# RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.5

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### RD Sharma Solutions for Class 8 Chapter 8 Division of Algebraic Expressions Ex 8.5Download PDF

Question 1.
Divide the first polynomial by the second polynomial in each of the following. Also, write the quotient and remainder.
(i) 3x2 + 4x + 5, x – 2
(ii) 10x2 – 7x + 8, 5x – 3
(iii) 5y3– 6y2 + 6y-1,5y-1
(iv)x4-x3 + 5x,x-1
(v) y4 +y2,y2-2
Solution:
(i) 3x2 + 4x + 5, x – 2
= 3x (x – 2) + 10x + 5
= 3x (x – 2) + 10 (x – 2) + 25
∴ Quotient = 3x + 10
Remainder = 25

(iii) 5y3 – 6y2 + 6y – 1, 5y – 1
= y(5y – 1) – 5y2 + 6y- 1
= y2 (5y – 1) -y (5y – 1) + 5y – 1
= y2 (5y- 1) -y (5y- 1) + 1 (5y- 1)
∴ Quotient = y2 – y + 1 and Remainder = 0
(iv) x4 – x3 + 5x, x – 1
= x3(x – 1) + 5x
= x3 (x – 1) + 5 (x – 1) + 5
∴ Quotient = x3 + 5, Remainder = 5
(v) y4+y2,y2– 2
= y2(y– 2) + 3y2
= y2 (y2 – 2) + 3 (y2 – 2) + 6
∴ Quotient =y2 + 3 and Remainder = 6

Question 2.
Find, whether or not the first polynomial is a factor of the second :

(i) x + 1, 2x2 + 5x + 4
(ii) y- 2, 3y3 + 5y2 + 5y + 2
(iii) 4x2 – 5, 4.x4 + 7x2 + 15
(iv) 4-z, 3z2 – 13z + 4
(v) 2a-3,10a2 – 9a – 5
(vi) 4y+1 ,8y2-2y + 1
Solution:
(i) x + 1, 2x2 + 5x + 4
2x2 + 5x + 4 = 2x (x + 1) + 3x + 4
= 2x (x + 1) + 3 (x + 1) + 1
∵ Remainder = 1
∴ x + 1 is not a factor of 2x2 + 5x + 4
(ii) y – 2, 3y3 + 5y2 + 5y + 2
3y3 + 5y2 + 5y + 2 = 3y2(y – 2)+11y2 + 5y + 2
= 3y2(y – 2)+11y (y – 2) + 27y + 2
= 3y2 (y – 2) + 11y (y – 2) + 27 (y – 2) + 56
∵ Remainder = 56
∴ y – 2 is not a factor of 3y3 + 5y2 + 5y + 2
(iii) 4x2 – 5, 4x4 + 7x2 + 15
4x4 + 7x2 + 15 = x2 (4x2 – 5) + 12x2 + 15
= x2 (4x2 – 5) + 3 (4x2 – 5) + 30
∵ Remainder = 30
∴ 4x2 – 5 is not a factor of 4x4 + 7x2 + 15
(iv) 4 – z, 3z2 – 13z + 4
3z2 – 13z + 4 = -3z (-z + 4) – z + 4
= -3z (-z + 4) + 1 (-z + 4)
∵ Remainder = 0
∴ 4 – z or – z + 4 is a factor of 3z2 – 13z + 4
(v) 2a – 3, 10a2 – 9a – 5
10a2 – 9a – 5 = 5a (2a – 3) + 6a – 5
= 5a (2a – 3) + 3 (2a – 3) + 4
∵ Remainder = 4
∴ 2a – 3 is not a factor of 10a2 – 9a – 5
(vi) 4y + 1, 8y2 – 2y + 1
8y2 – 2y + 1 = 2y (4y + 1) – 4y + 1
= 2y (4y + 1) – 1 (4y + 1) + 2
∵ Remainder = 2
∴ 4y + 1 is not a factor of 8y2 – 2y + 1

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