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.
- What is Factorisation?
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.
Now the question arises What is Regrouping?
- Rearranging the expression allows us to form groups leading to factorisation. This is called regrouping.
- Division of Algebraic Expressions: This section is divided into the following sub-sections:
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?
- 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.
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.
Page No 220:
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
Answer:
(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
Page No 220:
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
Answer:
(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) + b − c]
= 4a (−a + b − c)
(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)
Page No 220:
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 + 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 + bx − ay − by = a × x + b × x − a × y − b × y
= x (a + b) − y (a + b)
= (a + b) (x − y)
(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 + 7xy − xyz = z − x × 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
Page No 223:
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
Answer:
(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 [(a − b)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 [(a − b)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 [(a − b)2 = a2 − 2ab + b2]
(vii) (l + m)2 − 4lm = l2 + 2lm + m2 − 4lm
= l2 − 2lm + m2
= (l − m)2 [(a − b)2 = a2 − 2ab + b2]
(viii) a4 + 2a2b2 + b4 = (a2)2 + 2 (a2) (b2) + (b2)2
= (a2 + b2)2 [(a + b)2 = a2 + 2ab + b2]
Page No 223:
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(9a2 − 16b2)
= 7[(3a)2 − (4b)2]
= 7(3a + 4b) (3a − 4b) [a2 − b2 = (a − b) (a + b)]
(iii) 49x2 − 36 = (7x)2 − (6)2
= (7x − 6) (7x + 6) [a2 − b2 = (a − b) (a + b)]
(iv) 16x5 − 144x3 = 16x3(x2 − 9)
= 16 x3 [(x)2 − (3)2]
= 16 x3(x − 3) (x + 3) [a2 − b2 = (a − b) (a + b)]
(v) (l + m)2 − (l − m)2 = [(l + m) − (l − m)] [(l + m) + (l − m)]
[Using identity a2 − b2 = (a − b) (a + b)]
= (l + m − l + m) (l + m + l − m)
= 2m × 2l
= 4ml
= 4lm
(vi) 9x2y2 − 16 = (3xy)2 − (4)2
= (3xy − 4) (3xy + 4) [a2 − b2 = (a − b) (a + b)]
(vii) (x2 − 2xy + y2) − z2 = (x − y)2 − (z)2 [(a − b)2 = a2 − 2ab + b2]
= (x − y − z) (x − y + z) [a2 − b2 = (a − b) (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 (a − b)2 = a2 − 2ab + b2]
= [5a + (2b − 7c)] [5a − (2b − 7c)]
[Using identity a2 − b2 = (a − b) (a + b)]
= (5a + 2b − 7c) (5a − 2b + 7c)
Page No 223:
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
Answer:
(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
Page No 224:
Question 4:
Factorise
(i) a4 − b4
(ii) p4 − 81
(iii) x4 − (y + z)4
(iv) x4 − (x − z)4
(v) a4 − 2a2b2 + b4
Answer:
(i) a4 − b4 = (a2)2 − (b2)2
= (a2 − b2) (a2 + b2)
= (a − b) (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]
= (x − y − z) (x + y + z) [x2 + (y + z)2]
(iv) x4 − (x − z)4 = (x2)2 − [(x − z)2]2
= [x2 − (x − z)2] [x2 + (x − z)2]
= [x − (x − z)] [x + (x − z)] [x2 + (x − z)2]
= z(2x − z) [x2 + x2 − 2xz + z2]
= z(2x − z) (2x2 − 2xz + z2)
(v) a4 − 2a2b2 + b4 = (a2)2 − 2 (a2) (b2) + (b2)2
= (a2 − b2)2
= [(a − b) (a + b)]2
= (a − b)2 (a + b)2
Page No 224:
Question 5:
Factorise the following expressions
(i) p2 + 6p + 8
(ii) q2 − 10q + 21
(iii) p2 + 6p − 16
Answer:
(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
Page No 227:
Question 1:
Carry out the following divisions.
(i) 28x4 ÷ 56x
(ii) −36y3 ÷ 9y2
(iii) 66pq2r3 ÷ 11qr2
(iv) 34x3y3z3 ÷ 51xy2z3
(v) 12a8b8 ÷ (−6a6b4)
Answer:
(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
Page No 227:
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) (p3q6 − p6q3) ÷ p3q3
Answer:
(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) p3q6− p6q3 = p3q3(q3 − p3)
Page No 227:
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)
Answer:
(i)
(ii)
(iii)
(iv)
(v) 96 abc(3a − 12) (5b − 30) ÷ 144 (a − 4) (b − 6)
Page No 227:
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)
Answer:
(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)
Page No 227:
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(p2 − q2) ÷ 2p(p + q)
(vi) 12xy(9x2 − 16y2) ÷ 4xy(3x + 4y)
(vii) 39y3(50y2− 98) ÷ 26y2(5y+ 7)
Answer:
(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[p2 − p − 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 (p − q) (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
Page No 228:
Question 1:
Find and correct the errors in the statement: 4(x − 5) = 4x − 5
Answer:
L.H.S. = 4(x − 5) = 4 × x − 4 × 5 = 4x − 20 ≠ R.H.S.
The correct statement is 4(x − 5) = 4x − 20
Page No 228:
Question 2:
Find and correct the errors in the statement: x(3x + 2) = 3x2 + 2
Answer:
L.H.S. = x(3x + 2) = x × 3x + x × 2 = 3x2 + 2x ≠ R.H.S.
The correct statement is x(3x + 2) = 3x2 + 2x
Page No 228:
Question 3:
Find and correct the errors in the statement: 2x + 3y = 5xy
Answer:
L.H.S = 2x + 3y ≠ R.H.S.
The correct statement is 2x + 3y = 2x + 3y
Page No 228:
Question 4:
Find and correct the errors in the statement: x + 2x + 3x = 5x
Answer:
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
Page No 228:
Question 5:
Find and correct the errors in the statement: 5y + 2y + y − 7y = 0
Answer:
L.H.S. = 5y + 2y + y − 7y = 8y − 7y = y ≠ R.H.S
The correct statement is 5y + 2y + y − 7y = y
Page No 228:
Question 6:
Find and correct the errors in the statement: 3x + 2x = 5x2
Answer:
L.H.S. = 3x + 2x = 5x ≠ R.H.S
The correct statement is 3x + 2x = 5x
Page No 228:
Question 7:
Find and correct the errors in the statement: (2x)2 + 4(2x) + 7 = 2x2 + 8x + 7
Answer:
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
Page No 228:
Question 8:
Find and correct the errors in the statement: (2x)2 + 5x = 4x + 5x = 9x
Answer:
L.H.S = (2x)2 + 5x = 4x2 + 5x ≠ R.H.S.
The correct statement is (2x)2 + 5x = 4x2 + 5x
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Question 9:
Find and correct the errors in the statement: (3x + 2)2 = 3x2 + 6x + 4
Answer:
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
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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
Answer:
(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
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Question 11:
Find and correct the errors in the statement: (y − 3)2 = y2 − 9
Answer:
L.H.S = (y − 3)2 = (y)2 − 2(y)(3) + (3)2 [(a − b)2 = a2 − 2ab + b2]
= y2 − 6y + 9 ≠ R.H.S
The correct statement is (y − 3)2 = y2 − 6y + 9
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Question 12:
Find and correct the errors in the statement: (z + 5)2 = z2 + 25
Answer:
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
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Question 13:
Find and correct the errors in the statement: (2a + 3b) (a − b) = 2a2 − 3b2
Answer:
L.H.S. = (2a + 3b) (a − b) = 2a × a + 3b × a − 2a × b − 3b × b
= 2a2 + 3ab − 2ab − 3b2 = 2a2 + ab − 3b2 ≠ R.H.S.
The correct statement is (2a + 3b) (a − b) = 2a2 + ab − 3b2
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Question 14:
Find and correct the errors in the statement: (a + 4) (a + 2) = a2 + 8
Answer:
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
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Question 15:
Find and correct the errors in the statement: (a − 4) (a − 2) = a2 − 8
Answer:
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
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Question 16:
Find and correct the errors in the statement:
Answer:
L.H.S =
The correct statement is
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Question 17:
Find and correct the errors in the statement:
Answer:
The correct statement is
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Question 18:
Find and correct the errors in the statement:
Answer:
L.H.S =
The correct statement is
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Question 19:
Find and correct the errors in the statement:
Answer:
L.H.S. =
≠ R.H.S.
The correct statement is
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Question 20:
Find and correct the errors in the statement:
Answer:
L.H.S. =
The correct statement is
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Question 21:
Find and correct the errors in the statement:
Answer:
L.H.S. =
The correct statement is
View NCERT Solutions for all chapters of Class 8