Lesson: Factorisation
Exercise 14.1
Question 1
Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p
2
q
2
(iv) 2x, 3x
2
, 4
(v) 6abc, 24ab
2
, 12 a
2
b
(vi) 16x
3
, 4x
2
, 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x
2
y
3
, 10x
3
y
2
, 6x
2
y
2
z
(i) 12 (ii) 2y (iii) 14pq (iv) 1
(v) 6ab (vi) 4x (vii) 10 (viii) x
2
y
2
Solution:
(i) 12x, 36
12 2 2 3xx
36 2 2 3 3
So, the common factor
2 2 3 12
(ii) 2y, 22xy
22yy
22 2 11xy x y
So, the common factor
22yy
(iii) 14pq, 28p
2
q
2
14 2 7pq p q
22
28 2 2 7p q p p q q
So, the common factor
(iv) 2x, 3x
2
, 4
22xx
2
33x x x
4 2 2
We find that there is no common factor except one, as 1 is a factor of every term.
(v) 6abc, 24ab
2
, 12 a
2
b
6 2 3abc a b c
2
24 2 2 2 3ab a b b
2
12 2 2 3a b a a b
The common factor
23ab
6ab
.
(vi) 16x
3
, 4x2, 32x
3
16 2 2 2 2x x x x
2
4 2 2 1x x x
32 2 2 2 2 2xx
The common factor
22x
4x
.
(vii) 10pq, 20qr, 30rp
10 2 5pq p q
20 2 2 5qr q r
30 2 3 5rp r p
The common factor
25
10
(viii) 3x2y3, 10x
3
y
2
, 6x
2
y
2
z
23
33x y x x y y y
32
10 2 5x y x x x y y
22
6 2 3x y z x x y y z
The common factor
22
xy
Question 2
Factorize the following expressions.
(i) 7x 42
(ii) 6p 12q
(iii) 7a
2
+ 14a
(iv) 16z + 20z3
(v) 20l
2
m + 30alm
(vi) 5x
2
y 15 xy
2
(vii) 10a
2
15b2
+ 20c2
(viii) 4a
2
+ 4ab 4ca
(ix) x
2
y z + xy
2
z + xyz
2
(x) ax
2
y + bxy
2
+ cxyz
(i) 7(x 6)
(ii) 6(p 2q)
(iii) 7a (a + 2)
(iv) 4z ( 4 + 5z2)
(v) 10 lm (2l + 3a)
(vi) 5xy(x 3y)
(vii) 5(2a
2
3b
2
+ 4c
2
)
(viii) 4a ( a + b c)
(ix) xyz (x + y + z)
(x) xy (ax + by + cz)
Solution:
(i)
7 42x
7 7 6x
76x
76x
.
(ii)
6 12pq
6 6 2pq
62pq
.
(iii)
2
7 14aa
7 2 7a a a
72aa
72aa
.
(iv)
3
16 20zz
2 2 2 2 2 2 5z z z z
2 2 2 2 5z z z



2
4 4 5zz
(v)
2
20 30l m alm
2 2 5 2 3 5l l m a l m
2 5 2 3l m l a
10 2 3lm l a
(vi)
22
5 15x y xy
5 5 3x x y x y y
53x y x y


53xy x y
.
(vii)
2 2 2
10 15 20a b c
2 5 3 5 2 2 5a a b b c c
5 2 3 2 2a a b b c c
2 2 2
5 2 3 4a b c
.
(viii)
2
4 4 4a ab ca
2 2 2 2 2 2a a a b c a
22a a b c
4a a b c
.
(ix)
2 2 2
x yz xy z xyz
x x y z x y y z x y z z
x y z x y z
xyz x y z
.
(x) ax
2
y + bxy
2
+ cxyz
a x x y b x y y c x y z
xy ax by cz
Question 3
Factorize.
(i) x
2
+ xy + 8x + 8y (ii) 15xy 6x + 5y 2 (iii) ax + bx ay by
(iv) 15pq + 15 + 9q + 25p (v) z 7 + 7xy xyz
(i) (x + 8) (x + y) (ii) (3x + 1) (5y 2) (iii) (a + b) (x y)(iv) (5p + 3) (3q + 5)
(v) (z 7) (1 xy)
Solution:
(i)
2
88x xy x y
2
( ) (8 8 )x xy x y
(Grouping the first two and the last two terms)
8x x y x y
8x y x
.
(ii)
15xy 6x + 5y 2
(15 6 ) 5 2xy x y
(Grouping the first two and the last two terms)
3 5 2 1 5 2x y y
5 2 3 1yx
.
(iii) ax + bx ay by
Grouping the first two terms,
ax bx
x a b
Hence,
ax bx ay by
()ax bx ay by
(Grouping the first two and the last two terms)
x a b y a b


()x a b y a b
a b x y
(iv)
15 15 9 25pq q p
15 25 15 9pq p q
(Rearranging and grouping the terms)
5 3 5 5 5 3 3p q p q
5 3 5 3 5 3p q q
5 3 3 5pq
(v) z 7 + 7xy xyz
77z xy xyz
(Grouping the first two and the last two terms)
77z xy z
7 1 1z xy
71z xy
.
Exercise 14.2
Question 1
Factorise the following expressions.
(i) a
2
+ 8a + 16 (ii) p
2
10p + 25 (iii) 25m
2
+ 30m + 9
(iv) 49y
2
+ 84yz + 36z
2
(v) 4x
2
8x + 4
(vi) 121b
2
88bc + 16c
2
(vii) (l + m)
2
4lm (Hint: Expand ( l + m)
2
first)
(viii) a
4
+ 2a
2
b
2
+ b
4
(i)
44aa
(ii)
2
5p
(iii)
2
53m
(iv)
2
76yz
(v)
2
41x
(vi)
2
11 4bc
(vii)
2
lm
(viii)
2
22
ab
Solution:
(i) a
2
+ 8a + 16
2
2
2 4 4aa
Applying the identity
2
22
2a ab b a b
,
The given expression reduces to
2
4a
44aa
.
(ii)
2
10 25pp
22
2 5 5pp
2
5p
(Applying the identity
2
22
2a b a ab b
)
(iii)
2
25 30 9mm
2
2
5 2 5 3 3mm
2
53m
.
(iv)
22
49 84 36y yz z
22
7 2 7 6 6y y z z
2
76yz
.
(v)
2
4 8 4xx
2
4 2 1xx
22
4 2 1 1xx


2
41x
.
(vi)
22
121 88 16b bc c
22
11 2 11 4 4b b c c
2
11 4bc
.
(vii)
2
4l m lm
22
24l lm m lm
22
2l lm m
2
lm
.
(viii)
4 2 2 4
2a a b b
22
2 2 2 2
2a a b b
2
22
ab
.
Question 2
Factorize.
(i) 4p
2
9q
2
(ii) 63a
2
112b
2
(iii) 49x
2
36
(iv) 16x
5
144x
3
(v) (l + m)
2
(l m)
2
(vi) 9x
2
y
2
16
(vii) (x
2
2xy + y
2
) z
2
(viii) 25a
2
4b
2
+ 28bc 49c
2
(i)
2 3 2 3p q p q
(ii)
7 3 4 3 4a b a b
(iii)
7 6 7 6xx
(iv)
3
16 3 3x x x
(v)
4lm
(vi)
3 4 3 4xy xy
(vii)
x y z x y z
(viii)
5 2 7 5 2 7a b c a b c
Solution:
(i)
22
49pq
22
23pq
Applying the identity
22
a b a b a b
,
The given expression becomes
2 3 2 3p q p q
.
(ii)
22
63 112ab
22
7 9 16ab
22
7 3 4ab



7 3 4 3 4a b a b
.
(iii)
2
49 36x
22
76x
7 6 7 6xx
(iv)
53
16 144xx
32
16 9xx
3 2 2
16 3xx
3
16 3 3x x x
(v)
22
l m l m
2 2 2 2
22l m lm l m lm
2 2 2 2
22l m lm l m lm
4lm
.
(vi)
22
9 16xy
22
34xy
3 4 3 4xy xy
.
(vii)
2 2 2
2x xy y z
2
2
x y z
x y z x y z
.
(viii)
2 2 2
25 4 28 49a b bc c
2 2 2
25 4 28 49a b bc c
22
2
25 2 2 2 7 7a b b c c


2
2
25 2 7a b c
2
2
5 2 7a b c
5 2 7 5 2 7a b c a b c
Question 3
Factories the expressions.
(i) ax
2
+ bx
(ii) 7p
2
+ 21q
2
(iii) 2x
3
+ 2xy
2
+ 2xz
2
(iv) am
2
+ bm
2
+ bn
2
+ an
2
(v) (lm + l) + m + 1
(vi) y (y + z) + 9 (y + z)
(vii) 5y
2
20y 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy 4y + 6 9x
(i)
x ax b
(ii)
22
73pq
(iii)
2 2 2
2x x y z
(iv)
22
m n a b
(v)
11lm
(vi)
9y y z
(vii)
5 2 4y z y
(viii)
2 1 5 2ab
(ix)
3 2 2 3xy
Solution:
(i)
a x x b x
x ax b
(ii)
22
7 21pq
7 7 3p p q q
22
73pq
.
(iii)
3 2 2
2 2 2x xy xz
2 2 2x x x x y y x z z
2 x x x y y z z
2 2 2
2x x y z
(iv)
2 2 2 2
am bm bn an
22
m a b n b a
22
m n a b
(v)
1lm l m
1 1 1l m m
11ml
(vi)
9y y z y z
9y z y
(vii)
2
5 20 8 2y y z yz
5 5 4 2 4 2y y y z y z
5 4 2 4y y z y
5 2 4y z y
.
(viii)
10 4 5 2ab a b
2 5 2 5 2 1a b b
5 2 2 1ba
.
(ix)
6 4 6 9xy y x
2 3 2 3 2 3y x x
2 3 2 3 1 3 2y x x
2 3 3 2yx
.
Question 4
Factorise.
(i) a
4
b
4
(ii) p
4
81 (iii) x
4
(y + z)
4
(iv) x
4
(x z)
4
(v) a
4
2a
2
b
2
+ b
4
(i)
22
a b a b a b
(ii)
2
9 3 3p p p
(iii)
2
2
x y z x y z x y z
(iv)
22
2 2 2z x z x xz z
(v)
22
a b a b
Solution:
(i)
44
ab
22
22
ab
2 2 2 2
a b a b
Factorizing
22
ab
again, the above expression becomes
22
a b a b a b
(ii)
4
81p
2
22
9p
22
99pp
2
9 3 3p p p
(iii)
4
4
x y z
2
22
2
x y z
22
22
x y z x y z
2
2
x y z x y z x y z
2
2
x y z x y z x y z
.
(iv)
4
4
x x z
2
22
2
x x z
22
22
x x z x x z
2
2
x x z x x z x x z
2 2 2
2x x z x x z x x xz z
22
2 2 2z x z x xz z
(v)
4 2 2 4
2a a b b
22
2 2 2 2
2a a b b
2
22
ab
2
a b a b


22
a b a b
.
Question 5
Factorise the following expressions.
(i) p
2
+ 6p + 8 (ii) q
2
10q + 21 (iii) p
2
+ 6p 16
(i)
24pp
(ii)
37qq
(iii)
82pp
Solution:
(i)
2
68pp
2
2 4 8p p p
2 4 2p p p
24pp
.
(ii)
2
10 21qq
2
3 7 21q q q
3 7 3q q q
37qq
.
(iii)
2
6 16pp
2
8 2 16p p p
8 2 8p p p
82pp
.
Exercise 14.3
Question 1
Carry out the following divisions.
(i) 28x
4
÷ 56x (ii) 36y
3
÷ 9y2 (iii) 66pq
2
r
3
÷ 11qr
2
(iv) 34x
3
y
3
z
3
÷ 51xy
2
z
3
(v)
8 8 6 4
12 6a b a b
(i)
3
2
x
(ii)
4y
(iii)
6pqr
(iv)
2
2
3
xy
(v)
24
2ab
Solution:
(i)
4
28 56xx
28
28 2
xxxx
x


3
2
x
.
(ii)
32
36 9yy
49
9
yyy
yy

4y
.
(iii)
2 3 2
66 11pq r qr
11 6
11
p q q r r r
q r r
6 p q r
6pqr
(iv)
3 3 3 2 3
34 51x y z xy z
2 2 3
23
17 2
17 3
x x y y z
x y z
2
2
3
xy
2
2
3
xy
.
(v)
8 8 6 4
12 6a b a b
6 2 4 4
64
62
6
a a b b
ab
24
2 ab
24
2ab
Question 2
Divide the given polynomial by the given monomial.
(i) (5x
2
6x) ÷ 3x (ii) (3y
8
4y
6
+ 5y
4
) ÷ y
4
(iii) 8(x
3
y
2
z
2
+ x
2
y
3
z
2
+ x
2
y
2
z
3
) ÷ 4x
2
y
2
z
2
(iv) (x
3
+ 2x
2
+ 3x) ÷ 2x
(v) (p
3
q
6
p
6
q
3
) ÷ p
3
q
3
(i)
1
56
3
x
(ii)
42
3 4 5yy
(iii)
2 x y z
(iv)
2
1
23
2
xx
(v)
33
qp
Solution:
(i)
2
5 6 3x x x
56
3
xx
x
1
56
3
x
.
(ii)
8 6 4 4
3 4 5y y y y
8 6 4
4 4 4
3 4 5y y y
y y y
42
3 4 5yy
.
(iii) 8(x
3
y
2
z
2
+ x2y3z2 + x
2
y
2
z
3
) ÷ 4x
2
y
2
z
2
3 2 2 2 3 2 2 2 3
2 2 2 2 2 2 2 2 2
8
4
x y z x y z x y z
x y z x y z x y z



2 x y z
.
(iv) (x
3
+ 2x
2
+ 3x) ÷ 2x
2
2 3 2x x x x
2
1
23
2
xx
.
(v) (p
3
q
6
p
6
q
3
) ÷ p
3
q
3
3 6 6 3
3 3 3 3
p q p q
p q p q

66
33
qp
qp

33
qp
.
Question 3
Work out the following divisions.
(i) (10x 25) ÷ 5
(ii) (10x 25) ÷ (2x 5)
(iii) 10y (6y + 21) ÷ 5(2y + 7)
(iv) 9x
2
y
2
(3z 24) ÷ 27xy (z 8)
(v) 96abc(3a 12) (5b 30) ÷ 144(a 4) (b 6)
(i)
25x
(ii) 5 (iii) 6y (iv) xy (v)
10abc
Solution:
(i)
10 25x
5 2 5x
So, (10x 25) ÷ 5 =
5 2 5 5x 
25x
(ii)
10 25x
5 2 5x
So, (10x 25) ÷ (2x 5)
5 2 5 2 5xx
5
(iii)
10 6 21yy
10 3 2 7yy
So, 10y (6y + 21) ÷ 5(2y + 7)
10 3 2 7
52
yy
yy

23y
6y
.
(iv)
22
9 3 24x y z
22
9 3 8x y z
9x
2
y
2
(3z 24) ÷ 27xy (z 8)
9 3 8
27 8
x x y y z
x y z
xy
.
(v)
96 3 12 5 30abc a b
96 3 4 5 6abc a b
96abc(3a 12) (5b 30) ÷ 144(a 4) (b 6)
96 3 4 5 6
144 4 6
abc a b
ab

96 3 5
10
144
abc
abc


.
Question 4
Divide as directed.
(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)
(ii) 26xy(x + 5) (y 4) ÷ 13x(y 4)
(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)
(iv) 20(y + 4) (y
2
+ 5y + 3) ÷ 5(y + 4)
(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)
(i)
5 3 5x
(ii)
25yx
(iii)
1
2
r p q
(iv)
2
4 5 3yy
(v)
23xx
Solution:
(i)
5 2 1 3 5 2 1x x x
5 2 1 3 5
21
xx
x
5 3 5x
(ii)
26 5 4
13 4
xy x y
xy

25yx
25yx
.
(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)
52
2 52
pqr p q q r r p
pq q r r p
1
2
r p q
.
(iv) 20(y + 4) (y
2
+ 5y + 3) ÷ 5(y + 4)
2
5 4 4 5 3
54
y y y
y

2
4 5 3yy
.
(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)
1 2 3
1
x x x x
xx

23xx
.
Question 5
Factorise the expressions and divide them as directed.
(i) (y
2
+ 7y + 10) ÷ (y + 5)
(ii) (m
2
14m 32) ÷ (m + 2)
(iii) (5p
2
25p + 20) ÷ (p 1)
(iv) 4yz (z
2
+ 6z 16) ÷ 2y(z + 8)
(v) 5pq (p
2
q
2
) ÷ 2p (p + q)
(vi) 12xy (9x
2
16y
2
) ÷ 4xy (3x + 4y)
(vii) 39y
3
(50y
2
98) ÷ 26y
2
(5y + 7)
(i)
2y
(ii)
16m
(iii)
54p
(iv)
22zz
(v)
5
2
q p q
(vi)
3 3 4xy
(vii)
3 5 7yy
Solution:
(i)
2
7 10yy
2
5 2 10y y y
5 2 5y y y
52yy
(y
2
+ 7y + 10) ÷ (y +5)
52
5
yy
y

2y
.
(ii)
2
14 32mm
2
16 2 32m m m
16 2 16m m m
16 2mm
(m
2
14m 32) ÷ (m + 2)
16 2
2
mm
m

16m
.
(iii)
2
5 25 20pp
2
5 20 5 20p p p
5 4 5 4p p p
5 5 4pp
(5p
2
25p + 20) ÷ (p 1)
5 1 4
1
pp
p

54p
.
(iv)
2
4 5 16yz z z
2
4 8 2 16yz z z z
4 8 2 8yz z z z
4 8 2yz z z
4yz(z
2
+ 6z 16) ÷ 2y(z + 8)
4 8 2
28
yz z z
yz

22zz
.
(v)
22
5pq p q
5pq p q p q
5pq (p
2
q
2
) ÷ 2p (p + q)
5
2
pq p q p q
p p q

5
2
q p q
.
(vi)
22
12 9 16xy x y
22
12 3 4xy x y
12 3 4 3 4xy x y x y
12xy (9x
2
16y
2
) ÷ 4xy (3x + 4y)
3 4 3 4 3 4
4 3 4
xy x y x y
xy x y
3 3 4xy
.
(vii)
32
39 50 98yy
32
39 2 25 49yy
2
32
39 2 5 7yy
3
39 2 5 7 5 7y y y
39y
3
(50y
2
98) ÷ 26y2 (5y + 7)
3
2
13 3 2 5 7 5 7
13 2 5 7
y y y
yy

3 5 7yy
.
Exercise 14.4
Find and correct the errors in the following mathematical statements.
Question 1
4(x 5) = 4x 5
4 20x
Solution:
Error: The error occurs in the 2
nd
term on the RHS, which is incorrectly written as -5.
This should have been
4 ( 5) 20  
.
As per the distributive law,
()a b c ab ac
Question 2
x(3x + 2) = 3x
2
+ 2
2
3 2 3 2x x x x
Solution:
Error:
The error occurs in the 2
nd
term on the RHS, which is incorrectly written as +2. This should have been
22xx
.
As per the distributive law,
()a b c ab ac
Question 3
2x + 3y = 5xy
2 3 2 3x y x y
Solution:
Error:
The error occurs on the RHS, which is incorrectly written as 5xy.
In fact, the LHS cannot be simplified further as the two terms are unlike terms.
.
Question 4
x + 2x + 3x = 5x
2 3 6x x x x
Solution:
Error: The error occurs on the RHS, which is incorrectly written as 5x.
In fact, the LHS can be written as (1 + 2 + 3) x, which simplifies to 6x.
Question 5
5y + 2y + y 7y = 0
5 2 7y y y y y
Solution:
Error:
The error occurs on the RHS, which is incorrectly written as 0. In fact, the LHS
can be written as
5 2 7 (5 2 1 7)y y y y y y
Question 6
3x + 2x = 5x
2
3 2 5x x x
Solution:
Error:
The error occurs on the RHS, which is incorrectly written as 5x
2
.
In fact, the LHS can be written as
3 2 5xx
Question 7
(2x)
2
+ 4(2x) + 7 = 2x
2
+ 8x + 7
2
2
2 4 2 7 4 8 7x x x x
Solution:
Error:
The error occurs on the RHS in the first term, which is incorrectly written as 2x
2
.
In fact
2
2 2 2
2 2 4x x x
Question 8
(2x)
2
+ 5x = 4x + 5x = 9x
2
2
2 5 4 5x x x x
Solution:
Error:
The error occurs on the RHS of the first equation in the first term, which is incorrectly
written as 4x.
In fact,
2
2 2 2
2 2 4x x x
second and the RHS of the first equation cannot be
further simplified to 9x.
Question 9
(3x + 2)
2
= 3x
2
+ 6x + 4
2
2
3 2 9 12 4x x x
Solution:
Error:
The error occurs on the RHS in the first term as well as in the second term, which
are incorrectly written as 3x
2
and 6x respectively.
In fact
2
2
2 2 2 2
3 2 3 2 3 2 2 3 12 4 9 12 4x x x x x x x
The Correct formula/ identity, which is
2
ab
22
2a ab b
,
should be applied.
Question 10
Substituting x = 3 in
(a)x
2
+ 5x + 4 gives ( 3)
2
+ 5 ( 3) + 4 = 9 + 2 + 4 = 15
(b)x
2
5x + 4 gives ( 3)
2
5 ( 3) + 4 = 9 15 + 4 = 2
(c)x
2
+ 5x gives ( 3)
2
+ 5 ( 3) = 9 15 = 24
(a)
2
(b) 28 (c)
6
Solution:
(a)
2
3 5 3 4
9 15 4
2
.
(b)
2
3 5 3 4
9 15 4
28
.
(c)
2
3 5 3 9 15
6
Question 11
(y 3)
2
= y
2
9
2
2
3 6 9y y y
Solution:
Error:
Here
2
ab
has been wrongly taken as
22
ab
, whereas
2
22
2a b a ab b
.
Question 12
(z + 5)
2
= z
2
+ 25
2
10 25zz
Solution:
2
2
2 5 10 25zz
Error:
Here
2
ab
has been wrongly taken as
22
ab
, whereas
2
22
2a b a ab b
.
Question 13
(2a + 3b) (a b) = 2a
2
3b
2
22
23a ab b
Solution:
22
23a ab b
Error:
22
2 3 2 3 2 3a b a b a ab ab b
Here, when the brackets are opened, each term of one bracket has not been multiplied
by every term of the other bracket.
Question 14
(a + 4) (a + 2) = a
2
+ 8
2
68aa
Solution:
2
4 2 4 2 8a a a a a
2
68aa
Error:
Here, when the brackets are opened, each term of one bracket has not been multiplied
by every term of the other bracket.
Question 15
(a 4) (a 2) = a
2
8
2
68aa
Solution:
2
4 2 6 8a a a a
Error:
Here, when the brackets are opened, each term of one bracket has not been multiplied
by every term of the other bracket.
Question 16
2
2
3
0
3
x
x
2
2
3
1
3
x
x
Solution:
2
2
3
1
3
x
x
Error:
Here
0
a
a
a
has been incorrectly taken as 0.
In fact, the value is 1. Also, it must be mentioned that
0x
Question 17
2
2
31
1 1 2
3
x
x

2
22
3 1 1
1
33
x
xx

Solution:
2
22
3 1 1
1
33
x
xx

Error:
There has been a wrong splitting of
0
ab
a
a
as
a
b
a
In fact,
0
a b a b
a
a a a
Question 18
31
3 2 2
x
x
33
3 2 3 2
xx
xx

Solution:
33
3 2 3 2
xx
xx

Error:
3 x and 3x have been wrongly canceled and in the numerator 3x has been wrongly
replaced by 1 and in the denominator 3x has been wrongly replaced by 0.
Question 19
31
4 3 4xx
33
4 3 4 3xx

Solution:
33
4 3 4 3xx

Error:
3 & 3 have been wrongly canceled and in the numerator 3 has been wrongly replaced
by 1 and in the denominator 3 has been wrongly replaced by 0.
Question 20
45
5
4
x
x
4 5 4 5 5
1
4 4 4 4
xx
x x x x
Solution:
4 5 4 5 5
1
4 4 4 4
xx
x x x x
Question 21
75
7
5
x
x