Factorisation

EXERCISE-14.1

P1. 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

Sol. (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

P2. 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

Sol. (i) 7x = 7 × x

42 = 2 × 3 × 7

The common factor is 7.

$\therefore$  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.

$\therefore$  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.

$\therefore$ 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.

$\therefore$ −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.

$\therefore$ 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.

$\therefore$  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.

$\therefore$ −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.

$\therefore$  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)

P3. 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

Sol. (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

P1. 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                      (viii) a⁴ + 2a²b² + b⁴

(Hint: Expand (l + m)² first)

Sol. (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]

P2. 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

Sol. (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)

P3. 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

Sol. (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)

P4. Factorise

(i) a⁴ − b⁴                       

(ii) p⁴ − 81

(iii) x⁴ − (y + z)⁴       

(iv) x⁴ − (x − z)⁴

(v) a⁴ − 2a²b² + b⁴

Sol. (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

P5. Factorise the following expressions

(i) p² + 6p + 8

(ii) q² − 10q + 21

(iii) p² + 6p − 16

Sol. (i) p² + 6p + 8

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

$\therefore$ 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

$\therefore$ 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

P1. Carry out the following divisions.

(i) 28x⁴ ÷ 56x

(ii) −36y³ ÷ 9y²

(iii) 66pq²r³ ÷ 11qr²

(iv) 34x³y³z³ ÷ 51xy²z³

(v) 12a⁸b⁸ ÷ (−6a⁶b⁴)

Sol. (i) 28x⁴ = 2 × 2 × 7 × x × x × x × x

56x = 2 × 2 × 2 × 7 × x

$28x^4÷56x=\frac{2\times 2\times 7\times x\times x\times x\times x}{2\times 2\times 2\times 7\times x}$

$=\frac{x^3}{2}=\frac{1}{2}{x}^3$

 (ii) 36y³ = 2 × 2 × 3 × 3 × y × y × y

9y² = 3 × 3 × y × y

$–36y^3÷9y^2=\frac{–2\times 2\times 3\times 3\times y\times y\times y}{3\times 3\times y\times y}$

$=–4 y$

(iii) 66 pq² r³ 

= 2 × 3 × 11 × p × q × q × r × r × r

11qr² = 11 × q × r × r

$66{\mathrm{pq}}^2{\mathrm{r}}^3÷11{\mathrm{qr}}^2$

$=\frac{2\times 3\times 11\times \mathrm{p}\times \mathrm{q}\times \mathrm{q}\times \mathrm{r}\times \mathrm{r}\times \mathrm{r}}{11\times \mathrm{q}\times \mathrm{r}\times \mathrm{r}}$

= 6pqr

(iv) 34 x³y³z³

 = 2 × 17 × x × x × x × y × y × y × z × z × z

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

$34x^3{y}^3{z}^3÷51x{y}^2{z}^3$

$=\frac{2\times 17\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}\times \mathrm{y}\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}}{3\times 17\times \mathrm{x}\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}}$

$=\frac{2}{3}{x}^{2}y$

(v) 12a⁸b⁸ = 2 × 2 × 3 × a⁸ × b⁸

6a⁶b⁴ = 2 × 3 × a⁶ × b⁴

$12a^8{b}^8÷\left(–6a^6{b}^4\right)=\frac{2\times 2\times 3\times a^8\times b^8}{–2\times 3\times a^6\times b^4}$

= −2a²b⁴

P2. 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

Sol. (i) 5x² − 6x = x(5x − 6)

(5x² – 6x)$÷$3x

$=\frac{x(5x–6)}{3x}=\frac{1}{3}(5x–6)$

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

(3y⁸ – 4y⁶ + 5y⁴)$÷$ y⁴

$=\frac{y^4\left(3y^4–4y^2+5\right)}{y^4}$

= 3y⁴ − 4y² + 5

(iii) 8(x³y²z² + x²y³z² + x²y²z³)

= 8x²y²z²(x + y + z)

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

$=\frac{8x^2{y}^2{z}^2(x+y+z)}{4x^2{y}^2{z}^2}$

(iv) x³ + 2x² + 3x = x(x² + 2x + 3)

(x³ – 2x² + 3x)$÷$2x

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

$=\frac{1}{2}$(x² + 2x + 3)

(v) p³q⁶− p⁶q³ = p³q³(q³ − p³)

(p³q⁶ – p⁶q³)$÷$p³q³

$=\frac{p^3{q}^3\left(q^3–p^3\right)}{p^3{q}^3}$

= q³ – p³

P3. 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)

Sol. (i) (10x − 25) $÷$ 5

$=\frac{2\times 5\times \mathrm{x}–5\times 5}{5}$

$=\frac{5(2x–5)}{5}=2x–5$

(ii) (10x − 25) $÷$(2x − 5)

$=\frac{2\times 5\times \mathrm{x}–5\times 5}{2\mathrm{x}–5}$

$=\frac{5(2x–5)}{2x–5}=5$

(iii) 10y(6y + 21) $÷$ 5(2y + 7)

$=\frac{2\times 5\times \mathrm{y}[2\times 3\times \mathrm{y}+3\times 7]}{5(2\mathrm{y}+7)}$

$=\frac{2\times 5\times \mathrm{y}\times 3(2\mathrm{y}+7)}{5(2\mathrm{y}+7)}=6\mathrm{y}$

(iv) 9x²y²(3z − 24) $÷$ 27xy(z − 8)

$=\frac{9x^2{y}^2[3\times z–2\times 2\times 2\times 3]}{27xy(z–8)}$

$=\frac{xy\times 3(z–8)}{3(z–8)}=xy$

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

$=\frac{96abc(3\times a–3\times 4)(5\times b–2\times 3\times 5)}{144(a–4)(b–6)}$

$=\frac{2abc\times 3(a–4)\times 5(b–6)}{3(a–4)(b–6)}$

= − 10abc

P4. 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)

Sol. (i) 5(2x + 1) (3x + 5) $÷$ (2x + 1)

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

= 5(3x + 1)

(ii) 26xy(x + 5) (y − 4) $÷$ 13x(y − 4)

$=\frac{2\times 13\times xy(x+5)(y–4)}{13x(y–4)}$

= 2y (x + 5)

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

$=\frac{2\times 2\times 13\times \mathrm{p}\times \mathrm{q}\times \mathrm{r}\times (\mathrm{p}+\mathrm{q})\times (\mathrm{q}+\mathrm{r})\times (\mathrm{r}+\mathrm{p})}{2\times 2\times 2\times 13\times \mathrm{p}\times \mathrm{q}\times (\mathrm{q}+\mathrm{r})\times (\mathrm{r}+\mathrm{p})}$

$=\frac{1}{2}\mathrm{r}(\mathrm{p}+\mathrm{q})$

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

20(y + 4) (y² + 5y + 3) $÷$ 5(y + 4)

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

= 4(y² + 5y + 3)

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

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

= (x + 2) (x + 3)

P5. 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)

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

= y (y + 2) + 5 (y + 2)

= (y + 2) (y + 5)

(y² + 7y + 10) $÷$ (y + 5)

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

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

= m (m + 2) − 16 (m + 2)

= (m + 2) (m − 16)

(m² − 14m − 32)$÷$(m + 2)

$=\frac{(m+2)(m–16)}{(m+2)}=m–16$

(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)

(5p² − 25p + 20) $÷$ (p – 1)

$=\frac{5(p–1)(p–4)}{(p–1)}=5\left(p–4\right)$

(iv) 4yz(z² + 6z −16)

= 4yz [z² − 2z + 8z − 16]

= 4yz [z(z − 2) + 8(z − 2)]

= 4yz(z − 2) (z + 8)

4yz(z²+6z−16)$÷$2y(z + 8)

$=\frac{4yz(z–2)(z+8)}{2y(z+8)}=2z\left(z–2\right)$

(v) 5pq(p² − q²)

= 5pq (p − q) (p + q)

5pq(p² − q²)$÷$2p(p + q)

$=\frac{5\mathrm{pq}(\mathrm{p}–\mathrm{q})(\mathrm{p}+\mathrm{q})}{2\mathrm{p}(\mathrm{p}+\mathrm{q})}$

$=\frac{5}{2}\mathrm{q}(\mathrm{p}–\mathrm{q})$

 (vi) 12xy(9x² − 16y²)

= 12xy[(3x)² − (4y)²]

= 12xy(3x − 4y) (3x + 4y)

12xy(9x²−16y²)$÷$ 4xy(3x + 4y)

$=\frac{2\times 2\times 3\times \mathrm{x}\times \mathrm{y}\times (3\mathrm{x}–4\mathrm{y})\times (3\mathrm{x}+4\mathrm{y})}{2\times 2\times \mathrm{x}\times \mathrm{y}\times (3\mathrm{x}+4\mathrm{y})}$

= 3(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

P1. Find and correct the errors in the statement: 4(x − 5) = 4x − 5

Sol. L.H.S. = 4(x − 5)

= 4 × x − 4 × 5

= 4x − 20 ≠ R.H.S.

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

P2. Find and correct the errors in the statement: x(3x + 2) = 3x² + 2

Sol. L.H.S. = x(3x + 2)

= x × 3x + x × 2

= 3x² + 2x ≠ R.H.S.

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

P3. Find and correct the errors in the statement: 2x + 3y = 5xy

Sol. L.H.S = 2x + 3y ≠ R.H.S.

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

P4. Find and correct the errors in the statement: x + 2x + 3x = 5x

Sol. 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

P5. Find and correct the errors in the statement: 5y + 2y + y − 7y = 0

Sol. L.H.S. = 5y + 2y + y − 7y 

= 8y − 7y 

= y ≠ R.H.S

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

P6. Find and correct the errors in the statement: 3x + 2x = 5x²

Sol. L.H.S. = 3x + 2x 

= 5x ≠ R.H.S

The correct statement is 3x + 2x = 5x

P7. Find and correct the errors in the statement: (2x)² + 4(2x) + 7 = 2x² + 8x + 7

Sol. 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

P8. Find and correct the errors in the statement: (2x)² + 5x = 4x + 5x = 9x

Sol. L.H.S = (2x)² + 5x 

= 4x² + 5x ≠ R.H.S.

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

P9. Find and correct the errors in the statement: (3x + 2)² = 3x² + 6x + 4

Sol. 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

P10 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

Sol. (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

P11. Find and correct the errors in the statement: (y − 3)² = y² − 9

Sol. 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

P12. Find and correct the errors in the statement: (z + 5)² = z² + 25

Sol. 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

P13. Find and correct the errors in the statement: (2a + 3b) (a − b) = 2a² − 3b²

Sol. 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

P14. Find and correct the errors in the statement: (a + 4) (a + 2) = a² + 8

Sol. 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

P15. Find and correct the errors in the statement: (a − 4) (a − 2) = a² − 8

Sol. 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

P16. Find and correct the errors in the statement: $\mathbf{=}\frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}}$=0

Sol. L.H.S $=\frac{3x^2}{3x^2}$

$=\frac{3\times \mathrm{x}\times \mathrm{x}}{3\times \mathrm{x}\times \mathrm{x}}$= 1 $\ne$ R.H.S

The correct statement is $\frac{3x^2}{3x^2}=1$

P17. Find and correct the errors in the statement: $\frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{2}$

Sol. L.H.S $=\frac{3x^2+1}{3x^2}$

$=\frac{3x^2}{3x^2}+\frac{1}{3x^2}$

$=1+\frac{1}{3{\mathrm{x}}^2}$

 = 1$\ne$R.H.S

The correct statement is $\frac{3x^2+1}{3x^2}=1+\frac{1}{3x^2}$

P18. Find and correct the errors in the statement: $\frac{\mathbf{3}\mathbf{x}}{\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$

Sol. L.H.S =$\frac{3x}{3x+2}\ne R·H·S$

The correct statement is $\frac{3x}{3x+2}=\frac{3x}{3x+2}$

P19. Find and correct the errors in the statement: $\frac{\mathbf{3}}{\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{x}}$

Sol. L.H.S. =$\frac{3}{4x+3}\ne R·H·S$

The correct statement is $\frac{3}{4x+3}=\frac{3}{4x+3}$

P20. Find and correct the errors in the statement: $\frac{\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{5}}{\mathbf{4}\mathbf{x}}\mathbf{=}\mathbf{5}$

Sol. L.H.S. = $\frac{4x+5}{4x}=\frac{4x}{4x}+=\frac{5}{4x}$

$=1+\frac{5}{4\mathrm{x}}\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.$

The correct statement is $\frac{4x+5}{4x}=1+\frac{5}{4x}$

P21. Find and correct the errors in the statement: $\frac{\mathbf{7}\mathbf{x}\mathbf{+}\mathbf{5}}{\mathbf{5}}\mathbf{=}\mathbf{7}\mathit{x}$

Sol. L.H.S.$=\frac{7x+5}{5}=\frac{7x}{5}+\frac{5}{5}$$=\frac{7\mathrm{x}}{5}+1\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.$

The correct statement is  $\frac{7x+5}{5}=\frac{7x}{5}+1$