Class 8 Maths Chapter 8 Important Question Answers - Algebraic Expressions and Identities

Q1: Classify the following polynomials as monomials, binomials, and trinomials. Also, state the  polynomials do not fall in any of these three categories?
x + y, 1000, x + x² + x³, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd + da, pqr, p²q + pq², 2p + 2q, 
Sol: The classified terms are given below,
Monomials- 1000, pqr
Binomials- x + y, 2y – 3y², 4z – 15z², p²q + pq², 2p + 2q
Trinomials- x + x²+ x³, 7 + y + 5x, 2y – 3y² + 4y³, 5x – 4y + 3xy
Polynomials that do not fall in any of these categories are  x + y, x + x²+ x³,  ab + bc + cd + da

Q2: Subtract the following.
(i) 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(ii 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(iii) 4p²q – 3pq + 5pq²– 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2+ 5p²q
Sol:
(i) ( 12a – 9ab + 5b – 3 ) – ( 4a – 7ab + 3b + 12 )
= 12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12
= 12a – 4a – 9ab + 7ab + 5b –3b – 3 – 12
= 8a – 2ab + 2b – 15
(ii) ( 5xy – 2yz – 2zx + 10xyz ) – ( 3xy + 5yz – 7zx )                        
= 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx                         
= 5xy – 3xy – 2yz – 5yz – 2zx + 7zx + 10xyz                        
= 2xy – 7yz + 5zx + 10xyz
(iii) ( 18 – 3p – 11q + 5pq – 2pq2+ 5p²q ) – ( 4p²q – 3pq + 5pq²– 8p + 7q – 10 )                         
= 18 – 3p – 11q + 5pq – 2pq2 + 5p²q – 4p²q + 3pq – 5pq² + 8p – 7q + 10                         
= 18 + 10 – 3p + 8p – 11q – 7q + 5pq + 3pq – 2pq2 – 5pq² + 5p²q – 4p²q                           
= 28 + 5p – 18q + 8pq – 7pq² + p²q

Q3: Which term is the like term similar to 24a²bc?
(a) 13 × 8a × 2b × c × a 
(b) 8 × 3 × a × b × c
(c) 3 × 8 × a × b × c × c 
(d) 3 × 8 × a × b × b × c
Ans: (a)
Sol: To find out the similar term as 24a²bc, let us find the product of each of the equations,

1. 13 × 8a × 2b × c × a =  208a²bc
2. 8 × 3 × a × b × c = 24abc
3. 3 × 8 × a × b × c × c = 24abc²
4. 3 × 8 × a × b × b × c = 24ab²c

Hence, we can get that option (a) is correct.

Q4. Fill in the blanks.
(a) (x + a) (x + b) = x² + (a + b)x + ________.
(b) The product of two terms with like signs is a  ________ term.
(c) The product of two terms with unlike signs is a  ________ term.
(d) (a – b) _________ = a² – 2ab + b²
(e)  a² – b² = (a + b ) __________.
(f) (a – b)² + ____________ = a² – b²
(g) (a + b)² – 2ab = ___________ + ____________
(h) The product of two polynomials is a ________
(i) The coefficient in – 37abc is __________. 
(j) Number of terms in the expression a2 + bc × d is ________
Ans: 
(a) ab
As per the standard identity 4, (x + a) (x + b) =  x² + (a + b)x + ab
(b) Positive
(c) Negative
(d) ( a – b)²
As per standard identity 2, (a – b)² = a² – 2ab + b²
(e) (a – b)
As per standard identity 3, (a + b ) ( a – b ) = a² – b²
(f) 2ab – 2b²
Let us solve the equation with x in the blank space. As per identity 2, (a – b)² = a² – 2ab + b². 
Hence, a² – 2ab + b² + x = a² – b²
x = a² – b² – a² + 2ab – b²
x = 2ab – 2b²  
(g) a² + b²
Using Identity 1 ( a + b )² = a² + 2ab + b²,
(a + b)² – 2ab = a² + 2ab + b² – 2ab = a² + b²
(h) Polynomial
(i) -37
(j) 2

Q5: The length of a rectangular box is  ( x + 9y) and the area is x² + 12xy + 27y². Find the breadth.
Sol.  Area of a rectangle =  length x breadth, hence breadth = area / length.
breadth = x² + 12xy + 27y²
( x + 9y )
=  x² + 9xy + 3xy + 27y²
( x + 9y )
=  x ( x + 9y ) + 3y (x + 9y)
( x + 9y )
breadth   =  x + 3y

Q6. The exponents of the variables in a polynomial are always
(a) Integers 
(b) Positive integers 
(c) Non-negative integers 
(d) Non-positive integers

Ans: (c)
A polynomial will have a non-zero coefficient and variables having non-negative integers as exponents. For example : a + b + r + q, 3ab, 5xyz – 10, 2a + 3b + 7z, etc.

Q7. Using identities, find products for the below.
(a) 71²
(b) 99²
(c) 102²
(d) 998²
(e) 5.2²
(f) 297 × 303
(g) 78 × 82 
(h) 8.9²
(i) 10.5 × 9.5
Sol: 

(a) 71² =  ( 70 + 1 )²                                      Identity applied ( a + b )² = a² + 2ab + b²
= 70² + 2 ( 70 x 1 ) + 1²
= 4900 + 140 + 1
= 5041

(b) 99² = ( 100 – 1 )²                                       Identity applied ( a – b )² = a² – 2ab + b² 
= 100² – 2 ( 100 x 1 ) + 1² 
= 10000 – 200 + 1
= 9801
(c) 102² = ( 100 + 2 )²                                      Identity applied ( a + b )² = a² + 2ab + b²
= 100² + 2 ( 100 x 2 ) + 2²
= 10000 + 400 + 4
= 10404      
(d) 998² = ( 1000 – 2 )²                                      Identity applied ( a – b )² = a² – 2ab + b² 
= 1000² – 2 ( 1000 x 2 ) + 2²
= 1000000 – 4000 + 4
= 996004
(e) 5.2² = ( 5 + 0.2 )²                                           Identity applied ( a + b )² = a² + 2ab + b²
= 5² + 2 ( 5 x 0.2 ) + 0.2²
= 25 + 2 +  0.04
= 27.04
(f) 297 × 303 = ( 300 – 3 ) ( 300 + 3 )                 Identity applied  ( a + b ) ( a – b ) = a² – b²
= 300² – 3²
= 90000 – 9
= 89991
(g) 78 × 82 = ( 80 – 2 ) ( 80 + 2 )                         Identity applied ( a + b ) ( a – b ) = a² – b²
=  80² – 2²
= 6400 – 4
= 6396     
(h) 8.9² = ( 9.0 – 0.1 )²                                           Identity applied ( a – b )² = a² – 2ab + b²
= 9.0² – 2 ( 9.0 x 0.1 ) + 0.1²
= 81 – 1.8 + 0.01
=  79.21
(i) 10.5 x 9.5 = ( 10 + 0.5 ) ( 10 – 0.5 )                 Identity applied ( a + b ) ( a – b ) = a² – b²
=  10² – 0.5²
= 100 – 0.25
= 99.75

Q8: Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(a) 5a, 3a², 7a4
(b) 2p, 4q, 8r
(c) xy, 2x²y, 2xy²
(d) a, 2b, 3c
Sol: The volume of a rectangular box is the product of its length, breadth and height, i.e Volume = length x breadth x height.
Volumes are calculated as below,
(a) length = 5a, breadth = 3a², height = 7a4
Volume = 5a x 3a² x 7a4
= 105a7
(b) length = 2p, breadth = 4q, height = 8r
Volume = 2p x 4q x 8
= 64pqr
(c) length = xy, breadth = 2x²y, height = 2xy²
Volume = xy x 2x²y x 2xy²
= 4x4y4
(d) length = a, breadth = 2b, height = 3c
Volume = a x 2b x 3c
= 6abc

Q9: Show that LHS = RHS for the below equations.
(a) ( 3x + 7 )² – 84x = ( 3x – 7 )²
(b) ( 9p – 5q )² + 180pq = ( 9p + 5q )²
(c) ( 4pq + 3q )² – ( 4pq – 3q )² = 48pq²
(d) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Answer 18. 
(a) LHS =  ( 3x + 7 )² – 84x 
= (3x)² + 2 ( 3x x 7 ) + 7² – 84x
=  9x² + 42x + 49 – 84x
= 9x² – 42x + 49
RHS = ( 3x – 7 )² 
= (3x)² – 2 ( 3x x 7 ) + 7² 
= 9x² – 42x + 49
Hence LHS = RHS
(b) LHS = ( 9p – 5q )² + 180pq 
= (9p)² – 2 ( 9p x 5q ) + (5q)² + 180pq
= 81p² + 90pq + 25q²
RHS = ( 9p + 5q )²
= (9p)² + 2 ( 9p x 5q ) + (5q)²
= 81p² + 90pq + 25q²
Hence LHS = RHS

(c) LHS = ( 4pq + 3q )² – ( 4pq – 3q )²
= (4pq)² + 2 ( 4pq x 3q ) + (3q)² – ( (4pq)² – 2 ( 4pq x 3q ) + (3q)²)
= 16p²q² + 24pq² + 9q² – 16p²q² + 24pq² – 9q²
= 48pq²
RHS = 48pq²
Hence LHS = RHS
(d) LHS =  (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= a² + ab – ba – b² + b² + bc – cb – c² + c² + ca – ac – a²
= 0
RHS = 0
Hence LHS = RHS

Q10: Show that LHS = RHS for the below equations.
(a) ( 3x + 7 )² – 84x = ( 3x – 7 )²
(b) ( 9p – 5q )² + 180pq = ( 9p + 5q )²
(c) ( 4pq + 3q )² – ( 4pq – 3q )² = 48pq²
(d) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Sol. 
(a) LHS =  ( 3x + 7 )² – 84x 
= (3x)² + 2 ( 3x x 7 ) + 7² – 84x
=  9x² + 42x + 49 – 84x
= 9x² – 42x + 49
RHS = ( 3x – 7 )² 
= (3x)² – 2 ( 3x x 7 ) + 7² 
= 9x² – 42x + 49
Hence LHS = RHS
(b) LHS = ( 9p – 5q )² + 180pq 
= (9p)² – 2 ( 9p x 5q ) + (5q)² + 180pq
= 81p² + 90pq + 25q²
RHS = ( 9p + 5q )²
= (9p)² + 2 ( 9p x 5q ) + (5q)²
= 81p² + 90pq + 25q²
Hence LHS = RHS
(c) LHS = ( 4pq + 3q )² – ( 4pq – 3q )²
= (4pq)² + 2 ( 4pq x 3q ) + (3q)² – ( (4pq)² – 2 ( 4pq x 3q ) + (3q)²)
= 16p²q² + 24pq² + 9q² – 16p²q² + 24pq² – 9q²
= 48pq²
RHS = 48pq²
Hence LHS = RHS

(d) LHS =  (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= a² + ab – ba – b² + b² + bc – cb – c² + c² + ca – ac – a²
= 0
RHS = 0
Hence LHS = RHS

Q11. Select the correct option of volume of a rectangular box  with length = 2ab, breadth = 3ac and height = 2ac 
(a) 12a³bc² 
(b) 12a³bc
(c) 12a²bc
(d) 2ab +3ac + 2ac
Ans: (a)
The formula for calculating the volume of a rectangular box is 
Volume = length x breadth x height 
With the length of the input = 2ab, breadth = 3ac and height = 2ac 
Volume = 2ab x 3ac x 2ac
= 12a³bc²

Q12: The product of a binomial and monomial is a 
(a) Monomial 
(b) Binomial 
(c) Trinomial 
(d) None of these
Ans: (b)
Explanation: This can be demonstrated through an example  below,
x ( y + z ) = xy + xz 
This expression contains two terms, x is a monomial and ( y + z ) is a binomial.
The product of multiplying these terms results in a binomial product xy + xz.

Q13. Which of the following is an identity? 
(a) (p + q)²  = p² + q² 
(b) p² – q² = (p – q)² 
(c) p² – q² = p² + 2pq – q² 
(d) (p + q)² = p² + 2pq + q²
Ans: (d) 
The equation  (p + q)² = p² + 2pq + q² follows the first standard algebraic identity
( a + b )² = a² + 2ab + b². The rest of the other options do not follow any of the standard identities. Hence option (d) is correct.

Q14: Add the following.
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²
(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl
Sol: 
(i) (ab – bc) + (bc – ca) + (ca – ab)         
= ab – bc + bc – ca + ca – ab
= ab-ab-bc+bc-ca+ca
= 0
(ii) (a – b + ab) + (b – c + bc) + (c – a + ac)
= a – b + ab + b – c + bc + c -a +ac
= a – a -b + b + ab – c + c + bc + ac
= ab + bc + ac
(iii) ( 2p²q² – 3pq + 4) + ( 5 + 7pq – 3p²q²)
= 2p²q² – 3pq + 4 + 5 + 7pq – 3p²q²
= 2p²q² – 3p²q² + 7pq – 3pq + 4 + 5
= -1p²q² + 4pq + 9
= 4pq + 9 – p²q²
(iv) ( l² + m²) + (m² + n²) + (n² + l²) + (2lm + 2mn + 2nl)
= l² + m² + m² + n² + n² + l² + 2lm + 2mn + 2nl
= l² +  l² + m² + m² + n² + n² + 2lm + 2mn + 2nl
= 2l² + 2m² + 2n² + 2lm + 2mn + 2nl
= 2( l² + m² + n² + lm + mn + nl)   

Q15: Solve the below using correct identities.
(a) (48)²
(b) 181² – 19²
(c) 497 × 505
(d) 2.07 × 1.93
Sol:
(a) (48)²
= (50 – 2)²
As (a – b)² = a² – 2ab + b² , hence
(50 – 2)² = (50)² – 2 × 50 × 2 + (2)²
= 2500 – 200 + 4
= 2300 + 4
= 2304
(b) As a² – b² = (a – b) (a + b)
181² – 19² = (181 – 19) (181 + 19)
= 162 × 200
= 32400

(c) By using the identity (x + a) (x + b) = x2 + (a + b) x + ab
497 x 505 = ( 500 – 3 ) (500 + 5 )
= 500² + (–3 + 5) × 500 + (–3) (5) 
= 250000 + 1000 – 15 
= 250985
(d) As (a + b) (a – b) = a² – b² 
2.07 × 1.93 = (2 + 0.07) (2 – 0.07)
= 2² – 0.07²
= 3.9951