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 8.5 Download 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
= y² (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(y² – 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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