Mathematics NCERT Grade 8, Chapter 14: Factorisation- The chapter lays emphasis on the concept of how to express algebraic expressions as the products of their factors.
In the introduction part following topics are discussed:
1.  Factors of natural numbers
2.  Factors of algebraic expressions
• An irreducible factor is a factor that cannot be expressed further as a product of factors.
The chapter gives detail about the following topics:
• What is Factorisation?
When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.
After this, the Method of Common Factors is explained.
• A systematic way of factorising an expression is the common factor method. It contains three steps.
Another section deals with a method called Factorising by regrouping terms. To practice questions based on this method, unsolved exercise 14.1 is given.
Now the question arises What is Regrouping?
• Rearranging the expression allows us to form groups leading to factorisation. This is called regrouping.
Factorisation using identities and factors in the form of (x +a) (x+b) are explained.
• Division of Algebraic Expressions: This section is divided into the following sub-sections:
a. Division of a monomial by another monomial
b. Division of a polynomial by a monomial
• Division of algebraic expressions continued ( Polynomial $÷$ Polynomial)
• Division is the inverse of multiplication.
• Can You Find the Error?
​Important statements mentioned in the section:

• Remember to make use of brackets, while substituting a negative value.
• Remember, when you multiply the expression enclosed in a bracket by a constant (or a variable) outside, each term of the expression has to be multiplied by the constant (or the variable).​
• Coefficient 1 of a term is usually not shown. But while adding like terms, we include it in the sum.
• Remember, when you square a monomial, the numerical coefficient and each factor has to be squared.
• While dividing a polynomial by a monomial, we divide each term of the polynomial in the numerator by the monomial in the denominator.​
All the topics are supplemented with examples and some short questions.
4 unsolved exercises are also given which contains questions in different patterns so that students can do enough practice of the same.
To make the content of the chapter- Factorisation more interesting some fun facts are also discussed in the chapter.
Recaptulisation of all important points is done at the end.

#### Question 1:

Find the common factors of the terms

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14pq, 28p2q2

(iv) 2x, 3x2, 4

(v) 6abc, 24ab2, 12a2b

(vi) 16x3, −4x2, 32x

(vii) 10pq, 20qr, 30rp

(viii) 3x2y3, 10x3y2, 6x2y2z

(i) 12x = 2 × 2 × 3 × x

36 = 2 × 2 × 3 × 3

The common factors are 2, 2, 3.

And, 2 × 2 × 3 = 12

(ii) 2y = 2 × y

22xy = 2 × 11 × x × y

The common factors are 2, y.

And, 2 × y = 2y

(iii) 14pq = 2 × 7 × p × q

28p2q2 = 2 × 2 × 7 × p × p × q × q

The common factors are 2, 7, p, q.

And, 2 × 7 × p × q = 14pq

(iv) 2x = 2 × x

3x2 = 3 × x × x

4 = 2 × 2

The common factor is 1.

(v) 6abc = 2 × 3 × a × b × c

24ab2 = 2 × 2 × 2 × 3 × a × b × b

12a2b = 2 × 2 × 3 × a × a × b

The common factors are 2, 3, a, b.

And, 2 × 3 × a × b = 6ab

(vi) 16x3 = 2 × 2 × 2 × 2 × x × x × x

−4x2 = −1 × 2 × 2 × x × x

32x = 2 × 2 × 2 × 2 × 2 × x

The common factors are 2, 2, x.

And, 2 × 2 × x = 4x

(vii) 10pq = 2 × 5 × p × q

20qr = 2 × 2 × 5 × q × r

30rp = 2 × 3 × 5 × r × p

The common factors are 2, 5.

And, 2 × 5 = 10

(viii) 3x2y3 = 3 × x × x × y × y × y

10x3y2 = 2 × 5 × x × x × x × y × y

6x2y2z = 2 × 3 × x × x × y × y × z

The common factors are x, x, y, y.

And,

x × x × y × y = x2y2

#### Question 2:

Factorise the following expressions

(i) 7x − 42

(ii) 6p − 12q

(iii) 7a2 + 14a

(iv) −16z + 20z3

(v) 20l2m + 30 alm

(vi) 5x2y − 15xy2

(vii) 10a2 − 15b2 + 20c2

(viii) −4a2 + 4ab − 4 ca

(ix) x2yz + xy2z + xyz2

(x) ax2y + bxy2 + cxyz

(i) 7x = 7 × x

42 = 2 × 3 × 7

The common factor is 7.

∴ 7x − 42 = (7 × x) − (2 × 3 × 7) = 7 (x − 6)

(ii) 6p = 2 × 3 × p

12q = 2 × 2 × 3 × q

The common factors are 2 and 3.

∴ 6p − 12q = (2 × 3 × p) − (2 × 2 × 3 × q)

= 2 × 3 [p − (2 × q)]

= 6 (p − 2q)

(iii) 7a2 = 7 × a × a

14a = 2 × 7 × a

The common factors are 7 and a.

∴ 7a2 + 14a = (7 × a × a) + (2 × 7 × a)

= 7 × a [a + 2] = 7a (a + 2)

(iv) 16z = 2 × 2 × 2 × 2 × z

20z3 = 2 × 2 × 5 × z × z × z

The common factors are 2, 2, and z.

∴ −16z + 20z3 = − (2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)

= (2 × 2 × z) [− (2 × 2) + (5 × z × z)]

= 4z (− 4 + 5z2)

(v) 20l2m = 2 × 2 × 5 × l × l × m

30alm = 2 × 3 × 5 × a × l × m

The common factors are 2, 5, l, and m.

∴ 20l2m + 30alm = (2 × 2 × 5 × l × l × m) + (2 × 3 × 5 × a × l × m)

= (2 × 5 × l × m) [(2 × l) + (3 × a)]

= 10lm (2l + 3a)

(vi) 5x2y = 5 × x × x × y

15xy2 = 3 × 5 × x × y × y

The common factors are 5, x, and y.

∴ 5x2y − 15xy2 = (5 × x × x × y) − (3 × 5 × x × y × y)

= 5 × x × y [x − (3 × y)]

= 5xy (x − 3y)

(vii) 10a2 = 2 × 5 × a × a

15b2 = 3 × 5 × b × b

20c2 = 2 × 2 × 5 × c × c

The common factor is 5.

10a2 − 15b2 + 20c2 = (2 × 5 × a × a) − (3 × 5 × b × b) + (2 × 2 × 5 × c × c)

= 5 [(2 × a × a) − (3 × b × b) + (2 × 2 × c × c)]

= 5 (2a2 − 3b2 + 4c2)

(viii) 4a2 = 2 × 2 × a × a

4ab = 2 × 2 × a × b

4ca = 2 × 2 × c × a

The common factors are 2, 2, and a.

∴ −4a2 + 4ab − 4ca = − (2 × 2 × a × a) + (2 × 2 × a × b) − (2 × 2 × c × a)

= 2 × 2 × a [− (a) + bc]

= 4a (−a + bc)

(ix) x2yz = x × x × y × z

xy2z = x × y × y × z

xyz2 = x × y × z × z

The common factors are x, y, and z.

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

= x × y × z [x + y + z]

= xyz (x + y + z)

(x) ax2y = a × x × x × y

bxy2 = b × x × y × y

cxyz = c × x × y × z

The common factors are x and y.

ax2y + bxy2 + cxyz = (a × x × x × y) + (b × x × y × y) + (c × x × y × z)

= (x × y) [(a × x) + (b × y) + (c × z)]

= xy (ax + by + cz)

#### Question 3:

Factorise

(i) x2 + xy + 8x + 8y

(ii) 15xy − 6x + 5y − 2

(iii) ax + bxayby

(iv) 15pq + 15 + 9q + 25p

(v) z − 7 + 7xyxyz

(i) x2 + xy + 8x + 8y = x × x + x × y + 8 × x + 8 × y

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

= (x + y) (x + 8)

(ii) 15xy − 6x + 5y − 2 = 3 × 5 × x × y − 3 × 2 × x + 5 × y − 2

= 3x (5y − 2) + 1 (5y − 2)

= (5y − 2) (3x + 1)
​​​​​​​​​​​​​​​​​​​​​

(iii) ax + bxayby = a × x + b × xa × yb × y

= x (a + b) − y (a + b)

= (a + b) (xy)

(iv) 15pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15

= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5

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

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

(v) z − 7 + 7xyxyz = zx × y × z − 7 + 7 × x × y

= z (1 − xy) − 7 (1 − xy)

= (1 − xy) (z − 7)

##### Video Solution for factorisation (Page: 220 , Q.No.: 3)

NCERT Solution for Class 8 math - factorisation 220 , Question 3

#### Question 1:

Factorise the following expressions.

(i) a2 + 8a + 16

(ii) p2 − 10p + 25

(iii) 25m2 + 30m + 9

(iv) 49y2 + 84yz + 36z2

(v) 4x2 − 8x + 4

(vi) 121b2 − 88bc + 16c2

(vii) (l + m)2 − 4lm (Hint: Expand (l + m)2 first)

(viii) a4 + 2a2b2 + b4

(i) a2 + 8a + 16 = (a)2 + 2 × a × 4 + (4)2

= (a + 4)2 [(x + y)2 = x2 + 2xy + y2]

(ii) p2 − 10p + 25 = (p)2 − 2 × p × 5 + (5)2

= (p − 5)2 [(ab)2 = a2 − 2ab + b2]

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

= (5m + 3)2 [(a + b)2 = a2 + 2ab + b2]

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

= (7y + 6z)2 [(a + b)2 = a2 + 2ab + b2]

(v) 4x2 − 8x + 4 = (2x)2 − 2 (2x) (2) + (2)2

= (2x − 2)2 [(ab)2 = a2 − 2ab + b2]

= [(2) (x − 1)]2 = 4(x − 1)2

(vi) 121b2 − 88bc + 16c2 = (11b)2 − 2 (11b) (4c) + (4c)2

= (11b − 4c)2 [(ab)2 = a2 − 2ab + b2]

(vii) (l + m)2 − 4lm = l2 + 2lm + m2 − 4lm

= l2 − 2lm + m2

= (l m)2 [(ab)2 = a2 − 2ab + b2]

(viii) a4 + 2a2b2 + b4 = (a2)2 + 2 (a2) (b2) + (b2)2

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

#### Question 2:

Factorise

(i) 4p2 − 9q2

(ii) 63a2 − 112b2

(iii) 49x2 − 36

(iv) 16x5 − 144x3

(v) (l + m)2 − (lm)2

(vi) 9x2y2 − 16

(vii) (x2 − 2xy + y2) − z2

(viii) 25a2 − 4b2 + 28bc − 49c2

(i) 4p2 − 9q2 = (2p)2 − (3q)2

= (2p + 3q) (2p − 3q) [a2b2 = (ab) (a + b)]

(ii) 63a2 − 112b2 = 7(9a2 − 16b2)

= 7[(3a)2 − (4b)2]

= 7(3a + 4b) (3a − 4b) [a2b2 = (ab) (a + b)]

(iii) 49x2 − 36 = (7x)2 − (6)2

= (7x − 6) (7x + 6) [a2b2 = (ab) (a + b)]

(iv) 16x5 − 144x3 = 16x3(x2 − 9)

= 16 x3 [(x)2 − (3)2]

= 16 x3(x − 3) (x + 3) [a2b2 = (ab) (a + b)]

(v) (l + m)2 − (lm)2 = [(l + m) − (lm)] [(l + m) + (lm)]

[Using identity a2b2 = (ab) (a + b)]

= (l + ml + m) (l + m + lm)

= 2m × 2l

= 4ml

= 4lm

(vi) 9x2y2 − 16 = (3xy)2 − (4)2

= (3xy − 4) (3xy + 4) [a2b2 = (ab) (a + b)]

(vii) (x2 − 2xy + y2) − z2 = (xy)2 − (z)2 [(ab)2 = a2 − 2ab + b2]

= (xyz) (xy + z) [a2b2 = (ab) (a + b)]

(viii) 25a2 − 4b2 + 28bc − 49c2 = 25a2 − (4b2 − 28bc + 49c2)

= (5a)2 − [(2b)2 − 2 × 2b × 7c + (7c)2]

= (5a)2 − [(2b − 7c)2]

[Using identity (ab)2 = a2 − 2ab + b2]

= [5a + (2b − 7c)] [5a − (2b − 7c)]

[Using identity a2b2 = (ab) (a + b)]

= (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 + l) + m + 1

(vi) y(y + z) + 9(y + z)

(vii) 5y2 − 20y − 8z + 2yz

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

(ix) 6xy − 4y + 6 − 9x

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

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

(iii) 2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2)

(iv) am2 + bm2 + bn2 + an2 = am2 + bm2 + an2 + bn2

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

= (a + b) (m2 + n2)

(v) (lm + l) + m + 1 = lm + m + l + 1

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

= (l + l) (m + 1)

(vi) y (y + z) + 9 (y + z) = (y + z) (y + 9)

(vii) 5y2 − 20y − 8z + 2yz = 5y2 − 20y + 2yz − 8z

= 5y(y − 4) + 2z(y − 4)

= (y − 4) (5y + 2z)

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

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

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

(ix) 6xy − 4y + 6 − 9x = 6xy − 9x − 4y + 6

= 3x(2y − 3) − 2(2y − 3)

= (2y − 3) (3x − 2)
​​​​​​​​​​​​​​​​​​​​​

##### Video Solution for factorisation (Page: 223 , Q.No.: 3)

NCERT Solution for Class 8 math - factorisation 223 , Question 3

#### Question 4:

Factorise

(i) a4b4

(ii) p4 − 81

(iii) x4 − (y + z)4

(iv) x4 − (xz)4

(v) a4 − 2a2b2 + b4

(i) a4b4 = (a2)2 − (b2)2

= (a2b2) (a2 + b2)

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

(ii) p4 − 81 = (p2)2 − (9)2

= (p2 − 9) (p2 + 9)

= [(p)2 − (3)2] (p2 + 9)

= (p − 3) (p + 3) (p2 + 9)

(iii) x4 − (y + z)4 = (x2)2 − [(y +z)2]2

= [x2 − (y + z)2] [x2 + (y + z)2]

= [x − (y + z)][ x + (y + z)] [x2 + (y + z)2]

= (xyz) (x + y + z) [x2 + (y + z)2]

(iv) x4 − (xz)4 = (x2)2 − [(xz)2]2

= [x2 − (xz)2] [x2 + (xz)2]

= [x − (xz)] [x + (xz)] [x2 + (xz)2]

= z(2xz) [x2 + x2 − 2xz + z2]

= z(2xz) (2x2 − 2xz + z2)

(v) a4 − 2a2b2 + b4 = (a2)2 − 2 (a2) (b2) + (b2)2

= (a2 b2)2

= [(ab) (a + b)]2

= (ab)2 (a + b)2

#### Question 5:

Factorise the following expressions

(i) p2 + 6p + 8

(ii) q2 − 10q + 21

(iii) p2 + 6p − 16

(i) p2 + 6p + 8

It can be observed that, 8 = 4 × 2 and 4 + 2 = 6

p2 + 6p + 8 = p2 + 2p + 4p + 8

= p(p + 2) + 4(p + 2)

= (p + 2) (p + 4)

(ii) q2 − 10q + 21

It can be observed that, 21 = (−7) × (−3) and (−7) + (−3) = − 10

q2 − 10q + 21 = q2 − 7q − 3q + 21

= q(q − 7) − 3(q − 7)

= (q − 7) (q − 3)

(iii) p2 + 6p − 16

It can be observed that, 16 = (−2) × 8 and 8 + (−2) = 6

p2 + 6p − 16 = p2 + 8p − 2p − 16

= p(p + 8) − 2(p + 8)

= (p + 8) (p − 2)
​​​​​​​​​​​​​​​​​​​​​

##### Video Solution for factorisation (Page: 224 , Q.No.: 5)

NCERT Solution for Class 8 math - factorisation 224 , Question 5

#### Question 1:

Carry out the following divisions.

(i) 28x4 ÷ 56x

(ii) −36y3 ÷ 9y2

(iii) 66pq2r3 ÷ 11qr2

(iv) 34x3y3z3 ÷ 51xy2z3

(v) 12a8b8 ÷ (−6a6b4)

(i) 28x4 = 2 × 2 × 7 × x × x × x × x

56x = 2 × 2 × 2 × 7 × x

(ii) 36y3 = 2 × 2 × 3 × 3 × y × y × y

9y2 = 3 × 3 × y × y

(iii) 66 pq2 r3 = 2 × 3 × 11 × p × q × q × r × r × r

11qr2 = 11 × q × r × r

(iv) 34 x3y3z3 = 2 × 17 × x × x × x × y × y × y × z × z × z

51 xy2z3 = 3 ×17 × x × y × y ×z × z × z

(v) 12a8b8 = 2 × 2 × 3 × a8 × b8

6a6b4 = 2 × 3 × a6 × b4

= −2a2b4

#### Question 2:

Divide the given polynomial by the given monomial.

(i) (5x2 − 6x) ÷ 3x

(ii) (3y8 − 4y6 + 5y4) ÷ y4

(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2

(iv) (x3 + 2x2 + 3x) ÷ 2x

(v) (p3q6p6q3) ÷ p3q3

(i) 5x2 − 6x = x(5x − 6)

(ii) 3y8 − 4y6 + 5y4 = y4(3y4 − 4y2 + 5)

(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) = 8x2y2z2(x + y + z)

(iv) x3 + 2x2 + 3x = x(x2 + 2x + 3)

(v) p3q6p6q3 = p3q3(q3p3)

#### Question 3:

Work out the following divisions.

(i) (10x − 25) ÷ 5

(ii) (10x − 25) ÷ (2x − 5)

(iii) 10y(6y + 21) ÷ 5(2y + 7)

(iv) 9x2y2(3z − 24) ÷ 27xy(z − 8)

(v) 96abc(3a − 12)(5b − 30) ÷ 144(a − 4) (b − 6)

(i)

(ii)

(iii)

(iv)

(v) 96 abc(3a − 12) (5b − 30) ÷ 144 (a − 4) (b − 6)

#### 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) (y2 + 5y + 3) ÷ 5(y + 4)

(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)

(i) = 5(3x + 1)

(ii) = 2y (x + 5)

(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)

(iv) 20(y + 4) (y2 + 5y + 3) = 2 × 2 × 5 × (y + 4) (y2 + 5y + 3)

(v)

= (x + 2) (x + 3)

#### Question 5:

Factorise the expressions and divide them as directed.

(i) (y2 + 7y + 10) ÷ (y + 5)

(ii) (m2 − 14m − 32) ÷ (m + 2)

(iii) (5p2 − 25p + 20) ÷ (p − 1)

(iv) 4yz(z2 + 6z − 16) ÷ 2y(z + 8)

(v) 5pq(p2q2) ÷ 2p(p + q)

(vi) 12xy(9x2 − 16y2) ÷ 4xy(3x + 4y)

(vii) 39y3(50y2− 98) ÷ 26y2(5y+ 7)

(i) (y2 + 7y + 10) = y2 + 2y + 5y + 10

= y (y + 2) + 5 (y + 2)

= (y + 2) (y + 5)

(ii) m2 − 14m − 32 = m2 + 2m − 16m − 32

= m (m + 2) − 16 (m + 2)

= (m + 2) (m − 16)

(iii) 5p2 − 25p + 20 = 5(p2 − 5p + 4)

= 5[p2p − 4p + 4]

= 5[p(p −1) − 4(p −1)]

= 5(p −1) (p − 4)

​​​​​​​​​​​​​​​​​​​​​

(iv) 4yz(z2 + 6z −16) = 4yz [z2 − 2z + 8z − 16]

= 4yz [z(z − 2) + 8(z − 2)]

= 4yz(z − 2) (z + 8)

(v) 5pq(p2 q2) = 5pq (pq) (p + q)

(vi) 12xy(9x2 − 16y2) = 12xy[(3x)2 − (4y)2] = 12xy(3x − 4y) (3x + 4y)

(vii) 39y3(50y2 − 98) = 3 × 13 × y × y × y × 2[(25y2 − 49)]

= 3 × 13 × 2 × y × y × y × [(5y)2 − (7)2]

= 3 × 13 × 2 × y × y × y (5y − 7) (5y + 7)

26y2(5y + 7) = 2 × 13 × y × y × (5y + 7)

39y3(50y2 − 98) ÷26y2 (5y + 7)

##### Video Solution for factorisation (Page: 227 , Q.No.: 5)

NCERT Solution for Class 8 math - factorisation 227 , Question 5

#### Question 1:

Find and correct the errors in the statement: 4(x − 5) = 4x − 5

L.H.S. = 4(x − 5) = 4 × x − 4 × 5 = 4x − 20 ≠ R.H.S.

The correct statement is 4(x − 5) = 4x − 20

#### Question 2:

Find and correct the errors in the statement: x(3x + 2) = 3x2 + 2

L.H.S. = x(3x + 2) = x × 3x + x × 2 = 3x2 + 2x ≠ R.H.S.

The correct statement is x(3x + 2) = 3x2 + 2x

#### Question 3:

Find and correct the errors in the statement: 2x + 3y = 5xy

L.H.S = 2x + 3y ≠ R.H.S.

The correct statement is 2x + 3y = 2x + 3y

#### Question 4:

Find and correct the errors in the statement: x + 2x + 3x = 5x

L.H.S = x + 2x + 3x =1x + 2x + 3x = x(1 + 2 + 3) = 6x ≠ R.H.S.

The correct statement is x + 2x + 3x = 6x

#### Question 5:

Find and correct the errors in the statement: 5y + 2y + y − 7y = 0

L.H.S. = 5y + 2y + y − 7y = 8y − 7y = y ≠ R.H.S

The correct statement is 5y + 2y + y − 7y = y

#### Question 6:

Find and correct the errors in the statement: 3x + 2x = 5x2

L.H.S. = 3x + 2x = 5x ≠ R.H.S

The correct statement is 3x + 2x = 5x

#### Question 7:

Find and correct the errors in the statement: (2x)2 + 4(2x) + 7 = 2x2 + 8x + 7

L.H.S = (2x)2 + 4(2x) + 7 = 4x2 + 8x + 7 ≠ R.H.S

The correct statement is (2x)2 + 4(2x) + 7 = 4x2 + 8x + 7

#### Question 8:

Find and correct the errors in the statement: (2x)2 + 5x = 4x + 5x = 9x

L.H.S = (2x)2 + 5x = 4x2 + 5x ≠ R.H.S.

The correct statement is (2x)2 + 5x = 4x2 + 5x

#### Question 9:

Find and correct the errors in the statement: (3x + 2)2 = 3x2 + 6x + 4

L.H.S. = (3x + 2)2 = (3x)2 + 2(3x)(2) + (2)2 [(a + b)2 = a2 + 2ab + b2]

= 9x2 + 12x + 4 ≠ R.H.S

The correct statement is (3x + 2)2 = 9x2 + 12x + 4

#### Question 10:

Find and correct the errors in the following mathematical statement. Substituting x = −3 in

(a) x2 + 5x + 4 gives (−3)2 + 5 (−3) + 4 = 9 + 2 + 4 = 15

(b) x2 − 5x + 4 gives (−3)2 − 5 (−3) + 4 = 9 − 15 + 4 = −2

(c) x2 + 5x gives (−3)2 + 5 (−3) = −9 − 15 = −24

(a) For x = −3,

x2 + 5x + 4 = (−3)2 + 5 (−3) + 4 = 9 − 15 + 4 = 13 − 15 = −2

(b) For x = −3,

x2 − 5x + 4 = (−3)2 − 5 (−3) + 4 = 9 + 15 + 4 = 28

(c) For x = −3,

x2 + 5x = (−3)2 + 5(−3) = 9 − 15 = −6

#### Question 11:

Find and correct the errors in the statement: (y − 3)2 = y2 − 9

L.H.S = (y − 3)2 = (y)2 − 2(y)(3) + (3)2 [(ab)2 = a2 − 2ab + b2]

= y2 − 6y + 9 ≠ R.H.S

The correct statement is (y − 3)2 = y2 − 6y + 9

#### Question 12:

Find and correct the errors in the statement: (z + 5)2 = z2 + 25

L.H.S = (z + 5)2 = (z)2 + 2(z)(5) + (5)2 [(a + b)2 = a2 + 2ab + b2]

= z2 + 10z + 25 ≠ R.H.S

The correct statement is (z + 5)2 = z2 + 10z + 25

#### Question 13:

Find and correct the errors in the statement: (2a + 3b) (ab) = 2a2 − 3b2

L.H.S. = (2a + 3b) (ab) = 2a × a + 3b × a − 2a × b − 3b × b

= 2a2 + 3ab − 2ab − 3b2 = 2a2 + ab − 3b2 ≠ R.H.S.

The correct statement is (2a + 3b) (ab) = 2a2 + ab − 3b2

#### Question 14:

Find and correct the errors in the statement: (a + 4) (a + 2) = a2 + 8

L.H.S. = (a + 4) (a + 2) = (a)2 + (4 + 2) (a) + 4 × 2

= a2 + 6a + 8 ≠ R.H.S

The correct statement is (a + 4) (a + 2) = a2 + 6a + 8

#### Question 15:

Find and correct the errors in the statement: (a − 4) (a − 2) = a2 − 8

L.H.S. = (a − 4) (a − 2) = (a)2 + [(− 4) + (− 2)] (a) + (− 4) (− 2)

= a2 − 6a + 8 ≠ R.H.S.

The correct statement is (a − 4) (a − 2) = a2 − 6a + 8

#### Question 16:

Find and correct the errors in the statement:

L.H.S =

The correct statement is

#### Question 17:

Find and correct the errors in the statement:

The correct statement is

#### Question 18:

Find and correct the errors in the statement:

L.H.S =

The correct statement is

#### Question 19:

Find and correct the errors in the statement:

L.H.S. = ≠ R.H.S.

The correct statement is

#### Question 20:

Find and correct the errors in the statement:

L.H.S. =

The correct statement is

#### Question 21:

Find and correct the errors in the statement: