NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.1 what are expressions?

Question: 1 Give five examples of expressions containing one variable and five examples of expressions containing two variables.

Answer:

Five examples of expressions containing one variable are:

$x^{^{4}}, y, 3z, p^{^{2}}, -2q^{3}$

Five examples of expressions containing two variables are:

$x + y, 3p-4q,ab,uv^{2},-z^{2}+x^{3}$

Question: 2(i) Show on the number line :

Answer:

x on the number line:

Question: 2(ii) Show on the number line :

Answer:

x-4 on the number line:

Question: 2(iii) Show on the number line :

2x+1

Answer:

2x+1 on the number line:

Question: 2(iv) Show on the number line:

Answer:

3x - 2 on the number line

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.2 terms, factors and coefficients

Question:1 Identify the coefficient of each term in the expression.

Answer:

coefficient of each term are given below

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.3 monomials, binomials and polynomials

Question: 1(i) Classify the following polynomials as monomials, binomials, trinomials.

Answer:

Binomial since there are two terms with non zero coefficients.

Question: 1(ii) Classify the following polynomials as monomials, binomials, trinomials.

Answer:

Trinomial since there are three terms with non zero coefficients.

Question:1(iii) Classify the following polynomials as monomials, binomials, trinomials.

Answer:

Trinomial since there are three terms with non zero coefficients.

Question: 1(iv) Classify the following polynomials as monomials, binomials, trinomials.

Answer:

Binomial since there are two terms with non zero coefficients.

Question: 1(v) Classify the following polynomials as monomials, binomials, trinomials.

Answer:

Monomial since there is only one term.

Question: 2(a) Construct 3 binomials with only as a variable;

Answer:

Three binomials with the only x as a variable are:

$\\ \\x+2,\ x +x^{2},\ 3x^{3}-5x^{4}$

Question: 2(b) Construct 3 binomials with and as variables;

Answer:

Three binomials with x and y as variables are:

$\\ \\x+y,\ x-7y, xy^{2} + 2xy$

Question: 2(c) Construct 3 monomials with and as variables;

Answer:

Three monomials with x and y as variables are

$\\ xy,\ 3xy^{4},\ -2x^{3}y^{2}$

Question: 2(d) Construct 2 polynomials with 4 or more terms .

Answer:

Two polynomials with 4 or more terms are:

$a+b+c+d, x-3xy+2y+4xy^{2}$

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.4 like and unlike terms

Question:(i) Write two terms which are like

Answer:

$\\Two\ terms\ like\ 7xy\ are:\\ -3xy\ and\ 5xy$

Question:(ii) Write two terms which are like

Answer:

$\\Two\ terms\ which\ are\ like\ 4mn^{2}\ are:\\ mn^{2}\ and -3mn^{2.}$

we can write more like terms

Question:(iii) Write two terms which are like

Answer:

$\\Two\ terms\ which\ are\ like\ 2l\ are:\\ l\ and\ -3l$

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.1

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

Answer

following are the terms and coefficient

The terms are $5xyz^{2}\ and\ -3zy$ and the coefficients are 5 and -3.

Question: 1(ii) Identify the terms, their coefficients for each of the following expressions.

Answer:

the following is the solution

$\\The\ terms\ are\ 1,\ x,\ and\ x^{2}\ and\ the\ coefficients\ are\ 1,\ 1,\ and\ 1\ respectively.$

Question:1(iii) Identify the terms, their coefficients for each of the following expressions.

Answer:

Question: 1(iv) Identify the terms, their coefficients for each of the following expressions.

Answer:

The terms are 3, -pq, qr,and -rp and the coefficients are 3, -1, 1 and -1 respectively.

Question:1(v) Identify the terms, their coefficients for each of the following expressions.

$\frac{x}{2}+\frac{y}{2}-xy$

Answer:

Above are the terms and coefficients

Question: 1(vi) Identify the terms, their coefficients for each of the following expressions.

Answer:

The terms are 0.3a, -0.6ab and 0.5b and the coefficients are 0.3, -0.6 and 0.5.

Question: 2(a) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

Binomial.

Question: 2(b) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

Monomial.

Question: 2(c) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

This polynomial does not fit in any of these three categories.

Question: 2(d) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

Trinomial.

Question: 2(e) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

Binomial.

Question: 2(f) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

Trinomial.

Question: 2(g) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

Trinomial.

Question: 2(h) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

Binomial.

Question: 2(i) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

This polynomial does not fit in any of these three categories.

Question:2(j) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

Monomial.

Question: 2(k) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

Binomial.

Question: 2(i) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Answer:

Binomial.

Question: 3(i) Add the following.

Answer:

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

Question:3 (ii) Add the following.

Answer:

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

Question:3 (iii) Add the following

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: 3(iv) Add the following.

Answer:

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

Question: 4(a) Subtract from

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: 4(b) Subtract from

Answer:

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

Question: 4(c) Subtract from

Answer:

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.7.2 multiplying three or more monomials

Question:1 Find $4x\times 5y\times 7z$ . First find $4x\times 5y$ and multiply it by ; or first find $5y \times 7z$ and multiply it by .

Answer:

We observe that the result is same in both cases and the result does not depend on the order in which multiplication has been carried out.

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.2

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

Answer:

$4\times 7p=28p$

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

Answer:

<img alt="\\-4p\times 7p\\=(-4\times 7)p\times p\\=-28p^{2}" height="61"

src="https://lh3.googleusercontent.com/ZQLwJFEoi48QnxUuK79XbVO1Ug-ZvBoOflH3wFRfnzdHsH_uWOVnOwg4kaF8g0o8RSXTEfDcwrA7mDeS0UuxfpygL0vqvj2Ip_Ntb3J0UEOs2WBYddLAm3Ff_WG8HAzyPrqB_V0" style="margin-left: 0px; margin-top: 0px;" width="128" />

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

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.

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.

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.

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.

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.

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.

Answer:

area of rectangles 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.

Answer:

The area is calculated as follows

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

Question:3 Complete the table of products.

First monomial $\rightarrow$ Second monomial $\downarrow$
		...	...	...	...	...
	...	...		...	...	...
	...	...	...	...	...	...
	...	...	...	...	...	...
	...	...	...	...	...	...
	...	...	...	...	...	...

Answer:

First monomial $\rightarrow$ Second monomial $\downarrow$			$3x^{2}$		$7x^{2}y$	$-9x^{2}y^{2}$
	$4x^{2}$		$6x^{3}$	$-8x^{2}y$	$14x^{3}y$	$-18x^{3}y^{2}$
		$25y^{2}$	$-15x^{2}y$	$20xy^{2}$	$-35x^{2}y^{2}$	$45x^{2}y^{3}$
$3x^{2}$	$6x^{3}$	$-15x^{2}y^{}$	$9x^{4}$	$-12x^{3}y$	$21x^{4}y$	$-27x^{4}y^{2}$
	$-8x^{2}y$	$20xy^{2}$	$-12x^{3}y$	$16x^{2}y^{2}$	$-28x^{3}y$	$36x^{3}y^{3}$
$7x^{2}y$	$14x^{3}y$	$-35x^{2}y^{2}$	$21x^{4}y$	$-28x^{3}y^{2}$	$49x^{4}y^{2}$	$-63x^{4}y^{3}$
$-9x^{2}y^{2}$	$-18x^{3}y^{2}$	$45x^{2}y^{3}$	$-27x^{4}y^{2}$	$36x^{3}y^{3}$	$-63x^{4}y^{3}$	$81x^{4}y^{4}$

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

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.

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.

Answer:

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

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

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

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

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

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

Answer:

the product

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

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.8.1 multiplying a monomial by a binomial

Question:(i) Find the product

Answer:

Using distributive law,

Question:(ii) Find the product

Answer:

Using distributive law,

We have :

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.8.2 multiplying a monomial by a trinomial

Question:1 Find the product:

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

Answer:

By using distributive law,

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

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.3

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

Answer:

Multiplication of the given expression gives :

By distributive law,

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

Answer:

We have ab, (a-b).

Using distributive law we get,

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

Answer:

Using distributive law we can obtain multiplication of given expression:

<img alt="(a + b)(7a^2b^2) = 7a^3b^2 + 7a^2b^3" height="20"

src="https://lh6.googleusercontent.com/zym-v6hoY_4dPgfdHnN35kmtdqAvE_B4CvAHBzx42tg4OdnWGpufkQNURYQP8yq9xk7qpbbNA_sAzgb0gc0AnG0Vxa5aU4snOBR80IU5GQ_6VDoRBqkDdZGjqZfE11JML0vyGTI" style="margin-left: 0px; margin-top: 0px;" width="232" />

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

Answer:

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

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

Answer:

Using distributive law :

Question:2 Complete the table

First expression	Second expression	Product
(i)			...
(ii)			...
(iii)			...
(iv)			...
(v)			...

Answer:

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

First expression	Second expression	Product
(i)
(ii)
(iii)
(iv)
(v)

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 and find its values for

(i) $\small x=3$

Answer:

(a) We have

Put x = 3,

We get :

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 :

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 :

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)

Question:5(a) Add: and

Answer:

(a)First we will solve each brackets individually.

; ;

Addind all we get :

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 :

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

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

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

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

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.4

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.

<img alt="\small (2pq+3q^2)" height="19"

src="https://lh4.googleusercontent.com/faqbYXt8AEy2TVr4J1p5vfUz45qyR0d4muq8Xtx5GLLq1Q_3ixfJjPaV3qC_ucxFJnviMKxD4XzZothad36SoNOFM4fgPw4kUMXa3b14Bf5s7YQDTAuZkKh5Nzh92lnwz2Fa810" style="margin-left: 0px; margin-top: 0px;" width="79" /> 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 $\small (4a^2-\frac{8}{3}b^2)$ =

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

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

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 .

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.

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.

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.

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.

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.

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.

Answer:

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

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.11 standerd identities

Question:1(i) Put -b in place of b in identity 1. Do you get identity 2?

Answer:

Identity 1 $\Rightarrow (a+b)^{2} = a^{2} + 2ab + b^{2}$
If we replace b with -b in identity 1
We get,
$a^{2} + 2a(-b) + (-b)^{2} = a^{2} - 2ab + b^{2}$
which is equal to
$(a-b)^{2}$ which is identity 2
So, we get identity 2 by replacing b with -b in identity 1

NCERT free solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.11 standard identities

Question:1 Verify Identity (IV), for .

Answer:

Identity IV
(a + x)(b + x) = $x^{2} + (a+b)x + ab$
So, it is given that a = 2, b = 3 and x = 5
Lets put these value in identity IV
(2 + 5)(3 + 5) = $5^{2}$ + (2 + 3)5 +2 X 3
7 X 8 = 25 + 5 X 5 + 6
56 = 25 + 25 + 6
= 56
L.H.S. = R.H.S.
So, by this we can say that identity IV satisfy with given value of a,b and x

Question:2 Consider, the special case of Identity (IV) with a = b, what do you get? Is it related to Identity

Answer:

Identity IV is $\Rightarrow (a +x)(b+x) = x^{2} + (a+b)x + ab$
If a =b than

(a + x)(a + x) = $x^{2} + (a+a)x + a\times a$
$(a+x)^{2} = x^{2} + 2ax + a^{2}$
Which is identity I

Question:3 Consider, the special case of Identity (IV) with and What do you get? Is it related to Identity ?

Answer:

Identity IV is $\Rightarrow (a+x )(b +x) = x^{2} + (a+b)x + ab$
If a = b = -c than,
(x - c)(x - c) = $x^{2} + (-c + (-c))x + (-c) \times (-c)$
$(x-c)^{2} = x^{2} + -2cx + c^{2}$
Which is identity II

Question:4 Consider the special case of Identity (IV) with . What do you get? Is it related to Identity?

Answer:

Identity IV is $\Rightarrow (a+x )(b +x) = x^{2} + (a+b)x + ab$
If b = -a than,

(x + a)(x - a) = $x^{2} + (a +(-a))x + (-a) \times a$
= $x^{2} - a^{2}$
Which is identity III

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.5

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

Answer:

(x + 3) X (x +3) = $(x +3)^{2}$
So, we use identity I for this which is
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a=x and b = x
$(x+3)^{2} = x^{2} + 2(x)(3)+ 3^{2}$
= $x^{2} + 6x+ 9$

Question:1(ii) Use a suitable identity to get each of the following products in bracket.

Answer:

(2y + 5) X ( 2y + 5) = $(2y +5)^{2}$
We use identity I for this which is
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
IN this a = 2y and b = 5
$(2y+5)^{2} = (2y)^{2} + 2(2y)(5) + 5^{2}$
= $(2y+5)^{2} = 4y^{2} + 20y + 25$

Question:1(iii) Use a suitable identity to get each of the following products in bracket.

Answer:

(2a -7) X (2a - 7) = $(2a - 7)^{2}$
We use identity II for this which is
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
in this a = 2a and b = 7
$(2a-7)^{2} = (2a)^{2} - 2(2a)(7) + 7^{2}$
= $4a^{2} - 28a + 49$

Question:1(iv) Use a suitable identity to get each of the following products in bracket.

$(3a - \frac{1}{2}) (3a -\frac{1}{2} )$

Answer:

$(3a - \frac{1}{2}) \times (3a -\frac{1}{2} ) = ((3a - \frac{1}{2}))^{2}$
We use identity II for this which is
$(a-b)^{2} = a^{2} -2ab + b^{2}$
in this a = 3a and b = -1/2
$(3a-\frac{1}{2})^{2} = (3a)^{2} -2(3a)(\frac{1}{2}) + (\frac{1}{2})^{2}$
= $9a^{2} -3a + \frac{1}{4}$

Question:1(v) Use a suitable identity to get each of the following products in bracket.

Answer:

We use identity III for this which is
(a - b)(a + b) = $a^{2} - b^{2}$
In this a = 1.1m and b = 4
= $(1.1m)^{2} - (4)^{2}$
= 1.21 $m^{2}$ - 16

Question:1(vi) Use a suitable identity to get each of the following products in bracket.

Answer:

take the (-)ve sign common so our question becomes
- $-(a^{2}+b^{2})(a^{2}-b^{2})$
We use identity III for this which is
(a - b)(a + b) = $a^{2} - b^{2}$
In this a = $a^{2}$ and b = $b^{2}$

$-(a^{2}+b^{2})(a^{2}-b^{2})$ = $-((a^{2})^{2} -(b^{2})^{2}) = -a^{4} + b^{4}$

Question:1(vii) Use a suitable identity to get each of the following.

Answer:

(6x -7) X (6x - 7) = $(6x-7)^{2}$
We use identity III for this which is
(a - b)(a + b) = $a^{2} - b^{2}$
In this a = 6x and b = 7
(6x -7) X (6x - 7) = $(6x)^{2} - (7)^{2} = 36x^{2} - 49$

Question:1(viii) Use a suitable identity to get each of the following product.

Answer:

take (-)ve sign common from both the brackets So, our question become
(a -c) X (a -c) = $(a -c)^{2}$
We use identity II for this which is
$(a-b)^{2} =a^{2} -2ab + b^{2}$
In this a = a and b = c
$(a-c)^{2} =a^{2} -2ac + c^{2}$

Question:1(ix) Use a suitable identity to get each of the following product.

$(\frac{x}{2}+ \frac{3y}{4})(\frac{x}{2}+ \frac{3y}{4})$

Answer:

$(\frac{x}{2}+ \frac{3y}{4}) \times (\frac{x}{2}+ \frac{3y}{4}) = (\frac{x}{2}+ \frac{3y}{4})^{2}$

We use identity I for this which is
$(a+b)^{2} =a^{2}+2ab + b^{2}$
In this a = $\frac{x}{2}$ and b = $\frac{3y}{4}$

= $\frac{x^{2}}{4} + \frac{3xy}{4} + \frac{9y^{2}}{16}$

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

Answer:

$(7a-9b) \times (7a-9b) = (7a-9b)^{2}$

We use identity II for this which is
$(a-b)^{2} =a^{2}-2ab + b^{2}$
In this a = 7a and b = 9b
$(7a-9b)^{2} =(7a)^{2}-2(7a)(9b) + (9b)^{2}$
= $49a^{2}-126ab + 81b^{2}$

Question:2(i) Use the identity to find the following products.

Answer:

We use identity
in this a = 3 and b = 7
= $x^2+(3+7)x+3 \times 7$
=

Question:2(ii) Use the identity to find the following products.

Answer:

We use identity
In this a= 5 , b = 1 and x = 4x
=
=

Question:2(iii) Use the identity to find the following products.

Answer:

We use identity
in this x = 4x , a = -5 and b = -1
=
=

Question:1(iv) Use the identity to find the following products.

Answer:

We use identity
In this a = 5 , b = -1 and x = 4x
=
=

Question:2(v) Use the identity to find the following products.

Answer:

We use identity
In this a = 5y , b = 3y and x = 2x
=
= $4x^2+16xy + 15y^{2}$

Question:2(vi) Use the identity to find the following products.

Answer:

We use identity
In this a = 9 , b = 5 and x = <img alt="2a^{2}" height="16"

src="https://lh5.googleusercontent.com/2qHQF6RatvqmP_IzaprdObn1TW7Qc6P5ahuw5OBgw979_B5ENMeSRnBhk9M65bsGhcIIOjA5Z-Tfx4RDq34UjzRxIxPbAH48O8Ack3-cw0InSk7MCot84kU1ew5darMTXngAh6s" style="margin-left: 0px; margin-top: 0px;" width="24" />
$(2a^{2}+9)(2a^{2}+5)$ = $(2a^{2})^2+(9+5)2a^{2}+(9)(5)$
= $4a^{4} + 28a^{2} + 45$

Question:2(vii) Use the identity to find the following products.

Answer:

We use identity
In this a = -4 , b = -2 and x = xyz
=
= $x^{2}y^{2}z^{2} -6xyz + 8$

Question:3(i) Find the following squares by using the identities.

Answer:

We use identity
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a =b and b = 7
$(b-7)^{2} = b^{2} - 2(b)(7) + 7^{2}$
= $b^{2} - 14b + 49$

Question:3(ii) Find the following squares by using the identities.

Answer:

We use
$(a+b)^{2} = a^{2} + 2ab + b^{2}$

In this a = xy and b = 3z
$(xy+3z)^{2} = (xy)^{2} + 2(xy)(3z) + (3z)^{2}$
= $x^{2}y^{2} + 6xyz+ 9z^{2}$

Question:3(iii) Find the following squares by using the identities.

Answer:

We use
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = $6x^{2}$ and b = $5y^{2}$
$(6x-5y)^{2} = (6x)^{2} - 2(6x)(5y) + (5y)^{2}$
= $36x^{2} - 60xy + 25y^{2}$

Question:3(iv) Find the following squares by using the identities.

$(\frac{2}{3}m+\frac{3}{2}n)^2$

Answer:

we use the identity
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a = $\frac{2m}{3}$ and b = $\frac{3n}{2}$

= $\frac{4m^{2}}{9} + 2mn + \frac{9n^{2}}{4}$

Question:3(v) Find the following squares by using the identities.

Answer:

we use
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = 0.4p and b =0.5q
$(0.4p-0.5q)^{2} = (0.4p)^{2} - 2(0.4p)(0.5q) + (0.5q)^{2}$
= $0.16p^{2} - 0.4pq + 0.25q^{2}$

Question:3(vi) Find the following squares by using the identities.

Answer:

we use the identity
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a = 2xy and b =5y
$(2xy+5y)^{2} = (2xy)^{2} + 2(2xy)(5y) + (5y)^{2}$
= $4x^{2}y^{2} + 20xy^{2} + 25y^{2}$

Question:4(i) Simplify:

Answer:

we use
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = $a^{2}$ and b = $b^{2}$
$(a^{2}-b^{2})^{2} = (a^{2})^{2} - 2(a^{2})(b^{2}) + (b^{2})^{2}$
= $a^{4} - 2a^{2}b^{2} + b^{4}$

Question:4(ii) Simplify.

Answer:

we use
$a^{2} - b^{2} = (a-b)(a+b)$
In this a = (2x + 5) and b = (2x - 5)
$(2x + 5)^{2} - (2x - 5)^{2} = ( (2x + 5)- (2x - 5))( (2x + 5)+ (2x - 5))$
=
= (4x)(10)
=40x

remember that

here a= 2x, b= 5

$4ab=4\times 2x \times 5=40x$

Question:4(iii) Simplify.

Answer:

we use
$(a-b)^{2} = a^{2} -2ab + b^{2}$ and $(a+b)^{2} = a^{2} +2ab + b^{2}$
In this a = 7m and b = 8n
$(7m-8n)^{2} = (7m)^{2} -2(7m)(8n) + (8n)^{2}$
= $49m^{2} -112mn + 64n^{2}$
and
$(7m+8n)^{2} = (7m)^{2} +2(7m)(8n) + (8n)^{2}$
= $49m^{2} +112mn + 64n^{2}$

So, $(7m - 8n)^{2} + (7m + 8n)^{2}$ = $49m^{2} -112mn + 64n^{2}$ + $49m^{2} +112mn + 64n^{2}$
= $2(49m^{2} + 64n^{2})$

remember that

Question: 4(iv) Simplify.

Answer:

we use
$(a+b)^{2} = a^{2} +2ab + b^{2}$
1 ) In this a = 4m and b = 5n

$(4m+5n)^{2} = (4m)^{2} +2(4m)(5n) + (5n)^{2}$
= $16m^{2} +40mn + 25n^{2}$
2 ) in this a = 5m and b = 4n
$(5m+4n)^{2} = (5m)^{2} +2(5m)(4n) + (4n)^{2}$
= $25m^{2} +40mn + 16n^{2}$

So, $(4m + 5n)^{2} + (5m + 4n)^{2}$ = $16m^{2} +40mn + 25n^{2}$ + $25m^{2} +40mn + 16n^{2}$
= $41m^{2} +80mn + 41n^{2}$

Question: 4(v) Simplify.

Answer:

we use
$a^{2}- b^{2} = (a-b)(a+b)$
1 ) In this a = (2.5p- 1.5q) and b = (1.5p - 2.5q)
$(2.5p- 1.5q)^{2}- (1.5p- 2.5q)^{2} = ( (2.5p- 1.5q)- (1.5p- 2.5q))( (2.5p- 1.5q)+ (1.5p- 2.5q))$
=
= 4(p + q ) (p - q)
= 4 $(p^{2} - q^{2})$

Question:4(vi) Simplify.

Answer:

We use identity
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a = ab and b = bc
$(ab+bc)^{2} = (ab)^{2} + 2(ab)(bc) + (bc)^{2}$
= $a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2}$
Now, $a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2}$ - $2ab^{2}c$
= $a^{2}b^{2} + b^{2}c^{2}$

Question:4(vii) Simplify.

Answer:

We use identity
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = $m^{2}$ and b = $n^{2}m$
$(m^{2}-n^{2}m)^{2} = (m^{2})^{2} - 2(m^{2})(n^{2}m) + (n^{2}m)^{2}$
= $m^{4} - 2m^{3}n^{2} + n^{4}m^{2}$
Now, $m^{4} - 2m^{3}n^{2} + n^{4}m^{2}$ + $2m^{3}n^{2}$
= $m^{4} + n^{4}m^{2}$

Question:5(i) Show that

Answer:

L.H.S. =

= R.H.S.

Hence it is prooved

Question:5(ii) Show that

Answer:

L.H.S. = (Using )

$\left ( (a+b)^2 = a^2 + 2ab + b^2 \right )$

= R.H.S.

Question:5(iii) Show that.

$(\frac{4}{3}m-\frac{3}{4}n)^2 +2mn=\frac{16}{9}m^2+\frac{9}{16}n^2$

Answer:

First we will solve the LHS :

$= (\frac{4}{3}m-\frac{3}{4}n)^2 +2mn = \frac{16}{9}m^2 - 2mn + \frac{9}{16}n^2 + 2mn$

or $= \frac{16}{9}m^2 + \frac{9}{16}n^2$

= RHS

Question:5(iv) Show that.

Answer:

Opening both brackets we get,

= R.H.S.

Question:5(v) Show that

Answer:

Opening all brackets from the LHS, we get :

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

= RHS

Question:6(i) Using identities, evaluate.

Answer:

We will use the identity:

So,

Question:6(ii) Using identities, evaluate.

Answer:

Here we will use the identity :

So :

Question:6(iii) Using identities, evaluate.

Answer:

Here we will use the identity :

So :

Question:6(iv) Using identities, evaluate.

Answer:

Here we will the identity :

Question:6(v) Using identities, evaluate.

Answer:

Here we will use :

Thus

Question:6(vi) Using identities, evaluate.

$297 \times 303$

Answer:

This can be written as :

$297\times303 = (300-3)\times(300+3)$

using

Question:6(vii) Using identities, evaluate.

$78 \times 82$

Answer:

This can be written in form of :

$78\times82 = (80 - 2) \times(80+2)$

or $\because \left ( a-b \right )\left ( a+b \right ) = a^2 - b^2$

Question:6(viii) Using identities, evaluate.

Answer:

Here we will use the identity :

Thus :

Question:6(ix) Using identities, evaluate.

$10.5\times9.5$

Answer:

This can be written as :

$10.5\times9.5 = (10 +0.5)\times(10-0.5)$

or $\because (a+b)(a-b) = a^2 - b^2$

Question:7(i) Using , find

Answer:

We know,

Using this formula,

= (51 + 49)(51 - 49)

= (100)(2)

= 200

Question:7(ii) Using , find

Answer:

We know,

Using this formula,

= (1.02 + 0.98)(1.02 - 0.98)
= (2.00)(0.04)

= 0.08

Question:7(iii) Using , find.

Answer:

We know,

Using this formula,

= (153 - 147)(153 +147)

=(6) (300)

= 1800

Question:7(iv) Using , find

Answer:

We know,

Using this formula,

= (1.02 + 0.98)(1.02 - 0.98)

= (2.00)(0.04)

= 0.08

Question:8(i) Using $103 \times 104$

Answer:

We know,

Using this formula,

$103 \times 104$ = (100 + 3)(100 + 4)

Here x =100, a = 3, b = 4

$\therefore$ $103 \times 104$ $= 100^2+(3+4)100+(3\times4)$

= 11212

Question:8(ii) Using , find

$5.1\times 5.2$

Answer:

We know,

Using this formula,

$5.1\times 5.2$ = (5 + 0.1)(5 + 0.2)

Here x =5, a = 0.1, b = 0.2

$\therefore$ $5.1\times 5.2$ $=5^2+(0.1 + 0.2)5+(0.1\times0.2)$

= 26.52

Question:8(iii) Using , find

$103 \times 98$

Answer:

We know,

Using this formula,

$103 \times 98$ = (100 + 3)(100 - 2) = (100 + 3){100 + (-2)}

Here x =100, a = 3, b = -2

$\therefore$ $103 \times 98$ $= 100^2+(3 + (-2))100+(3\times(-2))$

= 10094

Question: 8(iv) Using , find

$9.7 \times 9.8$

Answer:

We know,

Using this formula,

$9.7 \times 9.8$ = (10 - 0.3)(10 - 0.2) = {10 + (-0.3)}{10 + (-0.2)}

Here x =10, a = -0.3, b = -0.2

$\therefore$ $9.7 \times 9.8$ $= 10^2+((-0.3) + (-0.2))10+((-0.3)\times(-0.2))$

= 95.

NCERT Class 8 Mathematics Solutions

Chapter 01 - Rational Numbers

Chapter 02 - Linear Equations in One Variable

Chapter 03 -Understanding Quadrilaterals

Chapter 04 - Practical Geometry

Chapter 05 - Data Handling

Chapter 06 - Squares and Square Roots

Chapter 07 - Cubes and Cube Roots

Chapter 08 - Comparing Quantities

Chapter 09 - Algebraic Expressions and Identities

Chapter 10 - Visualising Solid Shapes

Chapter 11 - Mensuration

Chapter 12 - Exponents and Powers

Chapter 13 - Direct and Indirect proportions

Chapter 14 - Factorisation

Chapter 15 - Introduction to Graphs

Chapter 16 - Playing with Numbers

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NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

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NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.4

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NCERT Class 8 Mathematics Solutions

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