CBSE Important Questions for Class 8 Maths Algebraic Expressions and Identities - 2025-26

Very Short Answer Questions (1 Mark)

1. Identify the coefficient of x in the expression

(a) 1 

(b) -1 

(c) -2 

(d) None of these

Ans: Given: \[x\]

We need to find the coefficient of the given expression.

We know that coefficient is the number before any variable. 

So, the expression \[x\] can be written as $ 1\times x $ . 

So, the coefficient will be $ 1. $ 

2. The expression for sum of numbers p and q subtracted from their product is

(a) p+q-pq  

(b) pq-p+q  

(c)  pq-(p+q)  

(d)  pq+p-q  

Ans: Given: variables $ p,q $ 

To find: the expression for sum of numbers $ p $ and $ q $ subtracted from their product

The product of the numbers will be

$ \begin{align} & p\times q \\ & =pq \\ \end{align} $

Sum of numbers will be

 $ p+q $ 

Therefore, the required answer will be

$pq-(p+q) $ 

3. Sum of \[\mathbf{ab,}\] \[\mathbf{a\text{ }+\text{ }b,}\] and \[\mathbf{b\text{ }+\text{ }ab}\] is

(a) 2ab+2a+b

(b)2ab+2b

(c)2ab+a+b

(d)2ab+a+2b

Ans: Given: $ab, a+b, b+ab $ 

We need to find the sum of the given terms.

Therefore, the sum will be

$ \begin{align} & ab+a+b+b+ab \\ & =2ab+a+2b \\ \end{align} $

4. The value of expression $ \mathbf{5{{x}^{2}}-2} $ when $\mathbf{ x=3 }$ is

 (a)-12 

 (b) 8 

 (c) 43 

 (d) 36 

Ans: Given: $ 5{{x}^{2}}-2 $ , $ x=3 $ 

We need to find the value of $ 5{{x}^{2}}-2 $ at $ x=3 $ 

We will put $ 3 $ in place of $ x $ in the expression $ 5{{x}^{2}}-2 $ 

Therefore, the value of the expression will be

$ \begin{align} & 5{{x}^{2}}-2 \\ & =5{{(3)}^{2}}-2 \\ & =5(3)-2 \\ & =45-2 \\ & =43 \\ \end{align} $

5. On simplifying the result is $\mathbf{ (a+b-2)-(b-a+2)+(a-b+2)} $

 (a) $\mathbf{ a-b+2} $ 

 (b) $ \mathbf{a-b-2 }$ 

 (c) $ \mathbf{3a-b-2 }$ 

 (d) $\mathbf{ 3a+b+2 }$ 

Ans: Given: $ (a+b-2)-(b-a+2)+(a-b+2) $ 

We need to simplify the given expression.

So, the expression will be simplified as

$ \begin{align} & (a+b-2)-(b-a+2)+(a-b+2) \\ & =a+b-2-b+a-2+a-b+2 \\ & =3a-b-2 \\ \end{align} $

6. What is the statement for the expression $\mathbf{ 2(y-9)} $ 

 (a)  2y  subtracted from  9  

 (b)  9  subtracted from  y  and multiplied by  2  

 (c)  9  subtracted from  9  

 (d) Three of  y  minus  9 $ 

Ans: Given: $ 2(y-9) $ 

We need to find the statement of the given expression.

We can write the expression as $ 9 $ subtracted from $ y $ and multiplied by $ 2 $.

7. The factors of $\mathbf{ 7{{a}^{2}}+14a} $ is

 (a) $ \mathbf{7(a+2) }$ 

 (b) $\mathbf{ 21a } $ 

 (c) $ \mathbf{7(a+1) }$ 

 (d) $\mathbf{ 7a(a+2) }$ 

Ans: Given: $ 7{{a}^{2}}+14a $ 

We need to factorize the given expression

The factors of $ 7{{a}^{2}}+14a $ will be

 $ 7{{a}^{2}}+14a $ 

Take $ 7a $ as a common factor, we get

 $ 7a(a+2) $ 

Therefore, these will be the factors of the given expression.

Short Answer Questions (2 Mark)

8. The volume of a rectangular box where length, breadth, and height are $ \mathbf{2a,4b,8c} $ respectively.

Ans: Given: length of rectangular box, $ l=2a $ 

Breadth of rectangular box, $ b=4b $ 

Height of rectangular box, $ h=8c $ 

We need to find the volume of the rectangular box with given dimensions.

We know, Volume of a cuboid $ =l\times b\times h $ 

Therefore, the volume of the rectangular box will be

$ \begin{align} & =2a\times 4b\times 8c \\ & =64abc \\ \end{align} $

9. Simplify $\mathbf{ (p+{{q}^{2}})({{p}^{2}}-q) } $ 

Ans: Given: $ (p+{{q}^{2}})({{p}^{2}}-q) $ 

We need to simplify the given expression.

To simplify, we will open the brackets by multiplying the terms in it with each other.

Therefore, the expression will become

\[\begin{align} & \left( p+{{q}^{2}} \right)\left( {{p}^{2}}-q \right) \\ & =p\left( {{p}^{2}}-q \right)+{{q}^{2}}\left( {{p}^{2}}-q \right) \\ & ={{p}^{3}}-pq+{{q}^{2}}{{p}^{2}}-{{q}^{3}} \\ \end{align}\]

10. If $ \mathbf{pq=3 } $ and $\mathbf{ p+q=6, } $ then $\mathbf{ ({{p}^{2}}+{{q}^{2}})} $ is

Ans: Given: $ pq=3 $ , $ p+q=6, $ 

We need to find $ ({{p}^{2}}+{{q}^{2}}) $ 

We know that, 

$ \begin{align} & {{(p+q)}^{2}}={{p}^{2}}+{{q}^{2}}+2pq \\ & \left( {{p}^{2}}+{{q}^{2}} \right)={{(p+q)}^{2}}-2pq \\ \end{align} $

Substituting the values, $ pq=3 $ , $ p+q=6, $ in above equation we get

$ \begin{align} & \left( {{p}^{2}}+{{q}^{2}} \right) \\ & ={{(6)}^{2}}-2(3) \\ & =36-6 \\ & =30 \\ \end{align} $

11. Simplify $\mathbf{ x(2x-1)+5 }$ and find its value at $ \mathbf{x=-3} $ 

Ans: Given: $ x(2x-1)+5 $ 

We need to find the value of the given expression at $ x=-3 $ 

We will substitute $ x=-3 $ in the given expression. Therefore, the expression after simplifying will be

$ \begin{align} & 2{{(-3)}^{2}}-(-3)+5 \\ & =2(9)+3+5 \\ & =18+8 \\ & =26 \\ \end{align} $

Short Answer Questions (3 Mark)

12. Find the value of $ \mathbf{{{\left( \frac{2}{3}x-\frac{3}{2}y \right)}^{2}}+2xy }$ 

Ans: Given: $ {{\left( \frac{2}{3}x-\frac{3}{2}y \right)}^{2}}+2xy $ 

We need to find the value of the given expression

We know that, $ {{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ 

Therefore, using the formula, we will get the value of the expression as:

\[\begin{align} & ={{\left( \frac{2}{3} \right)}^{2}}{{x}^{2}}+{{\left( \frac{3}{2} \right)}^{2}}{{y}^{2}}-2\left( \frac{2}{3} \right)x\left( \frac{3}{2} \right)y+2xy \\ & =\frac{4}{9}{{x}^{2}}+\frac{9}{4}{{y}^{2}}-2xy+2xy \\ & =\frac{4}{9}{{x}^{2}}+\frac{9}{4}{{y}^{2}} \\ \end{align}\]

13. Find $\mathbf{ (m+n)(m-n)\left( {{m}^{2}}+{{n}^{2}} \right)\left( {{m}^{4}}+{{n}^{4}} \right)} $ 

Ans: Given: $ (m+n)(m-n)\left( {{m}^{2}}+{{n}^{2}} \right)\left( {{m}^{4}}+{{n}^{4}} \right) $ 

We need to simplify the given expression. 

Formula Used: $ (a+b)(a-b)={{a}^{2}}-{{b}^{2}} $ 

Simplifying the expression, we get

$ \begin{align} & (m+n)(m-n)\left( {{m}^{2}}+{{n}^{2}} \right)\left( {{m}^{4}}+{{n}^{4}} \right) \\ & =\left( {{m}^{2}}-{{n}^{2}} \right)\left( {{m}^{2}}+{{n}^{2}} \right)\left( {{m}^{4}}+{{n}^{4}} \right) \\ & =\left[ {{\left( {{m}^{2}} \right)}^{2}}-{{\left( {{n}^{2}} \right)}^{2}} \right]\left( {{m}^{4}}+{{n}^{4}} \right) \\ & =\left[ {{m}^{4}}-{{n}^{4}} \right]\left[ {{m}^{4}}+{{n}^{4}} \right] \\ & ={{m}^{8}}-{{n}^{8}} \\ \end{align} $

14. From the sum of $\mathbf{ 3a-b+9} $ and $\mathbf{ -b-9} $ , subtract $\mathbf{ 3a-b-9} $ 

Ans: Given: expressions $ 3a-b+9 $ , $ -b-9 $ , $ 3a-b-9 $ 

We need to subtract $ 3a-b-9 $ from the sum of $ 3a-b+9 $ and $ -b-9 $ 

The sum of the first two terms, $ -b-9 $ and $ 3a-b+9 $ will be

$ \begin{align} & 3a-b+9+(-b-9) \\ & =3a-b+9-b-9 \\ & =3a-2b \\ \end{align} $

Now subtracting $ 3a-b+9 $ from $ 3a-2b $ , we get

$ \begin{align} & 3a-2b-(3a-b-9) \\ & =3a-2b-3a+b=9 \\ & =-b+9 \\ \end{align} $

15. Find $ \mathbf{194\times 206 }$ using suitable identity

Ans: Given: $ 194\times 206 $ 

We need to find the value of the given expression using an identity.

We can write,

$ \begin{align} & 194=(200-6) \\ & 206=(200+6) \\ \end{align} $

Using identity, $ (a-b)(a+b)={{a}^{2}}-{{b}^{2}} $ , the given expression can be simplified as:

$ \begin{align} & (200-6)(200+6) \\ & ={{(200)}^{2}}-{{(6)}^{2}} \\ & =40000-36 \\ & =39964 \\ \end{align} $

16. If $\mathbf{ x+\frac{1}{x}=9, }$ find the value of $\mathbf{ \left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right). }$ 

Ans: Given: $ x+\frac{1}{x}=9 $ 

To find: $ \left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right) $ 

Let $ x+\frac{1}{x}=9\text{ }..............(1) $ 

Square both sides, we will get

$ \begin{align} & \left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)={{(9)}^{2}} \\ & \left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=81 \\ \end{align} $

Now, we know that $ {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab $ 

So,

$ \begin{align} & {{\left( x+\frac{1}{x} \right)}^{2}}={{x}^{2}}+{{\left( \frac{1}{x} \right)}^{2}}+2x\times \frac{1}{x} \\ & {{\left( x+\frac{1}{x} \right)}^{2}}={{x}^{2}}+\frac{1}{{{x}^{2}}}+2 \\ & 81={{x}^{2}}+\frac{1}{{{x}^{2}}}+2 \\ & {{x}^{2}}+\frac{1}{{{x}^{2}}}=81-2 \\ & {{x}^{2}}+\frac{1}{{{x}^{2}}}=79 \\ \end{align} $

17. Simplify $\mathbf{ \left( {{x}^{2}}-3x+2 \right)(5x-2)-\left( 3{{x}^{2}}+4x-5 \right)(2x-1) }$ 

Ans: Given: $ \left( {{x}^{2}}-3x+2 \right)(5x-2)-\left( 3{{x}^{2}}+4x-5 \right)(2x-1) $ 

We need to simplify the given expression.

First simplifying, $ \left( {{x}^{2}}-3x+2 \right)(5x-2), $ we will get

$ \begin{align} & \left( {{x}^{2}}-3x+2 \right)(5x-2) \\ & =5{{x}^{3}}-15{{x}^{2}}+10x-2{{x}^{2}}+6x-4 \\ & =5{{x}^{3}}-17{{x}^{2}}+16x-4\text{ }...................(1) \\ \end{align} $

Now simplifying, $ \left( 3{{x}^{2}}+4x-5 \right)(2x-1) $ , we will get

$ \begin{align} & \left( 3{{x}^{2}}+4x-5 \right)(2x-1) \\ & =6{{x}^{3}}+8{{x}^{2}}-10x-3{{x}^{2}}-4x+5 \\ & =6{{x}^{3}}+5{{x}^{2}}-14x+5\text{ }..................(2) \\ \end{align} $

Subtract $ (1)-(2) $ to get the result

$ \begin{align} & \left( {{x}^{2}}-3x+2 \right)(5x-2)-\left( 3{{x}^{2}}+4x-5 \right)(2x-1) \\ & =5{{x}^{3}}-17{{x}^{2}}+16x-4-\left[ 6{{x}^{3}}+5{{x}^{2}}-14x+5 \right] \\ & =5{{x}^{3}}-17{{x}^{2}}+16x-4-6{{x}^{3}}-5{{x}^{2}}+14x-5 \\ & =-{{x}^{3}}-22{{x}^{2}}+30x-9 \\ \end{align} $

18. Find the following product $\mathbf{ (a-3b)(a+3b)\left( {{a}^{2}}+9{{b}^{2}} \right) }$ 

Ans: Given: $ (a-3b)(a+3b)\left( {{a}^{2}}+9{{b}^{2}} \right) $ 

We need to find the product of the given expression.

Formula Used: $ (a-b)(a+b)={{a}^{2}}-{{b}^{2}} $ 

Solve the given expression using the identity, we will get

$ \begin{align} & (a-3b)(a+3b)\left( {{a}^{2}}+9{{b}^{2}} \right) \\ & =\left[ {{(a)}^{2}}-{{(3b)}^{2}} \right]\left( {{a}^{2}}+9{{b}^{2}} \right) \\ & =\left( {{a}^{2}}-9{{b}^{2}} \right)\left( {{a}^{2}}+9{{b}^{2}} \right) \\ & ={{\left( {{a}^{2}} \right)}^{2}}-{{\left( 9{{b}^{2}} \right)}^{2}} \\ & ={{a}^{4}}-81{{b}^{4}} \\ \end{align} $

Long Answer Questions (5 Mark)

19. Find $\mathbf{ {{(3st+4t)}^{2}}-{{(3st-4t)}^{2}} }$ 

Ans: Given: $ {{(3st+4t)}^{2}}-{{(3st-4t)}^{2}} $ 

We need to simplify the given expression.

We know that,

$ \begin{align} & {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\ & \therefore {{(3st+4t)}^{2}}={{(3st)}^{2}}+{{(4t)}^{2}}+2(3st)(4t) \\ & =9{{s}^{2}}{{t}^{2}}+16{{t}^{2}}+24s{{t}^{2}}.................(1) \\ \end{align} $

We also know that,

$ \begin{align} & {{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\ & \therefore {{(3st-4t)}^{2}}={{(3st)}^{2}}+{{(4t)}^{2}}-2(3st)(4t) \\ & {{(3st-4t)}^{2}}=9{{s}^{2}}{{t}^{2}}+16{{t}^{2}}-24s{{t}^{2}}............(2) \\ \end{align} $

Therefore, using equations $ (1) $ and $ (2) $ , the given expression will become

$ \begin{align} & =9{{s}^{2}}{{t}^{2}}+16{{t}^{2}}+24s{{t}^{2}}-\left[ 9{{s}^{2}}{{t}^{2}}+16{{t}^{2}}-24s{{t}^{2}} \right] \\ & =9{{s}^{2}}{{t}^{2}}+16{{t}^{2}}+24s{{t}^{2}}-9{{s}^{2}}{{t}^{2}}-16{{t}^{2}}+24s{{t}^{2}} \\ & =48s{{t}^{2}} \\ \end{align} $

20. The area of a rectangle is uv where u is length and v is breadth. If the length of rectangle is increased by \[\mathbf{5\text{ units}}\] and breadth is decreased by \[\mathbf{3\text{ units}\text{.}}\]. The new area of rectangle is?

Ans: Given: Area of the old rectangle $ =uv $ 

Length of the old rectangle $ =u $ 

Breadth of told rectangle $ =v $ 

Length of the ew rectangle $ =u+5 $ 

Breadth of the new rectangle $ =v-3 $ 

We need to find the area of the new rectangle.

We know that, Area of rectangle $ =length\times breadth $ 

Therefore, the area of the new rectangle will be 

$ \begin{align} & =(u+5)(v-3) \\ & =uv-3u+5v-15 \\ \end{align} $

21. If \[\mathbf{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=27}\], find $\mathbf{ x-\frac{1}{x} }$ 

Ans: Given: \[\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=27\]

We need to find $ x-\frac{1}{x} $
Use identity, $ {{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ 

Substitute, $ a=x,b=\frac{1}{x} $ , we get

$ \begin{align} & {{\left( x-\frac{1}{x} \right)}^{2}}={{x}^{2}}+\frac{1}{{{x}^{2}}}-2x\times \frac{1}{x} \\ & \Rightarrow {{\left( x-\frac{1}{x} \right)}^{2}}={{x}^{2}}+\frac{1}{{{x}^{2}}}-2 \\ \end{align} $

We know that, \[\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=27\], therefore,

$ \begin{align} & \Rightarrow {{\left( x-\frac{1}{x} \right)}^{2}}=27-2 \\ & \Rightarrow {{\left( x-\frac{1}{x} \right)}^{2}}=25 \\ & \Rightarrow \left( x-\frac{1}{x} \right)=\sqrt{25} \\ & \Rightarrow \left( x-\frac{1}{x} \right)=5 \\ \end{align} $

22. Simplify using identity, $\mathbf{ \frac{2.3\times 2.3-0.3\times 0.3}{2.3\times 2.3-2\times 2.3\times 0.3+0.3\times 0.3} }$ 

Ans: Given: 0 $ \frac{2.3\times 2.3-0.3\times 0.3}{2.3\times 2.3-2\times 2.3\times 0.3+0.3\times 0.3} $ 

We need to simplify the given expression using an identity.

Identity used: 

$ \begin{align} & 1.{{a}^{2}}-{{b}^{2}}=(a+b)(a-b) \\ & 2.{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\ \end{align} $

Using these identities, the given expression can be written as:

$ \begin{align} & =\frac{2.3\times 2.3-0.3\times 0.3}{2.3\times 2.3-2\times 2.3\times 0.3+0.3\times 0.3} \\ & =\frac{{{(2.3)}^{2}}-{{(0.3)}^{2}}}{{{(2.3)}^{2}}-2(2.3)(0.3)+{{(0.3)}^{2}}} \\ & =\frac{(2.3+0.3)(2.3-0.3)}{{{(2.3-0.3)}^{2}}} \\ & =\frac{(2.3+0.3)}{(2.3-0.3)} \\ & =\frac{2.6}{2} \\ & =1.3 \\ \end{align} $

23. The sum of \[\mathbf{\left( x\text{ }+\text{ }3 \right)}\] observations is $\mathbf{ ({{x}^{4}}-81).} $ Find the mean of the observations.

Ans: Given: 

Number of observations $ =\left( x\text{ }+\text{ }3 \right) $ 

Sum of observations $ =({{x}^{4}}-81) $ 

We know that $ \text{Mean=}\frac{\text{Sum of all observations}}{\text{Total number of observations}} $ 

Therefore, mean of the observations will be

$ \begin{align} & =\frac{\left( {{x}^{4}}-81 \right)}{(x+3)} \\ & =\frac{{{x}^{4}}-{{3}^{4}}}{x+3} \\ & =\frac{{{\left( {{x}^{2}} \right)}^{2}}-{{\left( {{3}^{2}} \right)}^{2}}}{(x+3)} \\ \end{align} $

We know that, $ {{a}^{2}}-{{b}^{2}}=(a+b)(a-b) $ 

$ \begin{align} & =\frac{\left( {{x}^{2}}+{{3}^{2}} \right)\left( {{x}^{2}}-{{3}^{2}} \right)}{(x+3)} \\ & =\frac{\left( {{x}^{2}}+{{3}^{2}} \right)(x-3)(x+3)}{(x+3)} \\ & =\left( {{x}^{2}}+9 \right)(x-3) \\ & ={{x}^{3}}+9x-3{{x}^{2}}-27 \\ & ={{x}^{3}}-3{{x}^{2}}+9x-27 \\ \end{align} $

24. Abhishek is \[\mathbf{6}\] years older than Jagadish. Six years ago Abhishek was twice as old as Jagadish. Write down the situation in algebraic expression and find how old is each now?

Ans: Given: Age of Abhishek \[=6\] years older than Jagadish

Six years ago Abhishek age $ =2 $ times of Jagadish age

We need to write the algebraic expression and find the age of both.

Let, present age of Jagadish $ =x $ 

Abhishek is \[6\] years older than Jagadish. 

Therefore, age of Abhishek $ =x+6 $ 

Jagadish age \[6\] years ago $ =x-6 $

Age of Abhishek \[6\] years ago $ =(x+6)-6 $ 

According to the question, Six years ago Abhishek was twice as old as Jagadish. So,

$ \begin{align} & (x+6)-6=2(x-6) \\ & \Rightarrow x+6-6=2x-12 \\ & \Rightarrow -x=-12 \\ & \Rightarrow x=12 \\ \end{align} $

We can conclude that present age of Jagadish $ =12\text{ years} $ 

The present age of Abhishek will be

$ \begin{align} & =x+6 \\ & =18\text{ years} \\ \end{align} $

25. Rohit mother gave him $ \mathbf{\text{Rs}\text{. }4x{{y}^{2}} }$ and his father gave him Rs. $\mathbf{ (5x{{y}^{2}}+3) }$ . Out of this total money he spent Rs. $\mathbf{ (10-3x{{y}^{2}}) }$ on his birthday party. How much money is left with him?

Ans: Given: Mother gives Rohit amount $ =4x{{y}^{2}} $ 

Father gives Rohit amount $ =(5x{{y}^{2}}+3) $ 

Total amount spent by Rohit $ =(10-3x{{y}^{2}}) $ 

We need to find how much money is left with him.

Total Amount Received by Rohit is,

$ \begin{align} & =4x{{y}^{2}}+(5x{{y}^{2}}+3) \\ & =9x{{y}^{2}}+3 \\ \end{align} $

Total amount spent by Rohit is

 $ =(10-3x{{y}^{2}}) $ 

Therefore, Amount left with him will be

$ \begin{align} & =\left( 9x{{y}^{2}}+3 \right)-\left( 10-3x{{y}^{2}} \right) \\ & =9x{{y}^{2}}+3-10+3x{{y}^{2}} \\ & =12x{{y}^{2}}-7 \\ \end{align} $

How the Chapter 8 Maths Class 8 Important Questions Are Going to Help Students 

The chapter 8 class 8 important questions are going to help the students immensely because-

• These solutions provide step-by-step explanations helping the students to answer questions better.

• These solutions contain a list of important formulas that will help the students remember them during exam time.

• The questions and solutions are written based on the current syllabus.

• These questions and solutions will further help students during competitive exam revisions.

• Questions are compiled by experts who have in-depth knowledge about the subject.

7 Important Formulas from Class 8 Maths Chapter 8 Algebraic Expressions and Identities

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Conclusion 

CBSE Class 8 Maths Chapter 8 Algebraic Expressions and Identities Important Questions are essential for learning this crucial topic. They help students understand key concepts like terms, factors, coefficients, and operations on algebraic expressions. Practising these questions ensures a solid grasp of standard identities, which are vital for solving problems efficiently. These questions not only prepare students for exams but also strengthen their foundational skills in algebra. By solving them regularly, students can boost their confidence, improve accuracy, and score well in their exams with ease.

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