Class 8 RD Sharma Solutions – Chapter 8 Division Of Algebraic Expressions – Exercise 8.4 | Set 2

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