CBSE Class 8

These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.2

Question 1.
Factorise the following expressions.
(i) a² + 8a + 16
(ii) p² – lOp + 25
(iii) 25m² + 30m+ 9
(iv) 49y² + 84yz + 36z²
(v) 4x² – 8x + 4
(vi) 121b² – 88bc + 16c²
(vii) (1 + m)² – 4lm
(viii) a⁴ + 2a²b² + b4
(Hint: Expand (1 + m)² first)
Answer:
(i) a² + 8a + 16
= a² + 2 x a x 4 + 4²
= (a + 4)²
= (a + 4) (a + 4)
(ii) p² – 10p + 25
= p² – 2 x 5p + 5²– (p – 5)²
= (p – 5) (p – 5)

(iii) 25m² + 30m + 9
= (5m)² + 2 x 5m x 3 + 3²
= (5m + 3)²
= (5m + 3) (5m +3)

(iv) 49y² + 84yz + 36z²
= (7y)²+ 2 x 7y x 6z + (6z)²
= (7y + 6z)² = (7y + 6z) (7y + 6z)

(v) 4x² – 8x + 4
= (2x)² – 2 x 2x x 2 + 2²
= (2x – 2)² = 2² (x – 1)²
= 4 (x – 1) (x – 1)

(vi) 12lb² – 88bc + 16c²
= (11b)² – 2 x 11b x 4c + (4c)²
= (11b – 4c)²
= (11b – 4c) (11b – 4c)

(vii) (l + m)² – 4lm
= l² + 2lm + m² – 4lm
= l² + (21m – 41m) + m²
= 1 – 21m + m²
= (1 – m)² = (1 – m) (1 – m)

(viii) a⁴ + 2a² b² + b⁴
= (a²)² + 2 x a² x b² + (b²)²
= (a² + b²)²
= (a² + b²) (a² + b²)

Question 2.
Factorise
(i) 4p² – 9q²
(ii) 63a² – 112b²
(iii) 49x² – 36
(iv) 16x⁵ – 144x³
(v) (l + m)² – (l – m)²
(vi) 9x²y² – 16
(vii) (x² – 2xy + y²) – z²
(viii) 25a² – 4b² + 28bc – 49c²
Answer:
(i) 4p² – 9q²
= (2p)² – (3q)²
= (2p + 3q) (2p – 3q)
[a² – b² = (a + b) (a – b)]

(ii) 63a²-11²b²
= 7 x (9a² – 16b²)
= 7[(3a)² – (4b)² ]
= 7 (3a + 4b) (3a – 4b)
[a² – b² = (a + b) (a – b)]

(iii) 49x² – 36
= (7x)² – 6² = (7x + 6) (7x – 6)
[Using a² – b² = (a + b) (a – b)]

(iv) 16x³ – 144x³
= x³ [16×2 – 144]
= x³ x 16 [x² – 9]
= 16x³ [x² – 3²]
= 16x³ (x + 3) (x – 3)

(v) (l + m)² – (l – m)²
= (l + m + l – m) [l + m – (l – m)]
= 2l [l + m – l + m]
= 2l x 2m = 4lm

(vi) 9x²y² – 16
= (3xy)² – 4²
= (3xy + 4) (3xy – 4)

(vii) (x² – 2xy + y²) – z²
= [x² – 2xy + y²] – z²
= (x – y)² – z²
= (x – y + z) (x – y – z)

(viii) 25a² – 4b² + 28bc – 49c²
= 25a² – [4b² – 28bc + 49c² ]
= (5a)² – [(2b)² – 2 x 2b x 7c + (7c)²]
= (5a)² – (2b – 7c)²
= (5a + 2b – 7c) [5a – (2b – 7c)]
= (5a + 2b – 7c)(5a – 2b + 7c)

Question 3.
Factorise the expressions.
(i) ax² + bx
(ii) 7p² + 21q²
(iii) 2x³ + 2xy² + 2xz²
(iv) am² + bm² + bn² + an²
(v) (lm + 1) + m + 1
(vi) y (y + z) + 9 (y + z)
(vii) 5y² – 20y – 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy – 4y + 6 – 9x
Answer:
(i) ax² + bx
= a x x x x + b x x
= x (ax + b)

(ii) 7p² + 21q²
= 7 x p x p + 3 x 7 x q x q – 7 (p² + 3q²)

(iii) 2x³ + 2xy² + 2xz²
= 2 x x x x x x + 2 x x x y x y + 2 x x x z x z
= 2x(x xx + yxy + zxz)
= 2x (x² + y² + z²)

(iv) am² + bm² + bn² + an²
= m² (a + b) + n² (b + a)
= m² (a + b) + n² (a + b)
= (a + b) (m² + n²)

(v) (lm + l) + m + l
= l(m + l) + l(m + l)
= (m + l) (l + l)

(vi) y (y + z) + 9 (y + z)
= (y + z) (y + 9)
[(y + z) is common for both terms)]

(vii) 5y² – 20y – 8z + 2yz
= 5y x y – 4 x 5y – 2 x 2 x 2 x z + 2 x y x z
= 5y (y – 4) + 2z (- 4 + y)
= 5y (y – 4) + 2z (y – 4)
= (y – 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2
= 2 x 5 x a x b + 2 x 2 x a + 5 x b + 2
= 2a (5b + 2) + 1 (5b + 2)
= (5b + 2) (2a + 1)

(ix) 6xy – 4y + 6 – 9x
= 2 x 3 x x x y – 2 x 2 x y + 2 x 3 – 3 x 3 x x
= 2y (3x – 2) + 3 (2 – 3x)
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)

Question 4.
Factorise
(i) a⁴ – b⁴
(ii) p⁴ – 81
(iii) x⁴ – (y + z)⁴
(iv) x²⁴ – (x – z)
(v) a⁴ – 2a²b² + b⁴
Answer:
(i) a⁴ – b⁴
= (a²)² – (b²)²
[(using a² – b² = (a + b) (a – b)]
= (a² + b²) (a² – b²)
= (a² + b²) (a + b) (a – b)

(ii) p⁴ – 81
= (p²)² – 9²
[Using a² – b² = (a + b) (a – b)]
= (p² + 9) (p² – 9)
= (P² + 9) [p² – 32 ]
= (p² + 9) (p + 3) (p – 3)

(iii) x² – (y + z)4
= (x²)² – [(y + z)²]²
[using a² – b² = (a + b) (a – b)
= [x² + (y + z)²] [x² – (y + z)²]
= [x² + (y + z²)] (x + y + z) [x – (y + z)
= [x² + (y + z)²] [x + y + z) (x – y – z)

(iv) x⁴ – (x – z)⁴
= (x²)² – [(x-z)²]²
= [x² + (x – z)²] [x² – (x – z)²]
[Using a² – b² = (a + b) (a – b)]
= [x² + (x – z)² ] [(x + x – z) (x – x + z)]
= [x² + (x – z)²] (2x – z) (z)
= (x² + x² + z² – 2xz) (2x – z) (z)
= (2x² + z² – 2xz) (2x – z) (z)
= z (2x – z) (2x² + z² – 2xz)

(v) a⁴ – 2a² b² + b⁴
= (a²)² – 2 x a² x b² + (b²)² = (a² – b² )²
= (a² – b²) x (a² – b²)
= (a + b) (a – b) (a + b) (a – b)
= (a + b)² (a – b)²

Question 5.
Factorise the following expressions.
(i) p² + 6p + 8
(ii) q² – lOq + 21
(iii) p² + 6p – 16
Answer:
(i) p² + 6p + 8
= p² + 4p + 2p + 8
[8 = 2 x 4 and 2 + 4 = 6]
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

(ii) q² – 10q + 21
= q² – 7q – 3q + 21
[21 = – 7 x – 3 and – 10 = – 7 – 3]
= q(q – 7) – 3(q – 7)
= (q – 7) (q – 3)

(iii) p² + 6p – 16
= p² + 8p – 2p – 16
[-16 = 8 x – 2 and 8 – 2 = 6]
= p(p + 8) – 2 (p + 8)
= (p + 8) (p – 2)2

These NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Exercise 10.2

Question 1.
Look at the given map on a city.
Answer the following.
(a) Colour the map as follows: Blue-water, red-fire station, orange-library, yellow- school. Green-park, Pink-college, Purple-Hospital, Brown-Cemetery.
(b) Mark a green ‘X’ at the intersection of Road ‘C’ and Nehru Road, Green ‘Y’ at the intersection of Gandhi Road and Road A.
(c) In red, draw a short street route from library to the bus depot.
(d) Which is further east, the city park or the market?
(e) Which is further south, the primary school or the Sr. Secondary School?
Answer:
The shaded (coloured) map according to the required direction is given in part (a). Here various colours are as under


(a) Do as directed
(b) Do as directed
(c) Do as directed
(d) City park
(e) Senior Secondary School

Question 2.
Draw a map of your class using proper scale and symbols for different objects.
Answer:
It is an activity, so do it yourself.

Question 3.
Draw a map of your school compound using proper scale and symbols for various features like play ground, main building, garden, etc.
Answer:
It is an activity. Please do it yourself.

Question 4.
Draw a map giving instructions to your friend so that she could reach your house without any difficulty.
Answer:
It is an activity. Please do it yourself.

Faces, Edges and Vertices
Note:
I. A polygon is a 2-D shape made of line segments only.
II. A polyhedron is a 3-D shape made of plane faces.
III. Plural of polyhedron is polyhedra.
IV. A cube, cuboid, pyramid, prism, etc. are polyhedra whereas spheres, cones and cylinders are not polyhydra.

A cuboid has 6 faces, 12 edges and 8 vertices.

Convex polyhedrons: You will recall the concept of convex polygons. The idea of convex polyhedron is similar.

Regular Polyhedron: A convex polyhedron is said to be regular if its faces are made up of regular polygons and the same number of faces meet at each vertex.

Prism: Prism are polyhedra whose lateral base and top are congruent polygons and the other faces are parallelograms.

Pyramids: Pyramids are polyhedra whose base is a polygon and whose lateral faces are in the shape of triangles at a common vertex.

Note: The prisms or a pyramids is named after its bases.

Euler’s Formula: It is a relation between the faces, edges and vertices of a polyhedra. It is given by
F + V = E + 2
is F + V – E = 2
where, F → Number of faces
V → Number of vertices
E → Number of edges.

These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.1

Question 1.
Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p²q²
(iv) 2x, 3x², 4
(v) 6 abc, 24ab², 12 a²b
(vi) 16x³, – 4x², 32x
(vii) 10 pq, 20 qr, 30rp
(viii) 3x²y³, 10x³, 6x²y²z
Answer:
(i) 12x = 2 x 2 x 3 x x
36 = 2 x 2 x 3 x 3
∴ The common factor = 2 x 2 x 3 = 12

(ii) 2y = 2 x y
22y = 2 x 11 x y
∴ The common factor = 2 x y = 2y

(iii) 14pq = 2 x 7 x p x q
28 p²q² = 2 x 2 x 7 x p x p x q x q
∴ The common factor
= 2 x 7 x p x q = 14pq

(iv) 2x = 2 x x
3x² = 3 x x x x
4 = 2 x 2
∴ Common factor = 1
(1 is a factor of every term)

(v) 6abc = 2 x 3 x a x b x c
24ab² = 2 x 2 x 2 x 3 x a x b x b
12a²b = 2 x 2 x 3 x a x a x b
∴ The common factor
= 2 x 3 x a x b = 6ab

(vi) 16x³ = 2 x 2 x 2x 2 x x x x x x
-4x² = -1 x 2 x 2 x x x x
32x = 2 x 2 x 2 x 2 x 2 x x
∴ The common factor = 2 x 2 x x = 4x

(vii) 10pq = 2 x 5 x p x q
20qr = 2 x 2 x 5x q x r
30rp = 2 x 3 x 5 x r x p
∴  The common factor = 2 x 5 = 10

(viii) 3x²y³ = 3 x x x x x y x y x y
10x³y² = 2 x 5 x x x x x x x y x y
6x²y²z = 2 x 3 x x x x x y x y x z
∴  The common factor = x x x x y x y
= x²y²

Question 2.
Factorise the following expressions.
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a² + 14a
(iv) – 16 z + 20 z³
(v) 20 l² m + 30 alm
(vi) 5x²y – 15xy²
(vii) 10a² + 15 b² + 20c²
(viii) – 4a² + 4ab – 4ca
(ix) x²yz + xyz² xy²z xyz²
(x) ax²y + bxy² + cxyz
Answer:
(i) 7x – 42
=7 x x – 2 x 3 x 7
= 7 (x – 2 x 3)
= 7 (x – 6)

(ii) 6p – 12q
= 2 x 3p – 2 x 2 x 3 x q
= 2 x 3 [p – 2 x q]
= 6 (p – 2q)

(iii) 7a² + 14a
=7 x a x a + 2 x 7 x a
= 7a (a + 2)

(iv) – 16z + 20 z³
= -2 x 2 x 2 x 2 x z + 2 x 2 x 5 x z x z x z
= 2 x 2 x z [-2x² + 5 x z x z]
= 4z (-4 + 5z²)

(v) 20 l²m + 30 alm
= 2 x 2 x 5 x l x l x m + 2 x 3 x 5 x a x l x m
= 2 x 5 x l x m[2 x l + 3a]
= 10lm (2l + 3a)

(vi) 5x²y-15xy²
=5 x x x x x y – 3 x 5 x x x y x x
= 5xy (x – 3y)

(vii) 10a² – 15b² + 20c²
= 2 x 5 x a x a – 3 x 5 x b x b + 2 x 2 x 5 x c x c
= 5[2 x a x a – 3 x b x b + 2 x 2 x c x c]
= 5 (2a² – 3b² + 4c²)

(viii) -4a² + 4ab – 4ca
= -2 x 2 x a x a + 2 x 2 x a x b – 2 x 2 x c x a
= 2 x 2 x a (-a + b – c)
= 4a (-a + b – c)

(ix) x²yz + xy²z + xyz²
= x x x x y x z + x x y x y x z + x x y x z x z
= x x y x z[x + y + z]
= xyz (x + y + z)

(x) ax²y + bxy² + cxyz
= a x x x x x y + b x x x y x y + c x x x y x z
= x x y[a x x + b x y + c x z]
= xy (ax + by + cz)

Question 3.
Factorise
(i) x² + xy + 8x + 8y
(ii) 15xy – 6x + 5y – 2
(iii) ax + bx – ay – by
(iv) 15pq + 15 + 9q – 25p
(v) z – 7 + 7xy – xyz
Answer:
(i) x² + xy + 8x + 8y
= x(x + y) + 8 (x + y)
= (x + y) (x + 8)

(ii) 15xy – 6x + 5y – 2
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

(iii) ax + bx – ay – by
= x (a + b) – y (a + b)
= (a + b) (x – y)

(iv) 15pq + 15 + 9q + 25p
= 15pq + 25p + 9q + 15
= 5p (3q + 5) + 3 (3q + 5)
(Re-arranging the terms)
= (3q + 5) (5p + 3)

(v) z – 7 + 7xy – xyz
= z – xyz + 7xy – 7
= z( 1 – xy) + (7 (xy – 1)
(Re-arranging the terms)
= z (1 – xy) – 7 (1 – xy)
= (1 – xy) (z- 7)

These NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Exercise 10.1

Question 1.
For each of the given solid, the two views are given. Match for each solid the corresponding top and front views. The first one is done for you.

Answer:
(a) (iii) → (iv)
(b) (i) → (v)
(c) (iv) → (ii)
(d) (v) → (iii)
(e) (ii) → (i)

Question 2.
For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.

(a) (i) Front (ii) Side (iii) Top
(b) (i) Side (ii) Front (iii) Top
(c) (i) Front (ii) Side (iii) Top
(d) (i) Front (ii) Side (iii) Top

Question 3.
For each given solid, identify the top view, front view and side view.


Answer:
(a) (i) Top (ii) Front (iii) Side
(b) (i) Side (ii) Front (iii) Top
(c) (i) Top (ii) Side (iii) Front
(d) (i) Side (ii) Front (iii) Top
(e) (i) Front (ii) Top (iii) Side

Question 4.
Draw the front view, side view and top view of the given objects.

Answer:

These NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs InText Questions

NCERT Intext Question Page No. 243

Question 1.
A machinery worth 10,500 depreciated by 5%. Find its value after one year
Answer:
(a) Here, P = 10, 500, R = 5% p.a.
₹ = 1 year, n = 1
A = p[1 + \(\frac{5}{100}\)]

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions

NCERT Intext Question Page No. 139
Question 1.
Write two terms which are like
(i) 7xy
(ii) 4mn²
(iii) 21
Answer:
(i) Two terms like 7xy are: -3xy and 8xy.
(ii) Two terms like 4mn² are: -6mn² and 2n²m.
(iii) Two terms like 21 are: -51 and -7b.

NCERT Intext Question Page No. 143
Question 1.
Find 4x × 5y × 7z. First find 4x × 5y and multiply it by 7z; or first find 5y × 7z and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter?
Answer:
We have:
4x × 5y × 7z = (4x × 5y) × 7z
= 20xy × 7z = 140xyz
Also 4x × 5y × 7z = 4x × (5y × 7z)
= 4x × 35yz = 140xyz
We observe that
(4x × 5y) × 7x = 4 × (5y × 7z)
The product of monomials is associative, i.e. the order in which we multiply the monomials does not matter.

NCERT Intext Question Page No. 144
Question 1.
Find the product
(i) 2x (3x + 5xy)
(ii) a² (2ab – 5c)
Answer:
(i) 2x(3x + 5xy) = 2x × 3x + 2x × 5xy
= (2 × 3) × x × x + (2 × 5) × x × xy
= 6 × x² + 10 × x²y = 6x² + 10x²y

(ii) a² (2ab – 5c) = a² x 2ab + a2 ² – 5c
= (1 × 2) × a² × ab + [1 × (-5)] × a² × c
= 2 × a³b + (-5) × a²c = 2a³b – 5a²c

NCERT Intext Question Page No. 145
Question 1.
Find the product: (4p² + 5p + 7) × 3p
Answer:
(4p² + 5p + 7) × 3p
= (4p² × 3p) + (5p + 3p) + (7 × 3p)
= [(4 × 3) × p² × p²] + [(5 × 3) × p × p] + (7 × 3) × p
= 12 × p³ + 15 × p² + 21 ² p
= 12p³ + 15p² + 21p

NCERT Intext Question Page No. 149
Question 1.
Verify Identity (IV), for a = 2, b = 3, x = 5.
Answer:
We have
(x + a)(x + b) = x² + (a + b)x + ab
Putting a = 2, b = 3 and x = 5 in the identity:
LHS= (x + a) (x + b)
= (5 + 2) (5 + 3)
= 7 × 8 = 56
RHS= x² + (a + b)x + ab
= (5)² + (2 + 3) x (2 x 3)
= 25 × (5) × 5 + 6
= 25 × (25) × 6
∴ LHS = RHS
∴ The given identity is true for the given values.

Question 2.
Consider, the special case of Identity (IV) with a = b, what do you get? Is it related to Identity (I)?
Answer:
When a = b (each = y)
(x + a)(x + b) = x² + (a + b)x + ab becomes
(x + y)(x + y) = x² + (y + y)x + (y + y)
= x² + (2y)x + y²
= x² + 2xy + y²
= Yes, it is the same as Identity I

Question 3.
Consider, the special case of Identity (IV) with a = -c and b -c. What do you get? Is it related to Identity (II)?
Answer:
Identity IV is given by
(x + a)(x + b) = x² + (a + b)x + ab
Replacing ‘a by (-c) and ‘b’ by (-c), we have (x – c)(x – c)
= x² + [(-c) + (-c)]x + [(-c) × (-c)]
= x² + [-2c]x + (c²)] = x² – 2cx + c²
which is same as Identity II.

Question 4.
Consider the special case of Identity (IV) with b = -a. What do you get? Is it related to Identity (III)?
Answer:
The identity IV is given by
(x + a)(x + b) = x² + (a + b)x + ab Replacing ‘b’ by (-a), we have:
(x + a)(x – a) = x² + [a + (-a)]x + [a × (-a)]
= x²+ [0]x + [a²]
= x²+ 0 + (-a²)
= x² -a²
which is same as the identity III.