Class 8 RD Sharma Solutions Chapter 8 Division Of Algebraic Expressions Exercise 8.4 | Set 2 (original) (raw)
Last Updated : 23 Jul, 2025
RD Sharma's Class 12 Math's Chapter8, covering Division of Algebraic Expressions Exercise 8.4 right here. These solutions are well designed to help the students in grasping the solving techniques for a wide array of problems. Additionally, they incorporate handy shortcuts and real-world examples to facilitate quick comprehension of concepts and expedite the problem-solving learning curve.
Division Of Algebraic Expressions
The division of algebraic expressions entails dividing one algebraic expression by another. This procedure includes dividing every term in the numerator expression by the denominator expression, much like long division in arithmetic. The outcome is a new algebraic expression representing the quotient derived from the division.
Determine the solution to the equation provided below
Question 19. Divide 30x4 + 11x3 - 82x2 – 12x + 48 by 3x2 + 2x - 4
**Solution:
We have to divide 30x4 + 11x3 - 82x2 – 12x + 48 by 3x2 + 2x - 4
So by using long division method we get
Quotient = 10x2 - 3x - 12
Remainder = 0
Question 20. Divide 9x4 – 4x2 + 4 by 3x2 – 4x + 2
**Solution:
We have to divide 9x4 – 4x2 + 4 by 3x2 – 4x + 2
So by using long division method we get
Quotient = 3x2 + 4x + 2
Remainder = 0
Question 21. Verify division algorithm i.e., Dividend = Divisor * Quotient + Remainder, in each of the following. Also, write the quotient and remainder :
(i) Dividend = 14x2 + 13x - 15, Divisor = 7x - 4
**Solution:
Dividing the Dividend by divisor, we get
Quotient = 2x + 3
Remainder = -3
(ii) Dividend = 15z3 - 20z2 + 13z - 12, Divisor = 3z - 6
**Solution:
Dividing the Dividend by divisor, we get
33z - 66
Quotient = 5z2 + (10/3)z + 11
Remainder = 54
(iii) Dividend = 6y5 - 28y3 + 3y2 + 30y - 9, Divisor = 2y2 - 6
**Solution:
Dividing the Dividend by divisor, we get
3y2 - 9
Quotient = 3y3 - 5y + (3/2)
Remainder = 0
(iv) Dividend = 34x - 22x3 - 12x4 - 10x2 - 75, Divisor = 3x + 7
**Solution:
Dividing the Dividend by divisor, we get
Quotient = -4x3 + 2x2 - 8x + 30
Remainder = -285
(v) Dividend = 15y4 - 16y3 + 9y2 - (10/3)y + 6, Divisor = 3y - 2
**Solution:
Dividing the Dividend by divisor, we get
Quotient = 5y3 - 2y2 + (5/3)y
Remainder = 6
(vi) Dividend = 4y3 + 8y + 8y2 + 7, Divisor = 2y2 - y + 1
**Solution:
Dividing the Dividend by divisor, we get
Quotient = 2y + 5
Remainder = 11y + 2
(vii) Dividend = 6y5 + 4y4 + 4y3 + 7y2 + 27y + 6, Divisor = 2y3 + 1
**Solution:
Dividing the Dividend by divisor, we get
Quotient = 3y2 + 2y + 2
Divisor = 4y2 + 25y + 4
Question 22. Divide 15y4 + 16y3 + (10/3)y – 9y2 – 6 by 3y – 2 . Write down the coefficients of the terms in the quotient.
**Solution:
We have to divide 15y4 + 16y3 + (10/3) y – 9y2 – 6 by 3y – 2
So by using long division method we get
Quotient = 5y3 + (26/3)y2 + (25/9)y + (80/27)
Remainder = (-2/27)
Co-efficient of y3 is 5
Co-efficient of y2 is 26/9
Co-efficient of y is 25/9 and,
Constant term = 80/27
Question 23. Using division of polynomials state whether.
(i) x + 6 is a factor of x2 – x – 42
**Solution:
Dividing x2 – x – 42 by x + 6, we get
-7x – 42
Remainder = 0
Therefore, x + 6 is a factor of x2 – x – 42
(ii) 4x – 1 is a factor of 4x2 – 13x – 12
**Solution:
On dividing 4x2 – 13x – 12 by 4x – 1
Remainder = -15
Therefore, 4x-1 is not a factor of 4x2 – 13x – 12
(iii) 2y – 5 is a factor of 4y4 – 10y3 – 10y2 + 30y - 15
**Solution:
On dividing 4y4 – 10y3 – 10y2 + 30y - 15 by 2y - 5, we get
Remainder = -5/2
Therefore, 2y - 5 is not a factor of 4y4 – 10y3 – 10y2 + 30y - 15
(iv) 3y2 + 5 is a factor of 6y5 + 15y4 + 16y3 + 4y2 + 10y – 35
**Solution:
On dividing 6y5 + 15y4 + 16y3 + 4y2 + 10y – 35 by 3y2 + 5, we get
Remainder = 0
Therefore, 3y2 + 5 is a factor of 6y5 + 15y4 + 16y3 + 4y2 + 10y – 35
(v) z2 + 3 is a factor of z5– 9z
**Solution:
On dividing z5 – 9z by z2 + 3, we get
-3z3 - 9z
Remainder = 0
Therefore, z2 + 3 is a factor of z5– 9z
(vi) 2x2 – x + 3 is a factor of 6x5- x4 + 4x3 – 5x2 - x - 15
**Solution:-
On dividing 6x5 - x4 + 4x3 - 5x2 - x - 15 by 2x2 - x + 3
-10x2 + 5x - 15
Remainder = 0
Therefore, 2x2 - x + 3 is a factor of 60x5 - x4 + 4x3 - 5x2 - x - 15
Question 24. Find the value of ‘a’, if x + 2 is a factor of 4x4 + 2x3 – 3x2 + 8x + 5a.
**Solution:
Given that, x + 2 is a factor of 4x4 + 2x3 – 3x2 + 8x + 5a,
On dividing 4x4 + 2x3 – 3x2 + 8x + 5a by x + 2, we get
Remainder = 5a + 20
5a + 20 = 0
a = -4
Question 25. What must be added to x4 + 2x3 – 2x2 + x – 1 so that the resulting polynomial is exactly divisible by x2 + 2x – 3.
**Solution:
On dividing x4 + 2x3 – 2x2 + x – 1 by x2 + 2x – 3, we get
Remainder = 0
The No. added to given polynomial to get remainder 0 will be :
x + 2 = 0
