NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities

Algebraic Expressions and Identities Class 8 Questions And Answers PDF Free Download

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Algebraic Expressions and Identities Class 8 NCERT Solutions

Question: 1(i) Add the following.

$ab-bc , bc -ca, ca-ab$

Answer:

ab-bc+bc-ca+ca-ab=0.

Question: 1(ii) Add the following.

$a-b+ab, b-c+bc, c-a+ac$

Answer:

$a-b+ab+b-c+bc+c-a+ac$

$=(a-a)+(b-b)+(c-c)+ab+bc+ac$

$=ab+bc+ca$

Question: 1(iii) Add the following

$2p^2q^2-3pq+4, 5+7pq-3p^2q^2$

Answer:

$\\2p^{2}q^{2}-3pq+4+5+7pq-3p^{2}q^{2}$

$=(2-3)p^{2}q^{2} +(-3+7)pq +4+5$

$=-p^{2}q^{2}+4pq+9$

Question: 1(iv) Add the following.

$l^2+m^2+n^2 , n^2+l^2, 2lm+2mn+2nl$

Answer:

$l^{2}+m^{2}+n^{2}+n^{2}+l^{2}+2lm+2mn+2nl $

$=2l^{2}+m^{2}+2n^{2}+2lm+2mn+2nl$

Question: 2(a) Subtract $4a-7ab+3b+12$ from $12a-9ab+5b-3$

Answer:

12a-9ab+5b-3-(4a-7ab+3b+12)
=(12-4)a +(-9+7)ab+(5-3)b +(-3-12)
=8a-2ab+2b-15

Question: 2(b) Subtract $3xy+5yz-7zx$ from $5xy-2yz-2zx+10xyz$

Answer:

$\\5xy-2yz-2zx+10xyz-(3xy+5yz-7zx)$

$=(5-3)xy+(-2-5)yz+(-2+7)zx+10xyz$

$=2xy-7yz+5zx+10xyz$

Question: 2(c) Subtract $4p^2q - 3pq + 5pq^2 - 8p + 7q - 10$ from $18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q$

Answer:

$18-3p-11q+5pq-2pq^{2}+5p^{2}q-(4p^{2}q-3pq+5pq^{2}-8p+7q-10)$

$=18-(-10)-3p-(-8p)-11q-7q+5pq-(-3pq)-2pq^{2}-5pq^{2}+5p^{2}q-4p^{2}q$

$=28+5p-18q+8pq-7pq^{2}+p^{2}q$

Question: 1(i) Find the product of the following pairs of monomials.

$4,7p$

Answer:

$4\times 7p=28p$

Question: 1(ii) Find the product of the following pairs of monomials.

$-4p,7p$

Answer:

$\\-4p\times 7p\\=(-4\times 7)p\times p\\=-28p^{2}$

Question: 1(iii) Find the product of the following pairs of monomials

$-4p,7pq$

Answer:

$-4p\times 7pq\\=(-4\times 7)p\times pq\\=-28p^{2}q$

Question: 1(iv) Find the product of the following pairs of monomials.

$4p^3,-3p$

Answer:

$\\4p^{3}\times (-3p)\\ =4\times (-3)p^{3}\times p\\=-12p^{4}$

Question: 1(v) Find the product of the following pairs of monomials.

$4p,0$

Answer:

$\\4p\times 0=0$

Question: 2(A) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths, respectively.

$(p,q)$

Answer:

The question can be solved as follows

$\\Area=length\times breadth\\ =(p\times q)\\ =pq$

Question: 2(B) Find the areas of rectangles with the following pairs of monomials as their lengths and breadth, respectively.

$(10m,5n)$

Answer:

The area is calculated as follows

$\\Area=length\times breadth\\ =10m\times 5n\\ =50mn$

Question: 2(C) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths, respectively.

$(20x^2,5y^2)$

Answer:

The following is the solution

$\\Area=length\times breadth\\ =20x^{2}\times 5y^{2}\\ =100x^{2}y^{2}$

Question: 2(D) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths, respectively.

$(4x,3x^2)$

Answer:

The area of a rectangle is

$\\Area=length\times breadth\\ =4x\times 3x^{2}\\ =12x^{3}$

Question: 2(E) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths, respectively.

$(3mn,4np)$

Answer:

The area is calculated as follows

$\\Area=length\times breadth\\ =3mn\times 4np\\ =12mn^{2}p$.

Question: 3. Complete the table of products.

Answer:

Question: 4(i) Obtain the volume of rectangular boxes with the following length, breadth and height, respectively.

$5a, 3a^2, 7a^4$

Answer:

$\\Volume=length\times breadth\times height\\ =5a\times 3a^{2}\times 7a^{4}\\ =15a^{3}\times 7a^{4}\\ =105a^{7}$

Question: 4(ii) Obtain the volume of rectangular boxes with the following length, breadth and height, respectively.

$2p,4q,8r$

Answer:

The volume of rectangular boxes with the following length, breadth and height is

$\\Volume=length\times breadth\times height\\ =2p\times 4q\times 8r\\ =8pq\times 8r\\ =64pqr$

Question: 4(iii) Obtain the volume of rectangular boxes with the following length, breadth and height, respectively.

$xy, 2x^2y, 2xy^2$

Answer:

The volume of rectangular boxes with the following length, breadth and height is

$\\Volume=length\times breadth\times height\\ =xy\times 2x^{2}y\times 2xy^{2}\\ =2x^{3}y^{2}\times 2xy^{2}\\ =4x^{4}y^{4}$

Question: 4(iv) Obtain the volume of rectangular boxes with the following length, breadth, and height, respectively.

$a, 2b, 3c$

Answer:

The volume of rectangular boxes with the following length, breadth and height is

$\\Volume=length\times breadth\times height\\ =a\times 2b\times 3c\\ =2ab\times 3c\\ =6abc$

Question: 5(i) Obtain the product of

$xy,yz,zx$

Answer:

The product

$\\xy\times yz\times zx\\ =xy^{2}z\times zx\\ =x^{2}y^{2}z^{2}$

Question: 5(ii) Obtain the product of

$a,-a^2,a^3$

Answer:

The product

$\\a\times (-a^{2})\times a^{3}\\ =-a^{^{3}}\times a^{3} =-a^{6}$

Question: 5(iii) Obtain the product of

$2,\ 4y,\ 8y^{2},\ 16y^{3}$

Answer:

The product

$\\2\times 4y\times 8y^{2}\times 16y^{3}\\ =8y\times 8y^{2}\times 16y^{3}\\ =64y^{3}\times 16y^{3}\\ =1024y^{6}$

Question: 5(iv) Obtain the product of

$a, 2b, 3c, 6abc$

Answer:

The product

$\\a\times 2b\times 3c\times 6abc\\ =2ab\times 3c\times 6abc\\ =6abc\times 6abc\\ =36a^{2}b^{2}c^{2}$

Question: 5(v) Obtain the product of

$m, -mn, mnp$

Answer:

The product

$\\m\times (-mn)\times mnp\\ =-m^{2}n\times mnp\\ =-m^{3}n^{2}p$

Question: (i) Find the product

$2x(3x+5xy)$

Answer:

Using distributive law,

$2x(3x + 5xy) = 6x^2 + 10x^2y$

Question: (ii) Find the product

$a^2(2ab-5c)$

Answer:

Using distributive law,

We have : $a^2(2ab-5c) = 2a^3b - 5a^2c$

Question: 1 Find the product:

$(4p^2+5p+7)\times 3p$

Answer:

By using distributive law,

$(4p^2+5p+7)\times 3p = 12p^3 + 15p^2 + 21p$

Class 8 maths chapter 8 question answer - exercise: 8.3
Page number: 100, Total questions: 5

Question: 1(i) Carry out the multiplication of the expressions in each of the following pairs.

$4p, q+r$

Answer:

Multiplication of the given expression gives :

By distributive law,

$(4p)(q+r) = 4pq + 4pr$

Question: 1(ii) Carry out the multiplication of the expressions in each of the following pairs.

$ab, a-b$

Answer:

We have ab, (a-b).

Using distributive law, we get,

$ab(a-b) = a^2b - ab^2$

Question: 1(iii) Carry out the multiplication of the expressions in each of the following pairs.

$a+b, 7a^2b^2$

Answer:

Using distributive law, we can obtain the multiplication of the given expression:

$(a + b)(7a^2b^2) = 7a^3b^2 + 7a^2b^3$

Question: 1(iv) Carry out the multiplication of the expressions in each of the following pairs.

$a^2-9,4a$

Answer:

We will obtain the multiplication of the given expression by using the distributive law :

$(a^2 - 9 )(4a) = 4a^3 - 36a$

Question: 1(v) Carry out the multiplication of the expressions in each of the following pairs.

$pq+qr+rp, 0$

Answer:

Using distributive law :

$(pq + qr + rp)(0) = pq(0) + qr(0) + rp(0) = 0$

Question: 2 Complete the table

Answer:

We will use the distributive law to find the product in each case.

Question: 3(i) Find the product.

$(a^2)\times (2a^{22})\times (4a^{26})$

Answer:

Opening brackets :

$(a^2)\times (2a^{22})\times (4a^{26}) = (a^2\times2a^{22})\times(4a^{26}) = 2a^{24}\times4a^{26}$

or $=8a^{50}$

Question: 3(ii) Find the product.

$(\frac{2}{3}xy)\times (\frac{-9}{10}x^2y^2)$

Answer:

We have,

$(\frac{2}{3}xy)\times (\frac{-9}{10}x^2y^2) = \frac{-3}{5}x^3y^3$

Question: 3(iii) Find the product.

$(\frac{-10}{3}pq^3) \times (\frac{6}{5}p^3q)$

Answer:

We have

$(\frac{-10}{3}pq^3) \times (\frac{6}{5}p^3q) = -4p^4q^4$

Question: 3(iv) Find the product.

$x \times x^2\times x^3\times x^4$

Answer:

We have $x \times x^2\times x^3\times x^4$

$x \times x^2\times x^3\times x^4 = (x \times x^2)\times x^3\times x^4$

or $(x^3)\times x^3\times x^4$

$= x^{10}$

Question: 4(a) Simplify $3x(4x-5)+3$ and find its values for

(i) $\small x=3$

Answer:

(a) We have

$3x(4x-5)+3 = 12x^2 - 15x + 3$

Put x = 3,

We get : $12(3)^2 - 15(3) + 3 = 12(9) - 45 + 3 = 108 - 42 = 66$

Question:4(a) Simplify $\small 3x(4x-5)+3$ and find its values for

(ii) $\small x=\frac{1}{2}$

Answer:

We have

$\small 3x(4x-5)+3 = 12x^2 -15x + 3$

Put

$x = \frac{1}{2}$

So we get,

$12x^2 -15x + 3 = 12(\frac{1}{2})^2 - 15(\frac{1}{2}) + 3 = 6 - \frac{15}{2} = \frac{-3}{2}$

Question: 4(b) Simplify $\small a(a^2+a+1) + 5$ and find its value for

(i) $\small a =0$

Answer:

We have : $\small a(a^2+a+1) +5 = a^3 + a^2 + a +5$

Put a = 0 : $= 0^3 + 0^2 + 0 + 5 = 5$

Question: 4(b) Simplify $\small a(a^2+a+1)+5$ and find its value for

(ii) $\small a=1$

Answer:

We have $\small a(a^2+a+1)+5 = a^3 + a^2 + a + 5$

Put a = 1 ,

we get : $1^3 + 1^2 + 1 + 5 = 1 + 1 + 1+ 5 = 8$

Question:4(b) Simplify $\small a(a^2+a+1)+5$ and find its value for

(iii) $\small a=-1$

Answer:

We have $\small a(a^2+a+1)+5$ .

or $\small a(a^2+a+1)+5 = a^3+a^2+a+5$

Put a = (-1)

$= (-1)^3+(-1)^2+(-1)+5 = -1 + 1 -1 +5 = 4$

Question: 5(a) Add: $p(p-q),q(q-r)$ and $r(r-p)$

Answer:

(a) First, we will solve each bracket individually.

$p(p-q) = p^2 - pq$ ; $q(q-r) = q^2 - qr$ ; $r(r-p) = r^2 - rp$

Addind all we get : $p^2 - pq + q^2 - qr + r^2 - rp$

$= p^2 + q^2 + r^2 -pq-qr-rp$

Question:5(b) Add: $\small 2x(z-x-y)$ and $\small 2y(z-y-x)$

Answer:

Firstly, open the brackets:

$\small 2x(z-x-y) = 2xz -2x^2-2xy$

and $\small 2y(z-y-x) = 2yz-2y^2-2xy$

Adding both, we get :

$\small 2xz -2x^2-2xy +2yz-2y^2-2xy$

or $\small = -2x^2-2y^2-4xy + 2xz+2yz$

Question: 5(c) Subtract: $\small 3l(l-4m+5n)$ from $\small 4l(10n-3m+2l)$

Answer:

At first we will solve each bracket individually,

$\small 3l(l-4m+5n) = 3l^2 - 12lm + 15ln$

and $\small 4l(10n-3m+2l) = 40ln - 12ml + 8l^2$

Subtracting:

$\small 40ln - 12ml + 8l^2 - (3l^2 - 12lm+15ln)$

or $\small = 40ln - 12ml + 8l^2 - 3l^2 + 12lm-15ln$

or $\small = 25ln + 5l^2$

Question: 5(d) Subtract: $\small 3a(a+b+c)-2b(a-b+c)$ from $\small 4c(-a+b+c)$

Answer:

Solving brackets :

$3a(a+b+c)-2b(a-b+c) = 3a^2+3ab+3ac - 2ab+2b^2-2bc$

$= 3a^2+ab+3ac+ 2b^2-2bc$

and $\small 4c(-a+b+c) = -4ac +4bc + 4c^2$

Subtracting : $\small -4ac +4bc + 4c^2 -(3a^2 + ab + 3ac+2b^2-2bc)$

$\small = -4ac + 4bc+4c^2-3a^2-ab-3ac-2b^2+2bc$

$\small =-3a^2 -2b^2+4c^2-ab+ 6bc-7ac$.

Question: 1(i) Multiply the binomials.

$\small (2x+5)$ and $\small (4x-3)$

Answer:

We have (2x + 5) and (4x - 3)
(2x + 5) X (4x - 3) = (2x)(4x) + (2x)(-3) + (5)(4x) + (5)(-3)
= 8 $x^{2}$ - 6x + 20x - 15
= 8 $x^{2}$ + 14x -15

Question: 1(ii) Multiply the binomials.

$\small (y-8)$ and $\small (3y-4)$

Answer:

We need to multiply (y - 8) and (3y - 4)
(y - 8) X (3y - 4) = (y)(3y) + (y)(-4) + (-8)(3y) + (-8)(-4)
= 3 $y^{2}$ - 4y - 24y + 32
= 3 $y^{2}$ - 28y + 32

Question: 1(iii) Multiply the binomials

$\small (2.5l-0.5m)$ and $\small (2.5l+0.5m)$

Answer:

We need to multiply (2.5l - 0.5m) and (2.5l + 0..5m)
(2.5l - 0.5m) X (2.5l + 0..5m) = $(2.5l)^{2} - (0.5m)^{2}$ using $(a-b)(a+b) = (a)^{2} - (b)^{2}$
= 6.25 $l^{2}$ - 0.25 $m^{2}$

Question: 1(iv) Multiply the binomials.

$\small (a+3b)$ and $\small (x+5)$

Answer:

(a + 3b) X (x + 5) = (a)(x) + (a)(5) + (3b)(x) + (3b)(5)
= ax + 5a + 3bx + 15b

Question:1(v) Multiply the binomials.

$\small (2pq+3q^2)$ and $\small (3pq-2q^2)$

Answer:

(2pq + 3q 2 ) X (3pq - 2q 2 ) = (2pq)(3pq) + (2pq)(-2q 2 ) + ( 3q 2 )(3pq) + (3q 2 )(-2q 2 )
= 6p 2 q 2 - 4pq 3 + 9pq 3 - 6q 4
= 6p 2 q 2 +5pq 3 - 6q 4

Question: 1(vi) Multiply the binomials.

$\small (\frac{3}{4}a^2+3b^2)$ and $\small 4(a^2-\frac{2}{3}b^2)$

Answer:

Multiplication can be done as follows

$\small (\frac{3}{4}a^2+3b^2)$ X $\small (4a^2-\frac{8}{3}b^2)$ = $\frac{3a^{2}}{4} \times 4a^{2} + \frac{3a^{2}}{4} \times (-\frac{8b^{2}}{3}) + 3b^{2} \times 4a^{2} + 3b^{2} \times (-\frac{8b^{2}}{3})$

= $3a^{4} - 2a^{2}b^{2} + 12a^{2}b^{2} - 8b^{4}$

= $3a^{4} + 10a^{2}b^{2} - 8b^{4}$

Question: 2(i) Find the product.

$\small (5-2x)$ $\small (3+x)$

Answer:

(5 - 2x) X (3 + x) = (5)(3) + (5)(x) +(-2x)(3) + (-2x)(x)
= 15 + 5x - 6x - 2 $x^{2}$
= 15 - x - 2 $x^{2}$

Question: 2(ii) Find the product.

$\small (x+7y)(7x-y)$

Answer:

(x + 7y) X (7x - y) = (x)(7x) + (x)(-y) + (7y)(7x) + (7y)(-y)
= 7 $x^{2}$ - xy + 49xy - 7 $y^{2}$
= 7 $x^{2}$ + 48xy - 7 $y^{2}$

Question: 2(iii) Find the product.

$\small (a^2+b)(a+b^2)$

Answer:

($a^{2}$ + b) X (a + $b^{2}$ ) = ( $a^{2}$ )(a) + ( $a^{2}$ )( $b^{2}$ ) + (b)(a) + (b)( $b^{2}$ )
= $a^{3 } + a^{2}b^{2} + ab + b^{3}$

Question: 2(iv) Find the product.

$\small (p^2-q^2)(2p+q)$

Answer:

Following is the solution

($p^{2}- q^{2}$ ) X (2p + q) = $(p^{2})(2p) + (p^{2})(q) + (-q^{2})(2p) + (-q^{2})(q)$
$2p^{3} + p^{2}q - 2q^{2}p - q^{3}$

Question: 3(i) Simplify.

$\small (x^2-5)(x+5)+25$

Answer:

This can be simplified as follows

($x^{2}$ -5) X (x + 5) + 25 = ( $x^{2}$ )(x) + ( $x^{2}$ )(5) + (-5)(x) + (-5)(5) + 25
= $x^{3} + 5x^{2} - 5x -25 + 25$
= $x^{3} + 5x^{2} - 5x$

Question: 3(ii) Simplify

$(a^2+5)(b^3+3)+5$

Answer:

This can be simplified as

($a^{2}$ + 5) X ( $b^{3}$ + 3) + 5 = ( $a^{2}$ )( $b^{3}$ ) + ( $a^{2}$ )(3) + (5)( $b^{3}$ ) + (5)(3) + 5
= $a^{2}b^{3} + 3a^{2} + 5b^{3} + 15+5$
= $a^{2}b^{3} + 3a^{2} + 5b^{3} + 20$

Question: 3(iii) Simplify

$(t+s^2)(t^2-s)$

Answer:

Simplifications can be

(t + $s^{2}$ )( $t^{2}$ - s) = (t)( $t^{2}$ ) + (t)(-s) + ( $s^{2}$ )( $t^{2}$ ) + ( $s^{2}$ )(-s)
= $t^{3} - ts + s^{2}t^{2} - s^{3}$

Question: 3(iv) Simplify

$(a+b)(c-d)+(a-b)(c+d)+2 (ac+bd)$

Answer:

(a + b) X ( c -d) + (a - b) X (c + d) + 2(ac + bd )
= (a)(c) + (a)(-d) + (b)(c) + (b)(-d) + (a)(c) + (a)(d) + (-b)(c) + (-b)(d) + 2(ac + bd )
= ac - ad + bc - bd + ac +ad -bc - bd + 2(ac + bd )
= 2(ac - bd ) + 2(ac +bd )
= 2ac - 2bd + 2ac + 2bd
= 4ac

Question: 3(v) Simplify

$(x+y)(2x+y)+(x+2y)(x-y)$

Answer:

(x + y) X ( 2x + y) + (x + 2y) X (x - y)
= (x)(2x) + (x)(y) + (y)(2x) + (y)(y) + (x)(x) + (x)(-y) + (2y)(x) + (2y)(-y)
= 2 $x^{2}$ + xy + 2xy + $y^{2}$ + $x^{2}$ - xy + 2xy - 2 $y^{2}$
= 3 $x^{2}$ + 4xy - $y^{2}$

Question: 3(vi) Simplify.

$(x+y)(x^2-xy+y^2)$

Answer:

Simplification is done as follows

(x + y) X ( $x^{2} -xy + y^{2}$ ) = x X ( $x^{2} -xy + y^{2}$ ) + y ( $x^{2} -xy + y^{2}$ )
= $x^{3} -x^{2}y + xy^{2} + yx^{2} - xy^{2} + y^{3}$
= $x^{3}+ y^{3}$

Question: 3(vii) Simplify.

$(1.5x-4y)(1.5x+4y+3)-4.5x+12y$

Answer:

(1.5x - 4y) X (1.5x + 4y + 3) - 4.5x + 12y = (1.5x) X (1.5x + 4y + 3) -4y X (1.5x + 4y + 3) - 4.5x + 12y
= 2.25 $x^{2}$ + 6xy + 4.5x - 6xy - 16 $y^{2}$ - 12y -4.5x + 12 y
= 2.25 $x^{2}$ - 16 $y^{2}$

Question: 3(viii) Simplify

$(a+b+c)(a+b-c)$

Answer:

(a + b + c) (a + b - c) = a (a + b - c) + b (a + b - c) + c (a + b - c)
= $a^{2}$ + ab - ac + ab + $b^{2}$ -bc + ac + bc - $c^{2}$
= $a^{2}$ + $b^{2}$ - $c^{2}$ + 2ab

Algebraic Expressions and Identities Class 8 NCERT Solutions - Topics

The topics discussed in the NCERT Solutions for class 8, chapter 8, Algebraic Expressions and Identities, are:

• Terms
• Factors and Coefficients
• Monomials
• Binomials and Polynomials
• Like and Unlike Terms
• Addition and Subtraction of Algebraic Expressions
• Multiplication of Algebraic Expressions
• What is an Identity
• Standard Identities
• Applying Identities

Algebraic Expressions and Identities Class 8 Solutions - Important Formulae

Here are some important formulas that will help students to solve problems related to Algebraic Expressions and Identities.

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• (a + b)(a - b) = a² - b²

• (x + a)(x + b) = x² + (a + b)x + ab

• (x + a)(x - b) = x² + (a - b)x - ab

• (x - a)(x + b) = x² + (b - a)x - ab

• (x - a)(x - b) = x² - (a + b)x + ab

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)

NCERT Solutions For Class 8 Maths - Chapter Wise

Given below are all the chapter-wise NCERT class 8 maths solutions given in one place for ease of access:

NCERT Class 8 Maths Solutions: Subject Wise

Students can use the links below to prepare efficiently in other subjects as well as Mathematics.

• NCERT solutions for class 8 maths
• NCERT solutions for class 8 science

NCERT Books and Syllabus for Class 8

The following links will give students access to the latest CBSE syllabus and some reference books.

• NCERT Books Class 8 Maths
• NCERT Syllabus Class 8 Maths
• NCERT Books Class 8
• NCERT Syllabus Class 8