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

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.5

Other Exercises

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
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.5 1
(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

Hope given RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.5 are helpful to complete your math homework.

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