class 8 maths chapter 14 exercise 14.3

class 8 maths chapter 14 exercise 14.3 (1)

1  Carry out the following divisions :

$\left(i\right) 28x^4÷56x \left(ii\right) –36y^3÷9y^2$

(iii) 66pq²r³ ÷ 11qr²    (iv) 34x³y³z³ ÷ 51xy²z³

$\left(v\right) 12 a^8{b}^8 ÷ (–6a^6{b}^4)$

Solution to class 8 maths chapter 14 exercise 14.3 (1)

$\left(i\right) 28x^4÷56x=\frac{28x^4}{56x}$

$=\frac{28}{56}\times \frac{x^4}{x}=\frac{1}{2}{x}^3$

$\left[\because \frac{x^m}{x^n}=x^{m–n}\right]$

(ii) -36y³ ÷ 9y² 

$=\frac{–36y^3}{9y^2}$

=-4y

(iii) 66pq²r³ ÷ 11qr²

$=\frac{66p{q}^2{r}^3}{11q{r}^2}=\frac{66}{11}\times \frac{p{q}^2{r}^3}{q{r}^2}$

=6pqr

(iv) 34x³y³z³÷51xy²z³

$=\frac{34x^3{y}^3{z}^3}{51x{y}^2{z}^3}=\frac{34}{51}\times \frac{x^3{y}^3{z}^3}{x{y}^2{z}^3}$

$=\frac{2}{3}{x}^{2}y \left[\because \frac{x^m}{x^n}=x^{m–n}\right]$

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

$=–2a^2{b}^4 \left[\because \frac{x^m}{x^n}=x^{m–n}\right]$

class 8 maths chapter 14 exercise 14.3 (2)

(i) (5x² – 6x) ÷ 3x 

$\left(ii\right) (3y^8–4y^6+5y^4) ÷ y^4$

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

(iv) (x³+2x²+3x)÷2x

$\left(v\right) (p^3{q}^6 – p^6{q}^3)÷p^3{q}^3$

Solution to class 8 maths chapter 14 exercise 14.3 (2)

(i) (5x² – 6x) ÷ 3x

$=\frac{5x^2–6x}{3x}$

$=\frac{5x^2}{3x}–\frac{6x}{3x}=\frac{5}{3}x–2=\frac{1}{3}(5x–6)$

$\left(ii\right) (3y^8–4y^6+5y^4) ÷ y^4 = \frac{3y^8–4y^6+5y^4}{y^4}$

$=\frac{3y^8}{y^4}–\frac{4y^6}{y^4}+\frac{5y^4}{y^4}=3y^4–4y^2+5$

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

$=\frac{8(x³y²z²+x²y³z²+x²y²z³)}{4x^2{y}^2{z}^2}$

$=\frac{8x^3{y}^2{z}^2}{4x^2{y}^2{z}^2}+\frac{8x^2{y}^3{z}^2}{4x^2{y}^2{z}^2}+\frac{8x^2{y}^2{z}^3}{4x^2{y}^2{z}^2}$

=2x+2y+2z

=2(x+y+z)

(iv) (x³+2x²+3x) ÷ 2x

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

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

$\left(v\right) (p^3{q}^6–p^6{q}^3) ÷p^3{q}^3=\frac{p^3{q}^6–p^6{q}^3}{p^3{q}^3}$

$=\frac{p^3{q}^6}{p^3{q}^3}–\frac{p^6{q}^3}{p^3{q}^3}=q^3–p^3$

class 8 maths chapter 14 exercise 14.3 (3)

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)

Solution to class 8 maths chapter 14 exercise 14.3 (3)

(i) (10x – 25) ÷ 5

$=\frac{10x–25}{5}=\frac{5(2x–5)}{5}$

=2x – 5

(ii) (10x – 25) ÷ (2x – 5)

$=\frac{10x–25}{2x–5}=\frac{5(2x–5)}{(2x–5)}$

=5

(iii) 10y(6y+21) ÷ 5(2y+7)

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

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

=2×y×3=6y

(iv) 9x²y²(3z – 24) ÷ 27xy(z – 8)

$=\frac{9x^2{y}^2(3z–24)}{27xy(z–8)}$

$=\frac{9}{27}\times \frac{xy\times xy\times 3(z–8)}{xy(z–8)}$

=xy

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

$=\frac{96abc(3a–12)(5b–30)}{144(a–4)(b–6)}$

$=\frac{12\times 4\times 2\times abc\times 3(a–4)\times 5(b–6)}{12\times 4\times 3(a–4)(b–6)}$

=10abc

class 8 maths chapter 14 exercise 14.3 (4)

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)

Solution to class 8 maths chapter 14 exercise 14.3 (4)

(i) 5(2x+1)(3x+5) ÷ (2x+1)

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

=5(3x+5)

(ii) 26xy(x+5)(y – 4) ÷ 13x(y – 4)

$=\frac{26xy(x+5)(y–4)}{13x(y–4)}$

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

=2y(x+5)

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

$=\frac{52pqr(p+q)(q+r)(r+p)}{104pq(q+r)(r+p)}$

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

(iv) 20(y+4)(y²+5y+3) ÷ 5(y+4)

$=\frac{20(y+4)(y^2+5y+3)}{5(y+4)}$

=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)

class 8 maths chapter 14 exercise 14.3 (5)

5  Factorize 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)

Solution to class 8 maths chapter 14 exercise 14.3 (5)

(i) (y²+7y+10) ÷ (y+5)

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

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

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

=y+2

(ii) (m² – 14m – 32) ÷ (m+2)

$=\frac{m^2–14m+32}{m+2}$

$=\frac{m^2–16m+2m+(–16)\times 2}{m+2}$

$=\frac{m(m–16)+2(m–16)}{m+2}$

$=\frac{(m–16)(m+2)}{m+2}$

=m+2

(iii) (5p² – 25p+20)÷(p – 1)

$=\frac{5p^2–25p+20}{p–1}$

$=\frac{5p^2–20p–5p+20}{p–1}$

$=\frac{5p(p–4)–5(p–4)}{p–1}$

$=\frac{5(p–1)(p–4)}{(p–1)}$

=5(p – 4)

(iv) 4yz(z²+6z – 16)÷2y(z+8)

$=\frac{4yz(z^2+6z–16)}{2y(z+8)}$

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

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

$=\frac{4yz(z–2)(z+8)}{2y(z+8)}$

=2z(z – 2)

(v) 5pq(p² – q²) ÷ 2p(p+q)

$=\frac{5pq\left(p^2–q^2\right)}{2p\left(p+q\right)}$

$=\frac{5pq\left(p–q\right)\left(p+q\right)}{2p\left(p+q\right)} \left[\because a^2–b^2=\left(a–b\right)\left(a+b\right)\right]$

$=\frac{5}{2}q\left(p–q\right)$

(vi) 12xy(9x² – 16y²) ÷ 4xy(3x+4y)

$=\frac{12xy\left(9x^2–16y^2\right)}{4xy\left(3x+4y\right)}$

$=\frac{12xy\left[{\left(3x\right)}^2–{\left(4x\right)}^2\right]}{4xy\left(3x+4y\right)}$

$=\frac{12xy\left(3x–4y\right)\left(3x+4y\right)}{4xy\left(3x+4y\right)} \left[\because a^2–b^2=\left(a–b\right)\left(a+b\right)\right]$

=3(3x – 4y)

(vii) 39y³(50y² – 98) ÷ 26y²(5y+7)

$=\frac{39y^3\left(50y^2–98\right)}{26y^2\left(5y+7\right)}$

$=\frac{39y^3\times 2\left(25y^2–49\right)}{26y^2\left(5y+7\right)}$

$=\frac{39y^2\times 2\left[{\left(5y\right)}^2–{\left(7\right)}^2\right]}{26y^2\left(5y+7\right)}$

$=\frac{39y^2\times 2\left(5y–7\right)\left(5y+7\right)}{26y^2\left(5y+7\right)}$

$\left[\because a^2–b^2=\left(a–b\right)\left(a+b\right)\right]$

=3y(5y – 7)