14- Factorisation

Exercise -14.1

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

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)

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)

Exercise -14.2

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]

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)

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)

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

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)

Exercise -14.3

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

^{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)

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)

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)

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)

Exercise -14.4

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

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

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

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

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

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

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

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

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

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

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

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

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

^{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

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

Question 16- Find and correct the errors in the statement: 

Answer - L.H.S = 

The correct statement is 

Question 17- Find and correct the errors in the statement: 

Answer -

The correct statement is 

Question 18- Find and correct the errors in the statement: 

Answer - L.H.S =

The correct statement is 

Question 19- Find and correct the errors in the statement: 

Answer - L.H.S. = ≠ R.H.S.

The correct statement is 

Question 20- Find and correct the errors in the statement: 

Answer - L.H.S. =

The correct statement is 

Question 21- Find and correct the errors in the statement: 

Answer - L.H.S. = 
The correct statement is