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class 8 maths exercise 14.3 solutions

In this page we have class 8 maths exercise 14.3 solutions for Factorisation.This exercise has questions on Division of Algebraic Expressions by factorisation method.It includes division of polynomials by monomial, binomial and polynomails . Hope you like them and do not forget to like , social share and comment at the end of the page.

class 8 maths exercise 14.3 solutions

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 ÷ 56x
= (2×2×7×x×x³ ) /( 2×2×2×7×x)
=x³/2
(ii) –36y³ ÷ 9y²
= (-2×2×3×3×y×y² ) /( 3×3×y²)
=-4y
(iii) 66pq²r³ ÷ 11qr²
= 6pqr
(iv) 34x³y³z³ ÷ 51xy²z³
=(2/3) x²y
(v) 12a⁸b⁸ ÷ (– 6a⁶b⁴)
=-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) ÷ 3x =[x(5x-6)] / 3x
Cancelling x
=(5x-6)/3
(ii)(3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴ = y⁴(3y⁴ -4y² +5)/ y⁴
=3y⁴-4y²+5
(iii)8(x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z² =8x² y²z²(x+y+z)/ 4x²y²z²
= 2(x+y+z)
(iv)(x³ + 2x² + 3x) ÷ 2x =x(x² +2x+3)/2x
=( x² +2x+3)/2
(v)(p³q⁶ – p⁶q³) ÷ p³q³ = p³ q³(q³ - p³) / 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) (10x – 25) ÷ 5
=5(2x-5)/5
=(2x-5)
(ii) (10x – 25) ÷ (2x – 5)
= 5(2x-5) /(2x – 5)
=-5
(iii) 10y(6y + 21) ÷ 5(2y + 7)
=30y(2y+7)/ 5(2y + 7)
= 6y
(iv) 9x²y² (3z – 24) ÷ 27xy(z – 8)
=27 x²y²(z-8)/ 27xy(z – 8)
=xy
(v) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4) (b – 6)
=(96×3×5)abc(a-4)(b-6)/ 144(a – 4) (b – 6)
=10abc

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(2x + 1) (3x + 5) ÷ (2x + 1)
=5(2x + 1) (3x + 5) / (2x + 1)
=5(3x+5)
(ii) 26xy(x + 5) (y - 4) ÷ 13x(y - 4)
=26xy(x + 5) (y - 4) / 13x(y - 4)
Cancelling 13x(y-4)
=2y(x+5)
(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)
=52pqr (p + q) (q + r) (r + p)/ 104pq(q + r) (r + p)
Cancelling 52pq(q + r) (r + p)
=r(p+q)/2
(iv) 20(y + 4) (y² + 5y + 3) ÷ 5(y + 4)
=4(y² + 5y + 3)
v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)
= x(x + 1) (x + 2) (x + 3) / x(x + 1)
Cancelling x(x + 1)
=(x + 2) (x + 3)

Question 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)
Answer
(i)(y² + 7y + 10) ÷ (y + 5)
=( y² + 5y+2y + 10)/(y+5)
=[y(y+5)+2(y+5)] /(y+5)
=(y+1)(y+5) /(y+5)
=(y+1)
(ii) (m² - 14m - 32) ÷ (m + 2)
= (m² - 16m+2m - 32) / (m + 2)
=[m(m-16)+2(m-16)] /(m+2)
=(m+2)(m-16) /(m+2
=(m-16)
(iii) (5p² - 25p + 20) ÷ (p - 1)
=5(p²-5p+4)/(p-1)
=5[p(p-1) -4(p-1)] /(p-1)
=5(p-1)(p-4)/(p-1)
=5(p-4)
(iv) 4yz(z² + 6z - 16) ÷ 2y(z + 8)
=4yz[z² -2z+8z-16]/ 2y(z + 8)
=4yz[z(z-2)+8(z-2)] / 2y(z + 8)
=4yz(z-2)(z+8) / 2y(z + 8)
=2z(z-2)
(v) 5pq(p² - q²) ÷ 2p(p + q)
=5pq(p² - q²) / 2p(p + q)
=5pq(p-q)(p+q) / 2p(p + q)
=5q(p-q)/2
(vi) 12xy(9x² - 16y²) ÷ 4xy(3x + 4y)
=12xy(9x² - 16y²) / 4xy(3x + 4y)
=12xy(3x+4y)(3x-4y) / 4xy(3x + 4y)
=3(3x-4y)
(vii) 39y³(50y²- 98) ÷ 26y²(5y+ 7)
=39y³(50y²- 98) /26y²(5y+ 7)
=78y³(25y²-49)/ 26y²(5y+ 7)
=3y(5y+7)(5y-7)/ (5y+ 7)
=3y(5y-7)

Summary

class 8 maths exercise 14.3 solutions has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also
Download Factorisation Exercise 14.3 pdf
This chapter 14 has total 4 Exercise 14.1,14.2,14.3 and 14.4. This is the Third exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below

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