Class 8 Maths - Algebraic Expressions and Identities CBSE Worksheets Solutions

Fill in the blanks

Q1: Terms with same algebraic factors are called ____________ terms.
Ans: Like

Q2: A ________________ can take any value and ________________ has a fixed value.
Ans: variable , constant 

Q3: An expression with one or more terms is called _____________
Ans: polynomial

Q4: An expression with one term is called __________________ with two terms is ______________ and with three terms is _______________
Ans: monomial ,binomial ,trinomial 

Q5: An algebraic expression with equality sign is called ______________
Ans: equation 

State True or False

Q1: The degree of a constant term is 0
Ans: True

Q2: The sum or difference of two like terms may not like the given terms
Ans: False

Q3: 1 is an algebriac expression
Ans: True

Q4: The expression x + y + 5 x is a trinomial.
Ans: False

Q5: In like terms, the numerical coefficients should also be the same
Ans: False

Answer the following questions

Q1: The volume of a rectangular box where length, breadth, and height are 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×b×h
Therefore, the volume of the rectangular box will be
=2a×4b×8c=64abc

Q2: Simplify (p+q²)(p²−q)
Ans: Given: (p+q²)(p²−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
(p+q²)(p²−q)

=p(p²−q)+q²(p²−q)

=p³−pq+q²p²−q³

Q3: If pq=3 and p+q=6, then (p²+q²) is
Ans: Given: pq=3
, p+q=6,
We need to find (p²+q²)
We know that,
(p+q)²=p²+q²+2pq

(p²+q²)=(p+q)²−2pq
Substituting the values, pq=3
, p+q=6,
in above equation we get
(p²+q²)=(6)²−2(3)=36−6=30

Q4: Simplify x(2x−1)+5 and find its value at 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
2(−3)²−(−3)+5

=2(9)+3+5

=18+8

=26

Q5: Find the value of
Ans: Given:
We need to find the value of the given expression
We know that, (a−b)²=a²+b²−2ab
Therefore, using the formula, we will get the value of the expression as:


Q6: Think of a number x. Multiply it by 3 and add 5 to the product and subtract y subsequently. Find the resulting number.

Ans: Required number is (3x + 5)
Now we have to subtract y from the result i.e., 3x + 5 – y

Q7: From the sum of 3a−b+9 and −b−9, subtract 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
3a−b+9+(−b−9)=3a−b+9−b−9=3a−2b
Now subtracting 3a−b+9
 from 3a−2b
 , we get
3a−2b−(3a−b−9)=3a−2b−3a+b=9=−b+9

Q8: Find 194×206 using suitable identity
Ans: Given: 194×206
We need to find the value of the given expression using an identity.
We can write,194=(200−6)²
Using identity, (a−b)(a+b)=a²−b², the given expression can be simplified as:
(200−6)(200+6)=(200)²−(6)²=40000−36=39964

Q9: If find the value of
Ans: Given:
To find:
Let
Square both sides, we will get

Now, we know that (a+b)²=a²+b²+2ab
So,


Q10: Simplify (x²−3x+2)(5x−2)−(3x2+4x−5)(2x−1)
Ans: Given: (x²−3x+2)(5x−2)−(3x²+4x−5)(2x−1)
We need to simplify the given expression.
First simplifying, (x²−3x+2)(5x−2),
we will get
(x²−3x+2)(5x−2)

=5x3−15x²+10x−2x²+6x−4

=5x3−17x²+16x−4 ...................(1)
Now simplifying, (3x²+4x−5)(2x−1), we will get
(3x²+4x−5)(2x−1)

=6x3+8x²−10x−3x²−4x+5

=6x3+5x²−14x+5 ..................(2)
Subtract (1)−(2) to get the result
(x²−3x+2)(5x−2)−(3x²+4x−5)(2x−1)

=5x3−17x²+16x−4−[6x3+5x²−14x+5]

=5x3−17x²+16x−4−6x3−5x²+14x−5

=−x3−22x²+30x−9

Q11: Find (3st+4t)²−(3st−4t)²
Ans: Given: (3st+4t)²−(3st−4t)²
We need to simplify the given expression.
We know that,
(a+b)²=a²+b²+2ab

∴(3st+4t)²=(3st)²+(4t)2+2(3st)(4t)

=9s²t²+16t²+24st².................(1)
We also know that,
(a−b)²=a²+b²−2ab

∴(3st−4t)²=(3st)²+(4t)²−2(3st)(4t)(3st−4t)²

=9s²t²+16t²−24st²............(2)
Therefore, using equations (1) and (2) , the given expression will become
=9s²t²+16t²+24st²−[9s²t²+16t²−24st²]

=9s²t²+16t²+24st²−9s²t²−16t²+24st²=48st²

Q12: The area of a rectangle is uv where u is length and v is breadth. If the length of rectangle is increased by 5 units and breadth is decreased by 3 units. 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×breadth
Therefore, the area of the new rectangle will be
=(u+5)(v−3)=uv−3u+5v−15

Q13: If =27, find
Ans: Given: =27
We need to find
Use identity, (a−b)²=a²+b²−2ab
Substitute, a=x, b= 1/x, we get

We know that, =27, therefore,


Q14: Simplify using identity,
Ans: Given: 0
We need to simplify the given expression using an identity.
Identity used:
1.a²−b²=(a+b)(a−b)².(a−b)²=a²+b²−2ab
Using these identities, the given expression can be written as:


Q15: The sum of (x + 3) observations is (x⁴−81). Find the mean of the observations.
Ans: Given:
Number of observations =(x + 3)
Sum of observations =(x⁴−81)
We know that Mean=
Therefore, mean of the observations will be

We know that, a²−b²=(a+b)(a−b)