)(x+b)=x2+(a+b)x+ab to find the following products.
(i) (x+3)(x+7)

(ii) (4x+5)(4x+1)

(iii)(4x5)(4x1)

(iv) (4x+5)(4x1)

(v) (2x+5y)(2x+3y)

(vi) (2a2+9)(2a2+5)

(vii) (xyz4)(xyz2)


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i)We have: (x+3)(x+7)

=x2+(3+7)x+3×7

=x2+10x+21

(ii)We have: (4x+5)(4x+1)

=(4x)2+(5+1)(4x)+5×1

=16x2+24x+5

(iii)We have: (4x?5)(4x?1)

=(4x)2?(5+1)(4x)+(?5)×(?1)

=16x2?24x+5

(iv)We have: (4x+5)(4x?1)

=(4x)2+(5?1)(4x)+(5)×(?1)

=16x2+16x?5

(v)We have: (2x+5y)(2x+3y)

=(2x)2+(5y+3y)(2x)+(5y)×(3y)

=(2x)2+(8y)(2x)+(5y)×(3y)

=4x2+16xy+15y2

(vi)We have: (2a2+9)(2a2+5)

=(2a2)2+(9+5)(2a2)+(5)×(9)

=(2a2)2+(14)(2a2)+45

=4a4+28a2+45

(vii)We have: (xyz?4)(xyz?2)

=(xyz)2?(4+2)(xyz)+(?4)×(?2)

=(xyz)2?(6)(xyz)+8

=x2y2z2?6xyz+8

3.Find the following squares by using the identities.
(i) (b7)2

(ii) (xy+3z)2

(iii) (6x25y)2

(iv) (23m+32n)2

(v) (0.4p0.5q)2

(vi) (2xy+5y)2


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i)Given: (b7)2

=(b)2?2(b)(7)+(7)2[using (a?b)2=a2?2ab+b2]

=(b)2?14b+49

(ii)Given: (xy+3z)2

=(xy)2+2(xy)(3z)+(3z)2[using (a+b)2=a2+2ab+b2]

=x2y2+6xyz+9y2

(iii)Given:(6x25y)2

=(6x2)2?2(6x2)(5y)+(5y)2[using (a?b)2=a2?2ab+b2]

=36x4?60x2y+25y2

(iv)Given: (23m+32n)2

=(23m)2+2(23m)(32n)+(32n)2[using (a+b)2=a2+2ab+b2]

=49m2+2mn+94n2

(v)Given: (0.4p0.5q)2

=(0.4p)2?2(0.4p)(0.5q)+(0.5q)2[using (a?b)2=a2?2ab+b2]

=0.16p2?0.4pq+0.25q2

(vi)Given: (2xy+5y)2

=(2xy)2+2(2xy)(5y)+(5y)2[using (a+b)2=a2+2ab+b2]

=4x2y2+20xy2+25y2

4.Simplify:
(i)(a2b2)2

(ii) (2x+5)2(2x5)2

(iii) (7m8n)2+(7m+8n)2

(iv) (4m+5n)2+(5m+4n)2

(v) (2.5p1.5q)2(1.5p2.5q)2

(vi) (ab+bc)22ab2c

(vii) (m2n2m)2+2m3n2


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Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions


Answer :

(i)Given:(a2?b2)2
=(a2)2?2a2b2+(b2)2[use (a?b)2=a2?2ab+b2]

=a4?2a2b2+b4

(ii)Given:(2x+5)2(2x5)2

=[(2x)2+2(2x)(5)+(5)2]?[(2x)2?2(2x)(5)+(5)2]

=[4x2+20x+25]?[4x2?20x+25]

=4x2+20x+25?4x2+20x?25]

=(4x2?4x2)+(+20x+20x)+(+25?25)[Placing same terms together]

=40x

(iii)Given:(7m8n)2+(7m+8n)2

=[(7m)2?2(7m)(8n)+(8n)2]+[(7m)2+2(7m)(8n)+(8n)2]

=[49m2?112mn+64n2]+[49m2+112mn+64n2]

=(49m2+49m2)+(?112mn+112mn)+(64n2+64n2)[Placing same terms together]

=98m2+128n2

(iv)Given:(4m+5n)2+(5m+4n)2

=[(4m)2+2(4m)(5n)+(5n)2]+[(5m)2+2(5m)(4n)+(4n)2]

=[16m2+40mn+25n2]+[25m2+40mn+16n2]

=(16m2+25m2)+(40mn+40mn)+(25n2+16n2)[Placing same terms together]

=41m2+80mn+41n2

(v)Given: (2.5p1.5q)2(1.5p2.5q)2

=[(2.5p)2?2(2.5p)(1.5q)+(1.5q)2]?[(1.5p)2?2(1.5p)(2.5q)+(2.5q)2]

=[6.25p2?7.5pq+2.25q2]?[2.25p2?7.5pq+6.25q2]

=6.25p2?7.5pq+2.25q2?2.25p2+7.5pq?6.25q2

=(6.25p2?2.25p2)+(?7.5pq+7.5pq)+(2.25q2?6.25q2)[Placing same terms together]

=4p2?4q2

(vi)Given:(ab+bc)22ab2c

=(ab)2+2(ab)(bc)+(bc)2?2ab2c

=a2b2+2ab2c+b2c2?2ab2c

=a2b2+b2c2

(vii)Given:(m2n2m)2+2m3n2
=(m2)2?2(m2)(n2m)+(n2m)2+2m3n2

=m4?2m3n2+n4m2+2m3n2

=m4+n4m2

5. Show that:
(i)(3x+7)284x=(3x7)2

(ii) (9p5q)2+180pq=(9p+5q)2

(iii)