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

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} $

Why is it Necessary for Students to Solve Important Questions of Algebraic Expressions Class 8?

Students must solve the important questions of algebraic expressions class 8. This will help them to keep up with what is being taught in class and what might appear in the exams. These questions are carefully curated based on the current syllabus, and curriculum, previous years question papers. They contain only the questions which are most likely to appear in the examinations. This allows the students to study in-depth and prepare well in advance.

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. 

Relevance of Chapter 8 Algebraic Expressions

This chapter is crucial for students because this chapter will create the base for many future chapters and problems. In higher studies and mathematics, algebraic expressions are utilised in several situations. Algebraic expressions play a vital role in the mathematics curriculum and mathematics in general. If any student wishes to pursue mathematics in higher studies it will be very important for them. Students will be required to understand, read, write expressions, as well as be skilled in computations and usage of algebraic expressions.  Apart from that, even students who do not pursue maths but have basic mathematics in their syllabus will have to use algebraic expressions in various fields. 

Algebraic expressions will be used to decipher relationships between variables and constants, between people, elements, thoughts, and structures. This chapter will help students see the relation and role of variables and constants in a real-life context and also help them implement these expressions in real life. 

Also, this chapter will help students in all competitive examinations later on in life. So it is important that students properly learn, clear their basics, and understand this chapter from a fairly young age. 

What Students will Learn from Class 8 Maths Chapter 8 Important Questions

• Students will learn how to identify variable and constants in algebraic expressions given to them

• It helps them to develop an idea about algebraic expressions and see their purpose.

• It helps them to construct algebraic expressions from given real-life situations and solve them.

• It helps students to learn about algebraic expression concepts which would help them in later years in various chapters and examinations. 

Introduction to Algebraic Expressions 

What are Algebraic Expressions?

Algebraic Expression refers to an expression that is constructed from integer constants, variables, and algebraic operations like addition, subtraction, multiplication, and division. 

For example, 5m + 8 is an algebraic expression where the two terms are 5m and 8 and m is the variable. The two terms are separated by the arithmetic sign +. 

Examples of Algebraic Expressions 

30x + 4y, 6x + 10y etc are basic examples of algebraic expressions. The expressions are being represented with the help of variables, constants, and arithmetic symbols. The combination of these terms is known as algebraic expressions.

This is different from algebraic equations because algebraic expressions do not have any sides or equal to signs. 

Types of Algebraic Expressions 

There are mainly three types of Algebraic Expressions, namely - 

• Monomial Expression

• Binomial Expression

• Polynomial Expression 

Monomial Expression - Monomial Algebraic Expression refers to the expression which has only one term in it. 

For example, 3x, 5x, 5xy, etc, are monomial expressions.

Binomial Expression - Binomial Algebraic Expression refers to the expression which has two terms in it. 

For example, 5x+y, 10z- 5x, 20x+ 10, etc, are binomial expressions.

Polynomial Expression - Polynomial Expression refers to the expression which has more than one term with non-negative integral exponents of a variable in it. 

For example, ax - by - cz, x + 2x - 3y, etc are polynomial expressions. 

Other Types of Algebraic Expressions 

Apart from the above mentioned three types of algebraic expressions, there are two more classifications, namely -

• Numeric Expression 

• Variable Expression

Numeric Expression - A Numeric Expression is an expression that consists of numbers and operations but does not contain any variables. 

For example, 10 + 2, 11 - 5, etc are examples of numeric expressions.

Variable Expression - A variable Expression is an expression that consists of numbers as well as variables.

For example, 4x + 4y, 5xy - 5y, etc are variable expressions.

Some Important Terms of Algebraic Expressions

Some important terms that students will come across while studying the chapter Algebraic Expressions are -

• Coefficient of the Term

• Variables

• Constant 

• Factors of a Term

• Terms of Equations

• Like and Unlike terms

For example, in the expression 2xy + 14y, 2xy and 14y are the two terms.

Coefficient of the term xy is 2 

Constant term is 2. 

Examples of Like Terms are 3x and 2x 

Examples of Unlike Terms are 2x and 5y.

Suppose, 4xy is a term, then 4,x, and y are its factors.

Addition and Subtraction of Algebraic Expressions 

We can also add or subtract algebraic expressions easily. 

For example, if we are asked to add 3x + 10y - 5z and x + 2y + 2z,

We add the expressions and get 

(3x + 10y - 5z) + ( x + 2y + 2z) and get 4x + 12y - 3z. 

Important Formulas 

Some of the important algebraic formulas that we need to keep in mind are - 

• (a + b)2 = a2 + 2ab + b2

• (a – b)2 = a2 – 2ab + b2

• a2 – b2 = (a – b)(a + b)

• (a + b)3 = a3 + b3 + 3ab(a + b)

• (a – b)3 = a3 – b3 – 3ab(a – b) 

• a3 – b3 = (a – b)(a2 + ab + b2)

• a3 + b3 = (a + b)(a2 – ab + b2)

How to Derive Algebraic Expressions

An algebraic expression is a combination of constants and variable and arithmetic symbols like + or - or * or /. We can use these combinations to derive algebraic expressions from given problems. 

Suppose it is given that Ram has thrice the number of apples as Rita does. The total number of apples between Ram and Rit is 40. Express this problem in an appropriate algebraic expression. 

We will write this as - 3x + x = 40 where x is the number of apples Rita has. 

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The notes, questions, and solutions are prepared by trained experts who have ample knowledge in those particular chapters. They come up with necessary notes, important questions that might appear in the examinations after careful and close analysis of the syllabus and guidelines as provided by the board.

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Conclusion 

To conclude, we would advise our beloved students to study hard, go through the given study and revision notes thoroughly, write down the important formulas to remember them better, and practice the given important questions of algebraic expressions class 8 from our website. This will help the students to get to know the chapter well, be familiar with the various types of questions thus saving valuable time during the exams and fetch good marks in the final examinations.