NCERT Solutions For Class 8 Maths Chapter 12 Factorisation

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Factorization Introduction

In this chapter, we shall learn about the factorization of algebraic expressions, methods of factorization, and division of algebraic expressions.

Factors Of Natural Numbers

A number when written as a product of its prime factors is said to be in the prime factor form. Similarly, we can express algebraic expressions as products of their factors.

Factors Of Algebraic Expressions

A fundamental factor cannot be expressed further as a product of factors.

What Is Factorisation

When we factorize an algebraic expression, we write it as a product of irreducible factors. These factors may be numbers, algebraic variables, or algebraic expressions. This process is called factorization.

Method Of Common Factors

We factorize each term of the given algebraic expression as a product of irreducible factors and separate the common factors. Then, we combine the remaining factors in each term using the distributive law.

Read and Learn More NCERT Solutions For Class 8 Maths

Question 1. Factorise:

12x + 36

22y – 33z

14pq + 35pqr.

Solution:

We have

12i = 2 x 2 x 3 x x

36 = 2 x 2 x 3 x 3

The two terms have 2, 2, and 3 as

common factors.

Therefore, 12x + 36

= (2 x 2 x 3 x x) + (2 x 2 x 3 x 3)

= 2 x 2 x 3 x (x + 3) using distributive law

= 12 x (x + 3)

= 12(x + 3)

which is the required factor form.

We have

22y = 2 x ll x y

33z = 3 x li x z

The two terms have 11 as a common factor.

Therefore,

22y – 33z

= (11 x 2 x y) – (11 x 3 x z)

= 11 X [(2 X y) – (3 X Z)]

I use the distributive law

= 11 x (2y – 3z)

= ll(2y – 3z)

which is the required factor form.

We have

14pq = 2 x 7 x p x q

35pgr = 5×7 x p x g x r

The two terms have 7, p, and q as common factors.

Therefore, 14pq + 35pqr

= 7 X p x q x 2 + 7 x p x q x 5 x r

= 7 x p x q x [2 + (5 x r)]

using distributive law

= 7pq x (2 + 5r)

= 7pq(2 + 5r).

which is the required factor form.

Note:  We notice that the factor form of an expression has only  one term

Factorization By Regrouping Terms

Sometimes it so happens that all the terms in a given algebraic expression do not have a common factor, but the terms can be grouped so that all the terms in each group have a common factor. In doing so, we get a common factor across all the groups formed. This leads to the required factorization of the given algebraic expression.

Factorization Exercise 12.1

Question 1. Find the common factors of the given terms:

  1. 12x, 36
  2. 2y, 22xy
  3. 14pq, 28p2q2
  4. 2x, 3x2, 4
  5. 6abc, 24ab2, 12a2b
  6. 16x3-4x2, 32x
  7. 10pq, 20qr, 30rp
  8. 3x2y3, 10x3y2, 6x2y2z.

Solution:

1.  12x = 2 × 2 × 3 × x

36 = 2 × 2 × 3 × 3

Common prime factors are 2 (occurs twice) and 3.

Required common factor

= 2 × 2 × 3 = 12

2.  2y = 2 x y

22xy = 2 × 11 × x × y

Common factors are 2 and y.

Required common factor = 2 × y = 2y

3.  14pq = 2 x 7 x p x q

28 p2q2 = 2 × 2 × 7 × p × p × q × q

Common factors are 2, 7, p, and g.

Required common factor

= 2 × 7 × p × q = 14pg

4.  2x = 1 × 2 × x

3ÿ = 1 × 3 × x × x

4 = 1× 2 × 2

Common factor is 1

Required common factor = 1

5.  6abc = 2 x 3 x a x b x c

24ab2 = 2 × 2 × 2 × 3 × a × b × b

12a2b = 2 × 2 × 3 × a × a × b

Common factors are 2, 3, a and b

Required common factor

= 2 × 3 × a × b = 6ab

6. 16x3 = 2 × 2 × 2 × 2 × x × y × x

-4X2 =-1 × 2 × 2 × y × X

32r = 2 ×2 × 2 × 2 × 2 × x

Common factors are 2 (occurs twice) and x (occurs once).

Required common factor = 2 × 2 × x = 4x

7.  10pq = 2 ×5 × p × q

20qr = 2 × 2 × 5 × q × r

30rp = 2 x 3 x 5 x r xp

Common factors are 2 and 5.

Required common factor = 2 x 5 = 10

8.  3x2y3 = 3 X x X x X y X y X y

10x3y2 = 2 x 5 x x x x x x y x y

6 x3 y2 z = 2 x 3 x z x x x y X y X z

Common factors are x (occurs twice) and y (occurs twice)

Required common factor = X X X X y X y = x2y2

Question 2. Factorize the following expressions:

  1. 7x – 42
  2. 6p – 12q
  3. 7a2 + 14a
  4. -16z + 20z3
  5. 20l2m + 30 alm
  6. 5x2y – 15xy2
  7. 10a2 – 15b2 + 20c2
  8. – 4a2 + 4ab – 4ca
  9. x2yz + xy2z + xyz2
  10. ax2y + bxy2 + cxyz.

Solution:

1.  7x = 7 × x

42 = 2 x 3 x 7

7x – 42

= l x x – 2 x 3 x 7

= 7(x-2 x 3)

using distributive law

= 7 (x – 6)

It is the required factor form.

2. 6p – 2 x 3 x p

12g = 2 x 2 x 3 x q

6p – 12q

= 2 x 3 x p – 2 x 2 x 3 x q

= 2 x 3(p – 2 x q)

using distributive law

= 6(p – 2q)

It is the required factor form.

3. 7a2 = 7 x a x a

14a = 2 x 7 x a

7a2 + 14a

= 7 x a x a + 2 x 7 x a

= 7 x a (a + 2)

I use the distributive law

= 7a (a + 2)

It is the required factor form.

4. -16z = (-1) x 2 x 2 x 2 x 2 x z

20Z3= 2 × 2 × 5 × Z × x × Z

A – 16z + 20 ×3 N

= (-1) x 2 x 2 x 2 x 2 x 2 + 2 x 2 x 5 x z x z x z

= (2 x 2 x z) [(-1) x 2 x 2 + 5 X z X z] 5

I am using the distributive law

= 4z(-4 + 5z2)

It is the required factor form.

5.  20l2m = 2 x 2 x 5 x l x l x m

30alm = 2 x 3 x 6 x a x l x m

A 20l2m + 30alm

= 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

x (2 x l + 3 x a)

= 10lm (2l + 3a)

| Using the distributive law

It is the required factor form.

6.  5x2y = 5 × x × x × y

15xy× = 3 × 5 × x × y × y

A 5x2y – 15xy2

= 5 × x × x × y

-3 × 5 × x × y × y

= (5 × x × y)(x – 3 x y)

| Using the distributive law

= 5xy (x – 3y)

It is the required factor form.

7.  10a2 = 2 × 5 × a × a

1562 = 3 × 5 × b × b

20c2= 2 × 2 × 5 × c × c

10a2 – 15b2 + 20c2

= 2 × 5  × a × a – 3 × 5 × b × b +2 × 2 × 5 × c × c

+2 × 2 × 5 × c × c

= 5 (2 × a × a -3 × b × b + 2 × 2 × c × c)

using distributive law

= 5(2a2 – 3b2 + 4c2)

It is the required factor form.

8. 4a2 = 2 x 2 x a x a

4ab = 2 x 2 x a x 6

4ca = 2 x 2 x c x a

– 4a2 + 4ab – 4ca

= (-1) × 2 ×2 ×a × a

+ 2 ×  2 × a × b

-2 × 2 × c × a

= 2 × 2 × a [(- 1) × a + 6 – c)]

I use the distributive law

= 4a (- a. + 6 – c)

It is the required factor form.

9.  x2yz = x × x × y × z

xy2z =x × y × y × z

xyz2 =x × y × z × z

x2yz + xyz + xyz2

= X × X × y × z

+ X × y × y × Z

+ X × y × Z × z

= X × y × z(x + y + z)

I use the distributive law

= xyz{x + y + z)

It is the required factor form.

10.  ax2y = a × x × x × y

bxy2 = b × x × y × y

Xyz = C × X × y × z

ax2y + bxy2 + XYZ

= a X × X × X y + b × x × y X y + C × X × y × z

= X X y (a × x+b × y + c × z)

I use the distributive law

= xy (ax + by + cz).

It is the required factor form.

Question 3. Factorise :

  1.  x2 + xy + 8x + 8y
  2. 15xy – 6x + 5y – 2
  3. ax + bx – ay – by
  4. 15pq + 15 + 9q + 25p
  5.  z – 7 + 7xy – xyz.

Soiution: 

x2 + xy + 8x + 8y

= x(x + y) + 8(x + y)

Taking x common in the first two terms

and 8 commons in the last two terms.

= (x + y)(x + 8)

Taking (x + y) common

It is the required form.

2. 15xy – 6x + 5y – 2

= 3x(5y – 2) + 1(5y – 2)

Taldng 3x common in first two terms and 1 common in last two terms.

= (5y – 2) (3x + 1)

Taking (5y- 2) common

It is the required factor form.

3.  ax + bx – ay – by

= x(a + b) – y(a + b)

I Taking .v common in the first two terms x and -y common in the last two terms.

= (a + b)(x – y)

| Taking (a + b) common

It is the required factor form.

4. 15pq + 15 + 9g + 25p

= 15pq + 9q + 25p + 15

Arranging the terms

= 3q(5p + 3) + 5(5p + 3)

Taking 3g common in the first two terms and 5 common in the last two terms.

= (5p + 3)(3q + 5)

Taking (5p + 3) common

It is the required factor form.

5.  z-7 + 7xy – xyz

= z – 7 – XYZ + Ixy

Arranging the terms

= 1(z – 7) – xy (z – 7)

Taking 1 common in the first two terms

and- xy common in the last two terms.

= (z – 7) (1 – xy).

Taking (z – 7) common

It is the required factor form.

Factorization Using Identities

The following identities prove to be quite helpful in factorization of an algebraic expression :

(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

(a + b) (a – b) = a2 – b2

12.2.4 Factors Of The Form (X + A) {X + B)

(x + a) (x + b) = x2 + (a. + b) x + ab

To factorize an algebraic expression of type x2 + px + q, we find two factors a and b of q such that

ab = q and a + b-p

Then, the given expression becomes

x2 + (a + b) x + ab = x2 + ax + bx + ab

= x (x + a) + b (x + a) = (x + a) (x + b)

which are the required factors.

Factorization Exercise 12.2

Question 1. Factorize the following expressions:

  1. a2 + 8a + 16
  2. p2 – lOp + 25
  3. 25m2 + 30m + 9
  4. 49y2 + 84yz + 36z2
  5. 4x2 – 8x + 4 K
  6. 121b2 – 88bc + 16c2
  7. (l + m)2 – 4lm      [Hint: Expand {l + m)2 first]
  8. a4 + 2a2b2 + b4

Solution:

1.  a2 + 8a+ 16

= (a)2 + 2 (a)(4) + (4)2

= (a + 4)2

Using Identity I

= (a + 4) (a + 4)

It is the required factor form.

2.   p2 – 10p + 25

= (p)2 – 2(p) (5) + (5)2

= (p-5)2

1 Using Identity II

= (p – 5) (p – 5)

It is the required factor form.

3.  25m2 + 30m + 9

= (5m)2 + 2(5m) (3) + (3)2

= (5m + 3)2

Using Identity I

= (5m + 3) (5m + 3)

It is the required factor form.

4.  49y2 + 84yz + 36z2

= (7y)2 + 2(7y) (6z) + (6z2

= (7y + 6z)2 I Using Identity I

= (7y + 6z) (7y + 6z)

It is the required factor form.

5.  4x2 – 8x + 4

= 4(x2 – 2x + 1)

= 4 [(x)2 – 2(x) (1) + (1)2]

= 4(x – 1)2

Using Identity II

= 4(x – 1) (x – 1)

It is the required factor form.

6.  121b2 – 88bc + 16c2

= (11b)2 – 2(11b) (4c) + (4c)2

= (11b – 4c)2

Using Identity II

= (11b – 4c) (11b – 4c)

It is the required factor form.

7.  (l + m)2 – 4lm

= (l2 + 2lm + m2) – 4lm

I Using Identity I

= l2 + (2lm – 4lm) + m2

I Combining the like terms

= l2 – 2lm + m2

= (l)2 – 2(l)(m) + (m2

= (l – m)2

Using Identity II

= (l – m) (l- m)

It is the required factor form.

8.  a4 + 2a2b2 + b4

= (a2)2 + 2(a2) (b2) + (b2)2

= (a2 + b2)2 I Using Identity I

= (a2 + b2) (a2 + b2)

It is the required factor form.

Question 2. Factorise:

  1. 4p2 – 9q2
  2. 63a2 – 112b2
  3. 49x2 – 36
  4. 16X5 – 144x3
  5. (l + m)2 – (l – m)2
  6. 9x2y2 – 16
  7. (x2 – 2xy + y2) – z2
  8. 25a2 – 4b2 + 28bc – 49c2.

Solution:

1.  4p2 – 9q2

= (2p)2– m2

= (2p – 3q) (2p + 3q)

I am Using Identity III

It is the required factor form.

2.  63a2 – 112b2

= 7 (9a2 – 1662) | Taking 7 common

= 7 {(3a)2 – (46)2}

= 7 (3a – 45) (3a + 45)

Using Identity III

It is the required factor form

3.  49x2 – 36

= (7x)2 – (6)2

= (7x – 6) (7x + 6)

1 Using Identity III

It is the required factor form.

4. 16x5 – 144x3

= 16x3(x2 – 9)

I Taking 16x3 common

= 16x3 {(x)2 – (3)2}

= 16x3 (x – 3) (x + 3)

Using Identity III

It is the required factor form.

5.  (l + m)2 – (l – m)2

= {(l + m) – (l -m)}

x {(l + m) + (l – m)}

Using Identity III

= (l + m – l + m) (l + m + l – m)

= (2m) (2l)

= 4lm

It is the required factor form.

6.  9x2y2 – 16

= (3xy)2 – (4)2

= (3xy – 4) (3xy + 4)

Using Identity II

It is the required factor form

7. (x2 – 2xy + y2) – z2

= (x- y)2 – z2

Using Identity II

= (x – y – z) (x – y + z)

Using Identity III

It is the required factor form.

8.  25a2 – 4b2 + 28b2 – 49c2

= 25a2 – (4b2 – 28bc + 49c2)

= 25a2 – {(2b)2 – 2(2b) (7c) + (7c)2}

= (5a)2 – (2b – 7c)2

Using Identity II

= {5a – (2b – 7c)} {5a + (2b – 7c)}

I Using Identity III

= (5a – 2b + 7c) (5a + 2b – 7c).

It is the required factor form.

Question 3. Factorize the expressions:

  1. ax2 + bx2
  2. 7p2 + 21q2
  3. 2x3 + 2xy2+ 2xz2
  4. am2 + bm2 + bn2 + an2
  5. (Im + l) + m + 1
  6. y(y + z) + 9(y + z)
  7. 5y2 – 20y – 8z + 2yz
  8. lOab + 4a + 5b + 2
  9. 6xy – 4y + 6 – 9x.

Solution:

1.  ax2 = a × x × x

bx = b × x

ax2 + bx = ax × x × +  x = x (a × x + b)

| Using the distributive law

= x (ax + b)

It is the required factor form

2.  7p2 = 7 × p × p

21q2 = 3 × 7 × q × q

7p2 + 21q2 =7 × p × p + 3 × 7 × q × q

= 7(p × p + 3 × q × q)

Using distributive law

= 7(p2 + 3g2)

It is the required factor form

3.  2x3 = 2 × x ×  x × x

2xy2 = 2 × x × y × y

2xz2 = 2 × x × z × z

2X2+ 2y2 + 2z2

= 2 × x × x × x + 2 × x × y × y + 2 × x × z × z

= (2 × x) (x  × X + y × y + Z × z)

Using distributive law

= 2x(x2 + y2 + z2)

It is the required factor form

4.  am2 + bm2 + bn2 + an2

= am2+ bm2 + an2 + bn2

= m2 (a + b) +n2 (a + b)

I Taking m2 commonly in the first two terms

‘ and n2 are common in the last two terms.

= (a + b) (m2 + n2)

I Taking (a + b) common

It is the required factor form.

5.  (lm + l) + m + 1

= l(m + 1) + 1 (m + 1)

Taking l common in the first two terms and 1 common in the last two terms.

= (HI + 1) (l+ 1)

Taking (m + 1) common

It is the required factor form.

6.  y(y + z) + 9(y + z)

= (y + z) (y + 9)

Taking (y + z) common

It is the required factor form.

7. 5y2 – 20y – 8z + 2yz

= 5y2 – 20y + 2yz – 8z

= 5y (y – 4) + 2z(y – 4)

distributive law

= (y – 4) (5y + 2z)

I Taking (y – 4) common

It is the required factor form.

8.  10ab + 4a + 5b+ 2

= (10ab + 4a) + (5b + 2)

I Grouping the terms

= 2a (5b + 2) + 1 (5b + 2)

Using distributive law

= (5b + 2) (2a + 1)

| Taking (56 + 2) common

It is the required factor form.

9.  6xy – 4y + 6 – 9x

= 6xy – 4y – 9x + 6

Grouping the terms

= 2y (3x – 2) – 3(3x – 2)

Using distributive law

= (3x – 2) (2y – 3).

Taking (x – 2) common

It is the required form.

Question 4. Factorise:

  1. a4 – b4
  2. p4– 81
  3. x4 – (y + z)4
  4. x4 – (x – z)4
  5. a4– 2a2b2 + b4

Solution:

1.  a4 – b4

= (a2)2 – (b2)2

= (a2 – b2) (a2 + b2)

Using Identity III

= (a – b) (a + b) (a2 + b2)

| Using Identity III

It is the required factor form.

2.  p4– 81= (p2)2 – (9)2

= (P2 – 9) (p2 + 9)

| Using Identity III

= (P)2-(3)2)(p2 + 9)

= (P – 3) (p + 3) (p2 + 9)

Using Identity III

It is the required factor form.

3.  x4 – (y + z)4

= (x2)2 -{(y + z)2}2

= {x2 – (y + 2)2} {x2+ (y + 2)2}

Using Identity III

= {x – (y + z)} {x + (y + z)} {x2 + (y + z)2}

Using Identity III

= (X – y – Z)(X + y + z) {x2 + (y + z)2}

It is the required factor form.

4.  x4 – (x – z)4

= (x2)2 – {(x – z)}2

= {x2 – (x – z)2}{x2+(x – z)2

Using Identity III

= {x – (x – z)}{x +(x- z)} {x2 + (x – z)2}

Using Identity III

= (x – x + z) (x + X – z)

{x2 + (X – 2)2}

= Z(2X – Z) {X2 + (X – 2)2}

= z(2x – z) (x2 + X2 – 2xz + z2)

Using Identity II

= z(2x – z) (2x2 – 2xz + z2)

It is the required factor form.

5.  a4 – 2a2b2 + b4

= (a2)2– 2(a2) (b2) + (b2)2

= (a2 – b2)2

| Using Identity II

= {(a – 6) ( a + b)}2

Using Identity III

= (a – b)2 (a + b)2

= (a-b) (a – b) (a + b) (a + 6).

It is the required factor form

Question  5. Factorize the following expressions:

  1. p2 + 6p + 8
  2. q2– lOq + 21
  3. p2 + 6p – 16

Solution:

1.  P2 +6p + 8

= p2 + 6p + 9 – 1

= {(p)2 + 2(p)(3) + (3)2}-(l)2

= (p + 3)2 – (l)2

I Using Identity I

= (p + 3 – l)(p + 3 + 1)

Using Identity III

= (p + 2) (p + 4)

It is the required factor form

2.  q2– lOq + 21

= q2 – lOq + 25-4

= {(q)2-2(q)(5) + (5)2}-4

= (q – 5)2 – (2)2

Using Identity II

= (q – 5 – 2) (q – 5 + 2)

Using Identity III

= (q – 7) (q – 3)

It is the required factor form

3.  P2+ 6p – 16

= p2 + 6p + 9 – 25

= (p)2 + 2(p)(3) + (3)2-(5)2

= (P + 3)2 – (5)2

Using Identity I

= (P + 3 – 5) (p + 3 + 5)

I Applying for Identity III

= (p – 2) (p + 8)

It is the required factor form.

Division Of Algebraic Expressions

Here, we shall divide one algebraic expression by another.

Division Of A Monomial By Another Monomial

We shall factorize the numerator and denominator into irreducible factors and cancel out the common factors from the numerator and the denominator.

Question 1. Divide:

  1. 24xy2z3 by 6yz2
  2. 63a2b4c6 by 7a2b2c3

Solution:

1.  24xy2z3+ 6yz2

⇒ \(\frac{2 \times 2 \times 2 \times 3 \times x \times y \times y \times z \times z \times z}{2 \times 3 \times y \times z \times z}\)

⇒ \(\frac{2 \times 2 \times x \times y \times z}{1}=4 x y z\)

2.  63a2b4c6 + 7a2b2c3

⇒ \(\frac{3 \times 3 \times 7 \times a \times a \times b \times b \times b \times b \times c \times c \times c \times c \times c \times c}{7 \times a \times a \times b \times b \times c \times c \times c}\)

⇒ \(\frac{3 \times 3 \times b \times b \times c \times c \times c}{1}\)

⇒ \(=9 b^2 c^3\)

Division Of A Polynomial By A Monomial

We divide each term of the polynomial in the numerator by the monomial in the denominator.

Division Of Algebraic Expressions Continued (Polynomial + Polynomial)

We factorize the algebraic expressions in the numerator and the denominator into irreducible factors and cancel the common factors from the numerator and the denominator.

Factorization Exercise 12.3

Question 1. Carry out the following divisions:

  1. 28x4 + 56x
  2. – 36y3 9y2
  3. 66pq2r3 + llqr2
  4. 34x2y2z2 – 51xy2z2
  5. 12a8b8 + (-6a6b4).

Solution:

1.  28x2 + 56x

⇒ \(\frac{28 x^4}{56 x}\)

⇒ \(\frac{2 \times 2 \times 7 \times x \times x \times x \times x}{2 \times 2 \times 2 \times 7 \times x}\)

⇒ \(=\frac{x \times x \times x}{2}=\frac{x^3}{2}\)

2.  – 36y3 + 9y2

⇒ \(\frac{-36 y^{\circ}}{9 y^2}\)

⇒ \(\frac{(-1) \times 2 \times 2 \times 3 \times 3 \times y \times y \times y}{3 \times 3 \times y \times y}\)

⇒ \(\frac{(-1) \times 2 \times 2 \times y}{1}=\frac{-4 y}{1}=-4 y\)

3.  66pq2P + llqr2

⇒ \(\frac{66 p q^2 r^5}{11 q r^2}\)

⇒ \(\frac{2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r}{11 \times q \times r \times r}\)

⇒ \(\frac{2 \times 3 \times p \times q \times r}{1}=\frac{6 p q r}{1}=6 p q r\)

4.   34x3y3z3 + 5lxy2z3

\(\frac{34 x^3 y^3 z^3}{51 x y^2 z^3}\)

⇒ \(\frac{2 \times 17 \times x \times x \times x \times y \times y \times y \times z \times z \times z}{3 \times 17 \times x \times y \times y \times z \times z \times z}\)

⇒ \(\frac{2 \times x \times x \times y}{3}=\frac{2}{3} x^2 y\)

5. 12a8b8 + (-6a6b4).

⇒ \(\frac{12 a^8 b^8}{-6 a^6 b^4}\)

⇒ \(\frac{2 \times 2 \times 3 \times a \times a \times a \times a \times a \times a \times a \times a \times b \times b \times b \times b \times b \times b \times b \times b}{(-1) \times 2 \times 3 \times a \times a \times a \times a \times a \times a \times b \times b \times b \times b}\)

⇒ \(\frac{-2 \times a \times a \times b \times b \times b \times b}{1}\)

⇒ \(\frac{-2 a^2 b^4}{1}=-2 a^2 b^4 \)

Question 2. Divide the given polynomial by the given monomial:

  1. (5x2 – 6x) + 3x
  2. (3y8 – 4y6 + 5y4) + y4
  3. 8(x3y2z2 + x2y2 + x2y2Z2) 4x2y2z2
  4. (x2+ 2x2 + 3x) x 2x
  5. (P3q2 – P6q3) +p3q3

Solution:

1.  (5x2 – 6x) + 3x

⇒ \(=\frac{5 x^2-6 x}{3 x}=\frac{5 x^2}{3 x}-\frac{6 x}{3 x}=\frac{5}{3} x-2=\frac{1}{3}(5 x-6)\)

2.  (3y8 – 4y6 + 5y4) + y4

⇒ \(=\frac{3 y^8-4 y^6+5 y^4}{y^4}=\frac{3 y^8}{y^4}-\frac{4 y^6}{y^4}+\frac{5 y^4}{y^4}\)

– 3y4 – 4y2 + 5

3.  8(x3y2z2 + x2y2 + x2y2Z2) 4x2y2z2

⇒ \(\frac{8\left(x^3 y^2 z^2+x^2 y^3 z^2+x^2 y^2 z^3\right)}{4 x^2 y^2 z^2}\)

⇒ \(\frac{8 x^2 y^2 z^2(x+y+z)}{4 x^2 y^2 z^2}\) Taking (x? y2 z2) common

⇒ \(\frac{2(x+y+z)}{1}=2(x+y+z)\)

4.  (x2+ 2x2 + 3x) x 2x

⇒ \(\frac{x^3+2 x^2+3 x}{2 x}=\frac{x \times\left(x^2+2 x+3\right)}{2 \times x}\)

⇒ \(\frac{1}{2}\left(x^2+2 x+3\right)\)

5.  (P3q2 – P6q3) +p3q3

⇒ \(\frac{p^3 q^6-p^6 q^3}{p^3 q^3}=\frac{p^3 q^3\left(q^3-p^3\right)}{p^3 q^3}\)

⇒ \(\frac{q^3-p^3}{1}=q^3-p^3\)

Question 3. Work out the following divisions:

  1. (10x – 25) 5
  2. (10x – 25) 4- (2x – 5)
  3. 10y(6y + 21) + 5(2y + 7)
  4. 9x2y2(3z – 24) + 27xy(z – 8)
  5. 96abc(3a – 12) (5b – 30) + 144(a – 4) (b – 6).

Solution:

1. (10x – 25) – 5

⇒ \(\frac{10 x-25}{5}=\frac{5(2 x-5)}{5}\)

⇒ \(\frac{2 x-5}{1}=2 x-5\)

⇒ \((10 x-25) \div(2 x-5)\)

⇒ \(\frac{10 x-25}{2 x-5}=\frac{5(2 x-5)}{2 x-5}=\frac{5}{1}=5\)

2. 10 y(6 y+21)+5(2 y+7)

\(\frac{10 y(6 y+21)}{5(2 y+7)}=\frac{10 y \times 3(2 y+7)}{5(2 y+7)}\)

⇒ \(\frac{2 y \times 3}{1}=\frac{6 y}{1}=6 y\)

⇒ \(9 x^2 y^2(3 z-24)+27 x y(z-8)\)

⇒ \(\frac{9 x^2 y^2(3 z-24)}{27 x y(z-8)}=\frac{9 x^2 y^2 \times 3(z-8)}{27 x y(z-8)}\)

⇒ \( \frac{x y}{1}=x y\)

3. 96abc(3a – 12) (5b – 30) + 144(a – 4) (b – 6)

⇒ \(\frac{96 a b c(3 a-12)(5 b-30)}{144(a-4)(b-6)}\)

⇒ \(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 3 \times a b c \times 3(a-4) \times 5(b-6)}{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times(a-4)(b-6)}\)

|Taking 3 common in (3 a-12) and 5 common in $(5 b-30)

⇒ \(\frac{2 \times 5 \times a b c}{1}=\frac{10 a b c}{1}\)

Question 4. Divide as directed.

  1. 5(2x + 1) (3x + 5) -5- (2x + 1)
  2. 26xy(x + 5) (y – 4) * 13x(y – 4)
  3. 52pqr (p + q)(q + r)(r + p) + 104pq(q + r)(r + p)
  4. 20(y + 4) (y2 + 5y + 3) + 5(y + 4)
  5. x(x + l)(x + 2) (x + 3) + x(x + 1).

Solution:

1.  5(2x + 1) (3x + 5) + (2x + 1)

⇒ \(=\frac{5(2 x+1)(3 x+5)}{2 x+1}=\frac{5(3 x+5)}{1}\)

= 5(3x + 5)

2. 26xy(x + 5) (y – 4) + 13x(y – 4)

⇒ \(=\frac{26 x y(x+5)(y-4)}{13 x(y-4)}=\frac{2 y(x+5)}{1}\)

= 2y(x + 5)

3.  52pqr (p + q) (q + r) (r + p) -s- 104pq(q + r) (r + p)

⇒ \(\frac{52 p q r(p+q)(q+r)(r+p)}{104 p q(q+r)(r+p)}\)

⇒ \(\frac{2 \times 2 \times 13 \times p q r(p+q)(q+r)(r+p)}{2 \times 2 \times 2 \times 13 \times p q(q+r)(r+p)}=\frac{1}{2} r(p+q)\)

4.  20(y + 4) (y2 + 5y + 3) + 5(y + 4)

⇒ \(\frac{20(y+4)\left(y^2+5 y+3\right)}{5(y+4)}\)

⇒ \(\frac{2 \times 2 \times 5 \times(y+4)\left(y^2+5 y+3\right)}{5 \times(y+4)}=\frac{2 \times 2 \times\left(y^2+5 y+3\right)}{1}\)

= 4(y2 + 5y + 3)

5.  x(x + 1) (x + 2) (x + 3) + x(x + 1)

⇒ \(=\frac{x(x+1)(x+2)(x+3)}{x(x+1)}=\frac{(x+2)(x+3)}{1}\)

= (x + 2) (x + 3)

Question 5. Factorize the expressions and divide them as directed,

  1. ( y2 + 7y + 10) + (y + 5)
  2. (m2 – 14m – 32) + (m + 2)
  3. (5p2 – 25p + 20) + (p – 1)
  4. 4yz(z2 + 6z – 16) + 2y(z + 8)
  5. 5pq(p2 – q2) + 2p(p + q)
  6. 12xy(9x2 – 16y2) + 4xy(3x + 4y)
  7. 39y3(50y2 – 98) + 26y2(5y + 7)

Solution:

1.  ( y2 + 7y + 10) + (y + 5)

⇒ \(\frac{y^2+7 y+10}{y+5}=\frac{y^2+2 y+5 y+10}{y+5}\)

Using Identity IV; 2y + 5y = 7y; (2y) (5y) = 10/

⇒ \(\frac{y(y+2)+5(y+2)}{y+5}=\frac{(y+2)(y+5)}{y+5}\)

Taking (y + 2) common

⇒ \(\frac{y+2}{1}=y+2\)

2.  (m2 – 14m – 32) + (m + 2)

⇒ \(\frac{m^2-14 m-32}{m+2}=\frac{m^2-16 m+2 m-32}{m+2}\)

⇒ \(\frac{m(m-16)+2(m-16)}{m+2}=\frac{(m-16)(m+2)}{m+2} \quad \text { |Taking }(m-16) \text { common }\)

⇒ \(\frac{m-16 m+2 m=-14 m ; \quad(-16 m)(2 m)=-32 m^2}{}=m-16\)

3.  (5p2 – 25p + 20) + (p – 1)

⇒ \(\frac{5\left(p^2-5 p+4\right)}{p-1}=\frac{5\left(p^2-p-4 p+4\right)}{p-1}\)

\(\frac{5\{p(p-1)-4(p-1)\}}{p-1}=\frac{5(p-1)(p-4)}{p-1}\)

⇒ \(\frac{5(p-4)}{1}=5(p-4)\)

4.  4yz(z2 + 6z – 16) + 2y(z + 8)

⇒ \(\frac{4 y z\left(z^2+6 z-16\right)}{2 y(z+8)}=\frac{2 \times 2 \times y z\left(z^2+6 z-16\right)}{2 \times y(z+8)}\)

⇒ \(\frac{2 z\left(z^2+6 z-16\right)}{z+8}=\frac{2 z\left(z^2+8 z-2 z-16\right)}{z+8}\)

Using Identity IV: 8z – 2z = 6z; (8z) (- 2z) = – 16z2

⇒ \(\frac{2 z[z(z+8)-2(z+8)]}{z+8}\) Taking (z + 8) common

⇒ \(\frac{2 z(z+8)(z-2)}{z+8}=\frac{2 z(z-2)}{1}\)

= 2z (z – 2)

5.  5pq(p2 – q2) + 2p(p + q)

⇒ \(\frac{5 p q\left(p^2-q^2\right)}{2 p(p+q)}=\frac{5 p q(p+q)(p-q)}{2 p(p+q)}\) Using Identity III

⇒ \(\frac{5}{2} q(p-q)\)

6.  12xy(9x2 – 16y2) + 4xy(3x + 4y)

⇒ \(\frac{12 x y\left(9 x^2-16 y^2\right)}{4 x y(3 x+4 y)}=\frac{3\left(9 x^2-16 y^2\right)}{3 x+4 y}\)

⇒ \(\frac{3\left\{(3 x)^2-\left(4 y^2\right)\right\}}{3 x+4 y}\)

⇒ \(\frac{3(3 x+4 y)(3 x-4 y)}{3 x+4 y}\)

⇒ \(\frac{3(3 x-4 y)}{1}=3(3 x-4 y)\)

7.  39y3(50y2 – 98) + 26y2(5y + 7)

⇒ \(\frac{39 y^3\left(50 y^2-98\right)}{26 y^2(5 y+7)}\)

⇒ \(\frac{3 \times 13 \times y^3 \times 2 \times\left(25 y^2-49\right)}{2 \times 13 \times y^2(5 y+7)}\) I Taking 2 common

⇒ \(\frac{\left.3 y(5 y)^2-(7)^2\right)}{(5 y+7)}=\frac{3 y(5 y+7)(5 y-7)}{(5 y+7)}\) Using Identity III

⇒ \(\frac{3 y(5 y-7)}{1}=3 y(5 y-7)\)

Factorization Multiple-Choice Questions and Solutions

Question 1. The common factor of x2y2 and x3y3 is

  1. x2Y2
  2. x3y3
  3. x2y3
  4. x3y2

Solution: 1. x2Y2

x2y2= X X X X y X y

x3y3 = X X X X x X y X y X y

Question 2. The common factor of x2y2 and x4y is

  1. x4y2
  2. x2y2
  3. x3y2
  4. x3y

Solution: 4. x3y

x3y2= x x x x x y x y

x4y = x X x X X x x x y

Question 3. The common factor of a2m4 and a4m2 is

  1. a4m4
  2. a2m2
  3. a2m4
  4. a4m2

Solution: 2. a2m

a2m4 = a x a x m x m x m x m

a4m2 = a x a x a x a x m x m

Question 4. The common factor of p3q4 and p4q3 is

  1. p4q4
  2. p4q3
  3. p3q4
  4. p3q3

Solution: 3. p3q4

p3q4=p x p x p x q x q x qx q

p4q3 = p x p x p x p x q x q x q,

Question 5. The common factor 12y and 30 is

  1. 6
  2. 12
  3. 30
  4. 6y

Solution: 1. 6

12y = 2 x 2 x 3 x y

30 = 2 x 3 x 5

Question 6. The common factor of 2x, 3x3, 4 is

  1. 1
  2. 2
  3. 3
  4. 4

Solution: 1. 1

2x = 2 x x

3x3 = 3 x T x x x x

4 = 2 x 2.

Question 7. The common factor of 10ab, 30bc, 50ca is

  1. 10
  2. 30
  3. 50
  4. abc

Solution: 1. 10

10a6 = 2 x 5 x a x b

306c = 2 x 3 x 5 x b x c

50ca = 2 x 5 x 5 x c x a

Question 8. The common factor of 14a2b and 35a4b2

  1. a4b2
  2. 35a4b2
  3. 14a2b
  4. 7a2b

Solution: 4. 7a2b

14a2b = 2 x 7 x a x a x b

35a4b2 = 5 x 7 x a x a x a x a x a x b

Question 9. The common factor of 8a2b4c2, 12a4bc4 and 20a3b4 is

  1. a4b4
  2. a2b2
  3. 4a2b2
  4. 4a2b.

Solution: 4. 4a2b.

8a2b4c2 = 2 x 2 x 2 x a x a x b x b x b x b x c x c

12a4bc2 = 2 x 2 x 3 x a X a X a x a x b x c x c

20a3b4 = 2 x 2 x 5 x a x a x a x b x b x b x b

Question 10. The common factor of 6 a3b4c2, 21a2b and 15a3 is

  1. 3a2
  2. 3a3
  3. 6a3
  4. 6a3

Solution: 1. 3a2

6a3b4c3 = 2 x 3 x a’x o x a x b x b x t x 6 x c x c

21a2b = 3 x 7 x o x a x 6

15a3 = 3 x 5 x a x a x a

Question 11. The common factor of 2a2b4c2, 8a4b3c4 and 6a3b4c2 is

  1. 2a2b3c2
  2. 6a2b3c2
  3. 8a2b3c2
  4. a4b4c4

Solution: 1. 2a2b3c2

2a2b4c2 = 2 x a x a X b X b X b X b X c x c

8a4b3c2 = 2 x 2 x 2 x 0 x a x n x a x b x b x b X c X c X c X c

6a3b4c2 = 2 X 3 X a x a X a X b X b X b X b X c X C

Question 12. The common factor of 3a2b4c2, 12b2c4 and 15a3b4c4 is

  1. b4c4
  2. 3b2c2
  3. 15b2c2
  4. 12b2c4

Solution: 2. 3b2c2

3a3b4c3 = 3 x o x a x b x b x 6 x 6 x c x c

12b3c4 = 2 x 2 x 3 x b x. b xcxcxcxc

15a3b4c4 = 3 x 5 x a x a x a x b x b x b x b X c x c x c x c.

Question 13. The common factor of 3a2b4c2, 12b3c4 and 15a3b4b4 is

  1. 12x3
  2. 24x3
  3. 36x3
  4. 48x3

Solution: 1. 12x3

243y4 = 2 x 2 x 2 x 3

x x x x x x x y x y x y x y

36x4z4 = 2 x 2 x 3 x 3 X X X X X X X X X z X z X z X z

48x3y2z = 2 x 2 x 2 x 2 x 3 x x x x x x x y x y x z.

Question 14. The common factor of 72x3y4z4, 120z2d4x4 and 96y3z4d4 is

  1. 96z3
  2. 120z3
  3. 72z3
  4. 24z3

Solution: 4. 24z3

72x3y4z4 = 2 x 2 x 2 x 3 x 3 x x x x x x X y X y X y X y x z x z x z x z

120z2d4x4 = 2 x 2 x 2 x 3 x 5 x z x z x d x d x d x d x x x x x x x x

96y3z4d4 = 2 x 2 x 2 x 2 x 2 x 3 x y X y X y x z X z x z X z x d x d x d x d,

Question 15. The common factor of 36p2q3x4, 48pq3x2, and 54p3q3x4 is

  1. 6pq3x2
  2. 36pq3x2
  3. 54pq3x2
  4. 48pq3x2

Solution: 1. 6pq3x2

36p2q3x4 = 2 x 2 x 3 x 3 x p x p x q x q x q x x x x x x x x

48pq3x2= 2 X 2 X 2 X 2 X 3 x p x q x q x q x x x x

P3q4x4 = p x p x p x q x q x q x x x x x x x x.

Question 16. The factorisation of 12a2b + 15ab2 is

  1. 3ab(4a + 5b)
  2. 3a2b(4a + 5b)
  3. 3ab2(4a + 5b)
  4. 3a2b2(4a + 5b).

Solution: 1. 3ab(4a+ 5b)

12a2b + 16ab2 = 3ab(4a + 5b)

Question 17. The factorisation of 10x2 – 18x3 + 14x4 is

  1. 2×2(7x2 – 9.x + 5)
  2. 2x(7x2 – 9x + 5)
  3. 2(7x2 – 9x + 5)
  4. 2×3(7x2 – 9x + 5).

Solution: 1. 2x2(7x2 – 9x + 5)

10x2 – 18x3+ 14x4 = 2x2 (6 – 9x + 7x2).

Question 18. The factorisation of 6x – 42 is

  1. 6(x – 7)
  2. 3(x – 7)
  3. 2(x – 7)
  4. 6(x + 7)

Solution: 1. 6(x – 7)

6x – 42 = 6(x – 7)

Question 19. The factorisation of 6x + 12y is

  1. 14a3b3(2b2 – 3a2)
  2. 3(3x + 4y)
  3. 2(3x + 12y)
  4. none of these.

Solution: 1. 14a3b3(2b2 – 3a2)

6x + 12y = 6(x + 2y)

Question 20. The factorization of 28a3b2 – 42a2b3 is

  1. 14a3b3(2b2 – 3a2)
  2. 14a2b3(2b2 – 3a2)
  3. 14a3b2(2b2 – 3a2)
  4. none of these.

Solution: 1. 14a3b2(2b2 – 3a2)

28a3b3 – 42a2b3 = 14a3b3(2b2 – 3a2)

Question 21. The factorisation of a3 + a2b + ab2 is

  1. a(a2 + ab + b2)
  2. b(a2 + ab + b2)
  3. ab(a2 + ab + b2)
  4. none of these.

Solution: 1. a(a2 + ab + b2)

a3 + a2b + ab2 = a(a2 + ab+ b2)

Question 22. The factorisation of x2yz + xy2z + xyz2 is

  1. XYZ(x + y + z)
  2. x2yz(x + y + z)
  3. xy2z(x + y + z)
  4. xyz2(x + y + z).

Solution: 1. xyz(x + y + z)

x2yz + xy2z + xyz2 = xyz (x+ y + z)

Question 23. The factorisation of ax2y + bxy2 + cxyz is

  1. xy(ax + by + cz)
  2. axy(ax + by + cz)
  3. bxy(ax + by + cz)
  4. cxy(ax + by + cz).

Solution: 1. xy(ax + by + cz)

ax2y + bxy2 + XYZ

= xy (ax + by + cz)

Question 24. The factorisation of a (x + y + z) + b(x + y + z) + c(x + y + z) is

  1. (a + b + c)(x + y + z)
  2. (ab + be + ca)(x + y + z)
  3. (xy + yz + zx)(a + b + c)
  4. none of these.

Solution: 1. (a + b + c)(x + y + z)

a(x + y + z) + b(x + y + z) + c(x + y + z)

= (x + y + z) (a + b + c)

Question 25. The factorisation of 6xy – 4y + 6 – 9xis

  1. (3x – 2)(2y – 3)
  2. (3x + 2)(2y – 3)
  3. (3x – 2)(2y + 3)
  4. (3x + 2)(2y + 3).

Solution: 1. (3x – 2)(2y – 3)

6xy – 4y + 6 – 9x

= 2y(3x – 2) – 3(-2 + 3x)

= (3x – 2)(2y – 3)

Question 26. The factorisation of x2 + xy + 2x + 2y is

  1. (x + 2)(x + y)
  2. (x + 2)(x – y)
  3. (x – 2)(x + y)
  4. (x – 2)(x – y).

Solution: 1. (x + 2)(x + y)

x2 + xy + 2x + 2y

= x(x + y) + 2(x + y)

= (x + 2) (x + y)

Question 27. The factorization of ax + bx – ay – by is

  1. (x – y)(a + b)
  2. (x + y)(o + b)
  3. (x – y)(a – b)
  4. (x + y)(a – b).

Solution: 1. (x – y)(a + b)

ax + bx – ay – by

= x(a. + b) – y(a + b)

= (x – y)(a + b)

Question 28. The factorization of ab – a – b + 1 is

  1. (a – 1)(b – 1)
  2. (a + 1)(6 + 1)
  3. (a – 1)(6 + 1)
  4. (a + 1)(6 – 1)

Solution: 1. (a – 1)(b – 1)

ab – a – b + 1

= a(b – 1) – 1(b – 1)

= (a – l)(b – 1).

Question 29. The factorisation of x2 + x + xy + y + zx + z is

  1. (x + y + z)(x + 1)
  2. (x + y + z)(x + y)
  3. (x + y + z)(y + z)
  4. (x + y + z)(z + x).

Solution: 1. (x + y + z)(x + 1)

x2+ x + xy + y + zx + z

= x(x + 1) + y(x + 1) + z(x + 1)

= (X + 1)(x + y + z)

Question 30. The factorisation of x2y2 + xy + xy2z + yz + x2yz + xz is

  1. (xy + yz + zx)(xy + 1)
  2. (xy + yz + zx)(yz + 1)
  3. (xy + yz + zx)(zx + 1)
  4. none of these.

Solution: 1. (xy + yz + zx)(xy + 1)

x2y2 + xy + xy2z + yz + x2yz + xz

= xy(xy + 1) + yz(xy + 1) + Zx(xy + 1)

= (xy + yz + zx)(xy + 1).

Question 31. The factorisation of x2 + 8x + 16 is

  1. (x + 2)2
  2. (x + 4)2
  3. (x – 2)2
  4. (x – 4)2

Solution: 2. (x + 4)2

x2 + 8x + 16

= (x)2 + 2 (x)(4) + (4)2 = (x+ 4)2.

Question 32. The factorisation of 4y2 – 12y + 9 is

  1. (2y + 3)2
  2. (2y – 3)2
  3. (3y + 2)2
  4. (3y- 2)2

Solution: 2. (2y – 3)2

4y2 – 12y + 9

= (2y)2 – 2(2y)(3) + (3)2

= (2y – 3)2

Question 33. The factorisation of 49p2 – 36 is

  1. (7p + 6)(7p – 6)
  2. (6p + 7)(6p – 7)
  3. (7p + 6)2
  4. (Ip – 6)2

Solution: 1. (Ip + Q)(lp – 6)

49p2 – 36

= (7p)2 – (6)2 = (7p – 6)(7p + 6)

Question 34. The factorisation of y2 – 7y + 12 is

  1. (y + 3)(y + 4)
  2. (y + 3)(y – 4)
  3. (y – 3)(y + 4)
  4. (y – 3)(y – 4).

Solution: 4. (y – 3)(y – 4).

y2– 7y + 12

= y2 – 3y – 4y + 12

= y(y – 3) – 4(y – 3)

= (y – 3)(y – 4)

Question 35. The factorisation of z2 -4z – 12 is

  1. (z + 6)(z + 2)
  2. (z – 6)(z – 2)
  3. (z – 6)(z + 2)
  4. (z + 6)(z – 2).

Solution: 3. (z – 6)(z + 2)

z2 – 4z – 12

= z2 – 6z + 2z – 12

= z(z – 6) + 2(z – 6)

= (z – 6)(z + 2).

Question 36. The factorisation of am2+ bm2 + bn2 + an2 is

  1. (a + b)(m2 – n2)
  2. (a + b)(m2 + n2)
  3. (a – b)(m2 + n2)
  4. (a – b)(m2 – n2).

Solution: 2. (a + b)(m2 + n2)

am2 + bm2 + bn2 + an2

= m2(a + b) + n2(b + a)

= (a + b)(m2 + n2).

Question 37. The factorisation of (Im + t) + m + 1 is X

  1. (l + l)(m + 1)
  2. (l – 1)(m – 1)
  3. (l + l)(m – 1)
  4. (l – 1)(m + 1).

Solution: 1. (l + 1)(m + 1)

Im + l + m + 1

= l(m + 1) + l(m + 1)

= (l + 1)(m + 1)

Question 38. The factorization of (I + m)2 – 4lm is

  1. (l- m)2
  2. (l+m-2)2
  3. (l + m + 2)2
  4. none of these

Solution: 1. (l- m)2

(I + m)2 – 4Im

= l2 + m2+ 2Im – 4lm

= l2– 2Im + m2 = (l- m)2

Question 39. The factorisation of 1 + p + q + r + pq + qr + pr + pqr is

  1. (1 +p)(1 + q)(1 + r)
  2. (1 – p)1- q)(1- r)
  3. (1 – p)(1 – q)(1 + r)
  4. (1 + p)(1 – q)(1 – r).

Solution: 1.(l +p)(1 + q)(1 + r)

(1 + p) + q + r + pq + qr + pr + pqr = 1 + p + q + pq + r (1 + p + q + pq)

= (1 + r)(1 + p + q + pq)

= (1 + r)(1 + p)(l + q).

Question 40. The value of 0.645 x 0.645 + 2 x 0.645 x 0.355 + 0.355 x 0.355 is

  1. l
  2. 0
  3. -l
  4. 2

Solution: 1. 1

Value = (0.645 + 0.355)2 = (l)2 = 1.

Question 41. The factorisation of 1 + 16x + 64×2 is

  1. (l – 8x)2
  2. (l + 8x)2
  3. (8 – x)2
  4. (8 + x)2

Solution: 2. (1 + 8x)2

1 + 16x + 64X2

= (l)2 + 2(1) (8x) + (8x)2 = (1 + 8x)2

Question 42. The factorisation of a:2 + x + \(\frac{1}{4}\) is

  1. \(\left(\frac{x}{2}-1\right)^2\)
  2. \(\left(\frac{x}{2}+1\right)^2\)
  3. \(\left(x+\frac{1}{2}\right)^2\)
  4. \(\left(x-\frac{1}{2}\right)^2.\)

Solution: 3. \(\left(x+\frac{1}{2}\right)^2\)

⇒ \( x^2+x+\frac{1}{4}=x^2+2(x)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2\)

⇒ \(\left(x+\frac{1}{2}\right)^2\)

Question 43. The value of 992 is

  1. (90)2 + 2(90)(9) + (9)2
  2. (90)2 _ 2(90)(9) + (9)2
  3. (90)2 + (9)2
  4. None of these

Solution: 1. (90)2 + 2(90)(9) + (9)2

992 = (90 + 9)2

= (90)2 + 2(90)(9) + (9)2

Question 44. The value of 492 is

  1. (50)2– 2(50)(1) + (l)2
  2. (50)2 + 2 (50) (1) + (l)2
  3. (50)2 -(1)2
  4. (50)2+ (1)2

Solution: 1. (50)2- 2(50)(1) + (1)2

492 = (50 – 1)2

= (50)2 _ 2(50)(l) + (1)2

Question 45. The factorisation of\(\left(\frac{x^2}{y^2}-2+\frac{y^2}{x^2}\right)\) x – 0, y * 0 is

  1. \(\left(\frac{x}{y}+\frac{y}{x}\right)^2\)
  2. \(\left(\frac{x}{y}-\frac{y}{x}\right)^2\)
  3. \(\left(\frac{x}{y}-1\right)^2\)
  4. \(\left(\frac{x}{y}+1\right)^2.\)

Solution: 2. \(\left(\frac{x}{y}-\frac{y}{x}\right)^2\)

⇒ \(\frac{x^2}{y^2}-2+\frac{y^2}{x^2}\)

⇒ \(\left(\frac{x}{y}\right)^2-2\left(\frac{x}{y}\right)\left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)^2 \)

⇒ \(\left(\frac{x}{y}-\frac{y}{x}\right)^2\)

Question 46. The value of \( \frac{0.73 \times 0.73-0.27 \times 0.27}{0.73-0.27} is\)

  1. 1
  2. 0
  3. 0.73
  4. 0.27

Solution: 1. 1

Value= \(\frac{(0.73+0.27)(0.73-0.27)}{0.73-0.27}\) = 1

Question 47. The factorisation of x2 – 9 is

  1. (x – 3)2
  2. (x + 3)2
  3. (a + 3)(a – 3)
  4. None of these

Solution: 3. (a + 3)(a – 3)

x2 – 9 = (x)2 – (3)2 = (x – 3) (x + 3).

Question 48. The factorisation of 36xy – 1 is

  1. (6xy – 1)(6xy + 1)
  2. (6xy – 1)2
  3. (6xy + 1)2
  4. ( 6 + xy)2

Solution: 1. (6xy – 1)(6xy + 1)

36x2y2 – 1 = (6xy)2 – (1)2

= (6xy – 1)(6xy + 1).

Question 49. The value of \(\frac{0.564 \times 0.564-0.436 \times 0.436}{0.564-0.436} \text { is }\)

  1. 0
  2. 1
  3. -1
  4. None of these

Solution: 2. 1

Value = \(\frac{(0.564+0.436)(0.564-0.436)}{0.564-0.436}\) = 1

Question 50. The value of (0.68)2 – (0.32)2 is

  1. -l
  2. 0
  3. 1
  4. 0.36.

Solution:  4. 0

Value = (0.68 + 0.32) (0.68 – 0.32) = 0.36.

Question 51. The factorisation of 3x2 + 10x + 8 is

  1. (3x + 4)x + 2)
  2. (3x – 4)(x – 2)
  3. (3x + 4)(x – 2)
  4. (3x – 4)(x + 2).

Solution: 1. (3x + 4)(x + 2)

3X2 + lOx + 8

= 3X2 + 6x + 4x + 8

= 3x(x + 2) + 4(x + 2)

= (x + 2)(3x + 4).

Question 52. The factorisation of 3a2 – 16a + 16 is

  1. (x – 4)(3a – 4
  2. (a + 4)(3a + 4)
  3. (a – 4)(3a + 4)
  4. (a + 4)(3a – 4).

Solution: 1. (x – 4)(3a – 4)

3×2 – 16x + 16

= 3s2 – 12x – 4x + 16

= 3x(x – 4) – 4(x – 4)

= (x – 4) (3x – 4).

Question 53. The factorisation of 6a2 – 5a – 6 is

  1. (2a – 3)(3a + 2)
  2. (2a + 3)(3a + 2)
  3. (2a – 3)(3a – 2)
  4. (2a + 3)(3a – 2).

Solution: 1. (2a – 3)(3a + 2)

6X2 – 5x – 6

= 6×2 – 9x + 4x – 6

= 3x(2x – 3) + 2(2x – 3)

= (2x – 3)(3x + 2).

Question 54. The factorisation of 6 – a – 2a2 is

  1. (2 + a)(3 – 2a)
  2. (2 + a)(3 + 2a)
  3. (2 – a)(3 – 2a)
  4. (2 – a)(3 + 2a).

Solution: 1. (2 + a)(3 – 2a)

6 – x – 2x2

= 6 + 3x – 4x – 2x2

= 3(2 + x) – 2x (2 + x)

= (2 + x)(3 – 2x)

Question 55. If x2 – x – 42 = (x + k)(x + 6), then k =

  1. 6
  2. -6
  3. 7
  4. -7

Solution: 4. -7

x2 – x – 42

= x2 – 7x + 6x – 42

= x(x — 7) + 6(x — 7)

= (x – 7)(x + 6)

= (x + k) (x + 6)

k = -7

Question 56. The value of 3.5 x 3.5 – 2.5 x 2.5 is

  1. -6
  2. 6
  3. 60
  4. 1

Solution: 2. 6

Value = (3.5 + 2.5)(3.5 – 2.5) = 6

Question 57. If \(\left(x-\frac{1}{x}\right)^2=x^2+a+\frac{1}{x^2}\) then a=

  1. -2
  2. 2
  3. 2x
  4. -2x

Solution: 1. -2

⇒ \(\left(x-\frac{1}{x}\right)^2 =x^2-2+\frac{1}{x^2}\)

⇒ \(x^2+a+\frac{1}{x^2} a=-2 .\)

Question 58. If x = 2, y = -1, then the value of x2 + 4xy + 4y2 is

  1. 0
  2. 1
  3. -1
  4. 2

Solution: 1. 0

x2 + 4xy + 4y2

= (x)2 + 2(x)(2y) + (2y)2

– (x + 2y)2 = {2 + 2(- l)}2 = 0

Question 59. The quotient of 28x2 ÷ 14r is

  1. 2
  2. 2x
  3. x
  4. x2

Solution: 2. 2x

⇒ \(\frac{28 x^2}{14 x}=\frac{2 \times 2 \times 7 \times x \times x}{2 \times 7 \times x}=2 x .\)

Question 60. The quotient of 12a8b8 ÷ (- 4a6b6) is

  1. 3a2b2
  2. 3a2b
  3. 3ab2
  4. -3a2b2

Solution: 4. -3a2b2

⇒  \(-\frac{12 a^8 b^8}{4 a^6 b^6}\)

⇒  \(2 \times 2 \times 3 \times a \times a \times a \times a \times a \times a \times a \times a \)

⇒  \(-\frac{\times b \times b \times b \times b \times b \times b \times b \times b}{2 \times 2 \times a \times a \times a \times a \times a \times a} \times b \times b \times b \times b \times b \times b\)

= – 3a2b2

Factorization True-False

Write whether the following statements are True or False:

1. An equation is true for all values of its variables: False

2. The difference of the square of two consecutive natural numbers is equal to their sum: True

3. An identity is true for all values of its variables: True

4. (x + 1) (x – 1) (x2 + 1) = x4 + 1: False

5. The difference between the areas of the two squares with sides 5o and 5b is 25 (a – b) (a + b): True

Factorization Fill In The Blanks

1. The name of the property a (b + c) = ab + ac is: Distributive Property

2. The greatest common factor of 5a. and 151) is: 5

3. The common factor of 2xy and 3zt is: 1

4. The quotient obtained on dividing (x2 – 1) (x – 2) by – (x – 2) is:- (x2 – 1)

5. On dividing (x4+ y4) (x – y) by (x – y), the remainder is: 0

6. Find the value of an in 9a = 502 – 412: 91

7. Ifa + b = 10 and a2 + b2 = 44, then find ab: 28

8. Factorise x3 – 64x: x (x – 8) (x + 8)

9. x – \(\frac{1}{x}\) = 5, then find the value of \(x^2+\frac{1}{x^2}\): 27

10. Simplify : (a + b)2 + (a – b)2: . 2 (a2 + b2)

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