Algebraic Expressions And Identities
Class 8th Mathematics CBSE Solution
- Identify the terms, their coefficients for each of the following expressions.…
- Classify the following polynomials as monomials, binomials, trinomials. Which…
- Add the following. (i) ab-bc , bc-ca , ca-ab (ii) a-b+ab , b-c+bc , c-a+ac (iii)…
- Subtract 4a-7ab+3b+12 from 12a-9ab+5b-3
- Subtract 3xy+5yz-7zx from 5xy-2yz-2zx+10xyz
- Subtract 4p^2q-3pq+5pq^2 - 8p+7q-10 from 18-3p-11q+5pq-2pq^2 + 5p^2q…
- Find the product of the following pairs of monomials. (i) 4 , 7p (ii) -4p , 7p…
- Find the areas of rectangles with the following pairs of monomials as their…
- Complete the table of products.
- Obtain the volume of rectangular boxes with the following length, breadth and…
- Obtain the product of (i) xy , yz , zx (ii) a_1-a^2 , a^3 (iii) 2 , 4y , 8y^2 ,…
- Carry out the multiplication of the expressions in each of the following pairs.…
- Complete the table.
- Find the product. (i) (a^2) x (2a^22) x (4a^25) (ii) (2/3 xy) x (-9/10 x^2y^2)…
- Simplify 3x (4x-5) + 3 and find its values for (i) x = 3 (ii) x = 1/2…
- Simplify (a^2 + a+1) + 5 and find its value for (i) s = 0 (ii) s = 1 (iii) a =…
- Add: p(p-q), q(q-r) and r(r-p)
- Add: 2x (z-x-y) and 2y (z-y-x)
- Subtract: 3 / (1-4m+5n) from 4 / (10n-3m+21)
- Subtract: 3l (l 4m + 5n) from4l (10n 3m + 2l)
- Multiply the binomials. (i) (2x+5) and (4x-3) (ii) (y-8) and (3y-4) (iii) (2 ,…
- Find the product. (i) (5-2x) (3+x) (ii) (x+7y) (7x-y) (iii) (a^2 + b) (a+b^2)…
- Simplify. (i) (x^2 - 5) (x+5) + 25 (ii) (a^2 + 5) (b^3 + 3) + 5 (iii) (t+s^2)…
- Use a suitable identity to get each of the following products. (i) (x+3) (x+3)…
- Use the identity (x+5) (x+b) = x^2 + (a+b) x+ab to find the following products.…
- Find the following squares by using the identities. (i)(b 7)^2 (ii)(xy + 3z)^2…
- Simplify. (i) (a^2 - b^2)^2 (ii) (2x+5)^2 - (2x-5)^2 (iii) (7m-8n)^2 + (7m+8n)^2…
- Show that. (i) (3x+7)^2 - 84x = (3x-7)^2 (ii) (9p-5q)^2 + 180pq = (9p+5q)^2…
- Using identities, evaluate. (i) 71^2 (ii) 99^2 (iii) 102^2 (iv) 998^2 (v) 5.2^2…
- Using a^2 - b^2 = (a+b) (a-b) find (i) 51^2 - 49^2 (ii) (1.02)^2 - (0.98)^2…
- Using (x+a) (x+b) = x^2 + (a+b) x+ab find (i) 1 03 x 104 (ii) 5.1 x 5.2 (iii)…
Exercises 9.1
Question 1.Identify the terms, their coefficients for each of the following expressions.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Answer:
For an expression, terms are the values that are separated by addition or substraction. And the coefficients are the values of term that are with the variables.
For example:
3 abc + 2 smk2 - 4 k is an expression.
So, the terms are 3 abc, 2 smk2 and -4k
And coefficients of these terms are : 3, 2, - 4
(i) Terms: 5xyz2 and -3zy
Coefficients: 5 and -3
(ii) Terms: 1, x and x2
Coefficients: 1, 1 and 1
(iii) Terms: 4x2y2, -4x2y2z2 and z2
Coefficients: 4, -4 and 1
(iv) Terms: 3, -pq, qr and -rp
Coefficients: 3, -1, 1 and -1
(v) Terms: 
Coefficients: 
(vi) Terms: 
Coefficients: 
Question 2.
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
Answer:
Monomials: Having one term only.
Therefore Monomials are: 1000 and pqr
Binomials: Having two terms only.
Therefore Binomials are: x+y, 2y-3y2 and 4z-15z2
Trinomials: Having three term only.
Therefore Trinomials are: 7 + y + 5x, 2y - 3y2 + 4y3 and 5x – 4y + 3xy
Polynomials that do not fit any of these categories are:
x+ x2 +x 3 +x 4 , ab + bc + cd + da
Question 3.
Add the following.
(i) 
(ii) 
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
(iv) 
Answer:
(i) 
(ii) 
(iii) 
(2p2q2 – 3pq + 4) + (5 + 7pq – 3p2q2)
= 2p2q2 – 3pq + 4 + 5 + 7pq – 3p2q2
= 2p2q2 – 3p2q2 – 3pq + 7pq + 4 + 5
= -p2q2 + 4pq + 9
(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)
Question 4.
Subtract 
from 
Answer:
To Find: (12a - 9ab + 5b - 3) - (4a - 7ab + 3b + 12)
Using the following identites:
(-) × (-) = (+)
(-) × (+) = (-)
(+) × (+) = (+)
(+) × (-) = (+)
(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
Question 5.
Subtract 
from 
Answer:
(5xy - 2yz - 2zx + 10xyz) - (3xy + 5yz - 7zx)
= 5xy - 2yz - 2zx + 10xyz - 3xy - 5yz + 7zx
= 2xy - 7yz + 5zx + 10xyz
Question 6.
Subtract
from 
Answer:
Exercises 9.2
Question 1.Find the product of the following pairs of monomials.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
Answer:
(i) Product of 
(ii) Product of 
(iii) Product of 
(iv) Product of 
(v) Product of 
Question 2.
Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
Answer:
Area of Rectangle = Length × Breadth = 
Square units
Area of Rectangle = Length × Breadth = 
Square units
Area of Rectangle = Length × Breadth = 
Square units
Area of Rectangle = Length × Breadth = 
Square units
Area of Rectangle = Length × Breadth = 
Square units
Question 3.
Complete the table of products.
Answer:
Question 4.
Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 
(ii) 
(iii) 
(iv) 
Answer:
(i) Volume of rectangular box = Length × Breadth × Height = 
(ii) Volume of rectangular box = Length × Breadth × Height = 
(iii) Volume of rectangular box = Length × Breadth × Height = 
(iv) Volume of rectangular box = Length × Breadth × Height = 
Question 5.
Obtain the product of
(i) 
(ii) 
(iii) 
(iv) 
(v) 
Answer:
(i) Product of 
(ii) Product of 
(iii) Product of 
(iv) Product of 
(v) Product of 
Exercises 9.3
Question 1.Carry out the multiplication of the expressions in each of the following pairs.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
Answer:
(i) 
(ii) 
(iii) 
(iv) 4a × (a2 – 9) = 4a3 – 36a
(v) 
Question 2.
Complete the table.
Answer:
Question 3.
Find the product.
(i) 
(ii) 
(iii) 
(iv) 
Answer:
(i) Product of : a2 x 2 a22 x 4 a26
= 8 a50
(ii) 
(iii)Product of 
(iv) Product of 
Question 4.
Simplify 
and find its values for
(i) 
(ii) 
Answer: 3x(4x - 5) + 3
= 3x(4x) - 3x(5) + 3
= 12x2 - 15x + 3
(i) x = 3
Putting x = 3 in above equation, we get
12(3)2 - 15(3) + 3
= 12(9) - 45 + 3
= 108 - 42
= 66
(ii)
Putting
in above equation, we get
Question 5.
Simplify 
and find its value for
(i) 
(ii) 
(iii) 
Answer:
(i) 
Substituting a = 0 in the expression
a(a2 + a + 1) + 5 = 0(02 + 0 + 1) + 5
a(a2 + a + 1) + 5 = 0 + 5
a(a2 + a + 1) + 5 = 5
(ii)
(iii) Substituting a = -1 in the expression
Question 6.
Add: p(p-q), q(q-r) and r(r-p)
Answer:
ADD: p (p - q) + q (q - r) + r (r - p)
On simplifying the terms we get,
p (p - q) = p2 - pq
q (q - r) = q2 - qr
r (r - p) = r2 - rp
Now, let us look at the way this addition is done. We will put the like terms together and then add. We will have,
p (p - q) + q (q - r) + r (r - p) = p2 + q2 + r2 - (pq + qr + rp)
Question 7.
Add: 
and 
Answer:
Add : 2 x (z - x - y) + 2 y (z - y - x)
2 x(z - x - y) = 2 zx - 2 x2 - 2 xy
2 y (z - y - x) = 2 yz - 2 y2 - 2 yx
2 x (z - x - y) + 2 y (z - y - x) = 2 zx - 2 x2 - 2 xy + 2 yz - 2 y2 - 2 yx
2 x (z - x - y) + 2 y (z - y - x) = 2 [ zx - 2 xy + yz - x2 - y2]
Question 8.
Subtract: 
from 
Answer:
Subtraction:
4l(10n - 3m + 2l) - 3l(l - 4m + 5n)= 40ln - 12lm + 8l2 - (3l2 - 12lm + 15ln)
= 40ln - 12lm + 8l2 - 3l2 + 12lm - 15ln
= 5l2 + 25ln
= 5l(l + 5n)
Question 9.
Subtract: 3l (l – 4m + 5n) from 4l (10n – 3m + 2l)
Answer:
Subtraction:
= 4 l (10 n - 3 m + 2 l) - 3 l (l - 4 m + 5 n)
Opening Brackets we get,
= 40 ln - 12 lm + 8 l2 - 3 l2 + 12 lm - 15 ln
=40 ln - 12 lm + 12 lm -15 ln + 8l² - 3l²
=25 ln + 5l²
Hence the value is 25 ln + 5l².
Exercises 9.4
Question 1.Multiply the binomials.
(i) 
and 
(ii) 
and 
(iii) 
and 
(iv) 
and 
(v) 
and 
(vi) 
Answer:
(i) Product of 
(ii) Product of 
(iii) Product of 
[Using identity: (a + b)(a - b) = a2 – b2]
(iv) Product of 
(v) Product of 
(vi) Product of 
Question 2.
Find the product.
(i) 
(ii) 
(iii) 
(iv) 
Answer:
(i) Product of
(5-2x)(3+x)=5(3+x)-2x(3+x)
=15+5x-6x-2x²
= 15-x-2x2
(ii) Product of 
(iii) Product of 
(iv) Product of 
Question 3.
Simplify.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii)
(viii)
Answer:
(i) (x2 - 5) (x + 5 ) + 25
Opening the brackets we get,
= x2(x + 5 ) - 5 (x + 5) + 25
= x3 + 5 x2 - 5 x - 25 + 25
= x3 + 5 x2 - 5 x
(ii) (a2 + 5) (b3 + 3) + 5
Opening the brackets we get,
= a2 (b3 + 3) + 5 (b3 + 3) + 5
= a2 b3 + 3 a2 + 5 b3 + 15 + 5
= a2 b3 + 3 a2 + 5 b3 + 20
(iii)
= t3 - st + s2 t2 - s3
(iv)
Find the product of 
……………………………………………………(i)
Find the product of 
……………………………………………………(ii)
On opening the bracket 
……………………………….(iii)
On adding equations (i), (ii) and (iii)
(v)
Find the product of 
……………………………………………………(i)
Find the product of 
……………………………………………………(ii)
On adding equations (i) and (ii)
(vi)
Find the product of 
(vii)
Find the product of 
…………………………………..(i)
Adding 
to equation (i)
(viii) 
Find the product of 
Exercises 9.5
Question 1.Use a suitable identity to get each of the following products.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii)
(viii)
(ix) 
(x) 
Answer:
(i) Using Identity (a + b)2 = a2 + 2ab + b2
In the given expression a = x, b = 3
On substituting these values in the above identity, we get
(ii) Using Identity (a + b)2 = a2 + 2ab + b2
In the given expression a = 2y, b = 5
On substituting these values in the above identity, we get
(iii) Using Identity (a - b) 2 = a2 - 2ab + b2
In the given expression a = 2a, b = 7
On substituting these values in the above identity, we get
(iv) Using Identity (a - b)2 = a2 - 2ab + b2
In the given expression a = 3a, b = 1/2
On substituting these values in the above identity, we get
(v) Using Identity (a + b)(a - b) = a2 - b2
In the given expression a = 1.1m, b = 0.4
On substituting these values in the above identity, we get
(vi) Using Identity (a + b)(a - b) = a2 - b2
In the given expression a = b2, b = a2
On substituting these values in the above identity, we get
(vii) Using Identity (a + b)(a - b) = a2 - b2
In the given expression a = 6x, b = 7
On substituting these values in the above identity, we get
(viii)Using Identity (a - b) 2 = a2 - 2ab + b2
On substituting these values in the above identity, we get
(ix) Using Identity (a + b) 2 = a2 + 2ab + b2
In the given expression a = 
On substituting these values in the above identity, we get
(x) Using Identity (a - b) 2 = a2 - 2ab + b2
In the given expression a = 
On substituting these values in the above identity, we get
Question 2.
Use the identity 
to find the following products.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii)
Answer:
(i) Using Identity 
In the given expression 
On substituting these values in the above identity, we get
(ii) Using Identity
In the given expression 
On substituting these values in the above identity, we get
=
(iii) Using Identity 
In the given expression 
On substituting these values in the above identity, we get
=
(iv) Using Identity 
In the given expression
On substituting these values in the above identity, we get
=
(v) Using Identity
In the given expression 
On substituting these values in the above identity, we get
=
(vi) Using Identity
In the given expression 
On substituting these values in the above identity, we get
=
(vii) Using Identity
In the given expression 
On substituting these values in the above identity, we get
=
Question 3.
Find the following squares by using the identities.
(i) (b – 7)2
(ii) (xy + 3z)2
(iii) (6x2 – 5y)2
(iv) 
(v) 
(vi)
Answer:
(i)
Using Identity (a - b)² = a² - 2ab + b² ,
(b – 7)2
On Comparing, a= b, b=7
On substituting these values in the above identity, we get
(b – 7)2 = (b)2-2×b×7+(7)2
=b2-14b+49
(ii)
Using Identity (a + b)² = a² + 2ab + b² ,
(xy + 3z)2
On Comparing, a=xy, b=3z
On substituting these values in the above identity, we get
(xy + 3z)2 =(xy)2+2×(xy)×(3z)+(3z)2
=x2y2+6xyz+9z2
(iii)
Using Identity (a - b)² = a² - 2ab + b² ,
(6x2 – 5y)2
On Comparing, a=6x2 , b=5y
On substituting these values in the above identity, we get
(6x2 – 5y)2 = (6x2)2-2×(6x2)×(5y)+(5y)2
= 36x4 -60x2y +25y2
(iv)
Using Identity (a + b)² = a² + 2ab + b² ,
On Comparing, a=
, b=
On substituting these values in the above identity, we get
= 
= 
(v) Using Identity (a - b)2 = a2 - 2ab + b2
In the given expression, a = 
On substituting these values in the above identity, we get
= 0.16p2 - 0.4pq + 0.25q2
(vi) Using Identity (a + b) 2 = a2 + 2ab + b2
In the given expression, a = 
On substituting these values in the above identity, we get
Question 4.
Simplify.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii)(m2 - n2m) 2 + 2m3n2
Answer:
(i) Using Identity (a - b) 2 = a2 - 2ab + b2
In the given expression a = 
On substituting these values in the above identity, we get
= a4 - 2a2b2 + b4
(ii) 
Using Identity a2 - b2 = (a + b)(a - b)
In the given expression a = 
On substituting these values in the above identity, we get
(iii) 
Using Identity (a - b)2 = a2 - 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
………………………………………………..(i)
Using Identity (a + b) 2 = a2 + 2ab + b2 for finding the value of 
In the given expression a =
On substituting these values in the above identity, we get
⇒ 
………………………………………………..(ii)
On Adding equations (i) and (ii), we get
(iv) 
Using Identity (a + b) 2 = a2 + 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
………………………………………………..(i)
Using Identity (a + b) 2 = a2 + 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
………………………………………………..(ii)
On Adding equations (i) and (ii), we get
(v) 
Using Identity (a - b) 2 = a2 - 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
………………………………………………..(i)
Using Identity (a - b) 2 = a2 - 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
………………………………………………..(i)
On Subtracting equations (ii) from (i), we get
(vi) 
Using Identity (a + b)2 = a2 + 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
………………………………………………..(i)
On Subtracting 
from (i), we get
(vii) (m2 - n2m)2 + 2m3n2
Using Identity (a - b) 2 = a2 - 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
………………………………………………..(i)
On adding 
to equation (i), we get
Question 5.
Show that.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
Answer:
(i) First Find the value of LHS
Using Identity (a + b) 2 = a2 + 2ab + b2 for finding the value of 
In the given expression a = 3x and b = 6
On substituting these values in the above identity, we get
⇒ 
(i)
On subtracting 84x in eqn (i), we get
Now Find the value of RHS
Using Identity (a - b) 2 = a2 - 2ab + b2 for finding the value of 
In the given expression a = 3x, b = 7
On substituting these values in the above identity, we get
⇒ 
(ii)
Therefore from eqn (i) and eqn (ii) we found that LHS = RHS
(ii) First Find the value of LHS
Using Identity (a - b) 2 = a2 - 2ab + b2 for finding the value of 
In the given expression a = 9p, b = 5q
On substituting these values in the above identity, we get
⇒ 
(i)
On Adding 180pq in eqn (i), we get
Now Find the value of RHS
Using Identity (a + b)2 = a2 + 2ab + b2 for finding the value of
In the given expression a = 9p, b = 5q
On substituting these values in the above identity, we get
⇒ 
(ii)
Therefore from eqn (i) and eqn (ii) we found that LHS = RHS
(iii) First Find the value of LHS
Using Identity (a - b) 2 = a2 - 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
(i)
On Adding 2mn in eqn (i), we get
Therefore we found that LHS = RHS
(iv) First Find the value of LHS
Using Identity (a + b) 2 = a2 + 2ab + b2 for finding the value of 
In the given expression a = 4pq and b = 3q
On substituting these values in the above identity, we get
⇒ 
(i)
Using Identity (a - b) 2 = a2 - 2ab + b2 for finding the value of 
In the given expression a = 4pq, b = 3q
On substituting these values in the above identity, we get
⇒ 
(ii)
On subtracting eqn (ii) from eqn (i), we get
Therefore we found that LHS = RHS
(v) Using Identity (a + b)(a - b) = a2 - b2 (i)
Applying above identity in (b - c) (b + c), we get
(ii)
On Applying above identity in (c - a) (c + a), we get
(iii)
On adding eqn (i), (ii) and (iii), we get
i.e. LHS = RHS
Hence Proved!
Question 6.
Using identities, evaluate.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii)
(viii)
(ix) 
Answer:
(i) We can write 70 as 70 + 1
Using Identity (a + b) 2 = a2 + 2 ab + b2 for finding the value of 
In the given expression a = 70, b=1
On substituting these values in the above identity, we get
⇒ 4900+140+1= 5041
(ii) We can write 99 as 100 - 1
Using Identity (a - b) 2 = a2 - 2 ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
(iii) We can write 102 as 100 + 2
Using Identity (a + b) 2 = a2 + 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
(iv) We can write 998 as 1000 - 2
Using Identity (a - b) 2 = a2 - 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
(v) We can write 5.2 as 5 + 0.2
Using Identity (a + b) 2 = a2 + 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
(vi) 297× 303
We can write 297 as 300 – 3 and 303 as 300 + 3
Using Identity 
In the given expression 
On substituting these values in the above identity, we get
=
(vii) 78 × 82
We can write 78 as 80 – 2 and 82 as 80 + 2
Using Identity
In the given expression 
On substituting these values in the above identity, we get
=
(viii) We can write 8.9 as 8 + 0.9
Using Identity (a + b) 2 = a2 + 2ab + b2 for finding the value of 
In the given expression a = 
On substituting these values in the above identity, we get
⇒ 
(ix) 1.05 × 9.5 = 1.05 × 0.95 × 10
We can write 1.05 as 1 + 0.05 and 0.95 × 10 as (1 – 0.05) × 10
Using Identity
Question 7.
Using 
find
(i) 
(ii) 
(iii) 
(iv) 
Answer:
(i) 512 - 492 =
Using Identity 
Here a = 51, b = 49
(ii) (1.02)2 - (0.98) 2
Using Identity
Here a = 1.02, b = 0.98
(iii) (153)2 - (147)2
Using Identity
Here a = 153, b = 147
(iv) (12.1) 2 - (7.9) 2
Using Identity
Here a = 12.1, b = 7.9
Question 8.
Using 
find
(i) 1
(ii) 
(iii) 
(iv) 
Answer:
(i) Using Identity
In the given expression 
On substituting these values in the above identity, we get
=
(ii) Using Identity
In the given expression 
On substituting these values in the above identity, we get
=
(iii) Using Identity
In the given expression 
On substituting these values in the above identity, we get
=
(iv) Using Identity
In the given expression 
On substituting these values in the above identity, we get
=
