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

### Question 19. Divide 30x^{4} + 11x^{3 }– 82x^{2 }– 12x + 48 by 3x^{2} + 2x – 4

**Solution:**

We have to divide 30x

^{4}+ 11x^{3 }– 82x^{2 }– 12x + 48 by 3x^{2}+ 2x – 4So by using long division method we get

Quotient = 10x

^{2 }– 3x – 12Remainder = 0

### Question 20. Divide 9x^{4} – 4x^{2} + 4 by 3x^{2} – 4x + 2

**Solution:**

We have to divide 9x

^{4}– 4x^{2}+ 4 by 3x^{2}– 4x + 2So by using long division method we get

Quotient = 3x

^{2 }+ 4x + 2Remainder = 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 = 14x^{2 }+ 13x – 15, Divisor = 7x – 4

**Solution:**

Dividing the Dividend by divisor, we get

Quotient = 2x + 3

Remainder = -3

### (ii) Dividend = 15z^{3 }– 20z^{2 }+ 13z – 12, Divisor = 3z – 6

**Solution:**

Dividing the Dividend by divisor, we get

33z – 66

Quotient = 5z

^{2 }+ (10/3)z + 11Remainder = 54

### (iii) Dividend = 6y^{5 }– 28y^{3 }+ 3y^{2 }+ 30y – 9, Divisor = 2y^{2} – 6

**Solution:**

Dividing the Dividend by divisor, we get

3y

^{2 }– 9Quotient = 3y

^{3 }– 5y + (3/2)Remainder = 0

### (iv) Dividend = 34x – 22x^{3 }– 12x^{4 }– 10x^{2 }– 75, Divisor = 3x + 7

**Solution:**

Dividing the Dividend by divisor, we get

Quotient = -4x

^{3}+ 2x^{2 }– 8x + 30Remainder = -285

### (v) Dividend = 15y^{4 }– 16y^{3 }+ 9y^{2 }– (10/3)y + 6, Divisor = 3y – 2

**Solution:**

Dividing the Dividend by divisor, we get

Quotient = 5y

^{3 }– 2y^{2 }+ (5/3)yRemainder = 6

### (vi) Dividend = 4y^{3 }+ 8y + 8y^{2 }+ 7, Divisor = 2y^{2 }– y + 1

**Solution:**

Dividing the Dividend by divisor, we get

Quotient = 2y + 5

Remainder = 11y + 2

### (vii) Dividend = 6y^{5 }+ 4y^{4 }+ 4y^{3 }+ 7y^{2 }+ 27y + 6, Divisor = 2y^{3 }+ 1

**Solution:**

Dividing the Dividend by divisor, we get

Quotient = 3y

^{2 }+ 2y + 2Divisor = 4y

^{2 }+ 25y + 4

### Question 22. Divide 15y^{4} + 16y^{3} + (10/3)y – 9y^{2} – 6 by 3y – 2 . Write down the coefficients of the terms in the quotient.

**Solution:**

We have to divide 15y

^{4}+ 16y^{3}+ (10/3) y – 9y^{2}– 6 by 3y – 2So by using long division method we get

Quotient = 5y

^{3 }+ (26/3)y^{2 }+ (25/9)y + (80/27)Remainder = (-2/27)

Co-efficient of y

^{3}is 5Co-efficient of y

^{2}is 26/9Co-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 x^{2} – x – 42

**Solution:**

Dividing x

^{2}– x – 42 by x + 6, we get-7x – 42

Remainder = 0

Therefore, x + 6 is a factor of x

^{2}– x – 42

### (ii) 4x – 1 is a factor of 4x^{2} – 13x – 12

**Solution:**

On dividing 4x

^{2 }– 13x – 12 by 4x – 1Remainder = -15

Therefore, 4x-1 is not a factor of 4x

^{2}– 13x – 12

### (iii) 2y – 5 is a factor of 4y^{4} – 10y^{3 }– 10y^{2} + 30y – 15

**Solution:**

On dividing 4y

^{4}– 10y^{3}– 10y^{2}+ 30y – 15 by 2y – 5, we getRemainder = -5/2

Therefore, 2y – 5 is not a factor of 4y

^{4}– 10y^{3 }– 10y^{2 }+ 30y – 15

### (iv) 3y^{2} + 5 is a factor of 6y^{5} + 15y^{4} + 16y^{3} + 4y^{2} + 10y – 35

**Solution:**

On dividing 6y

^{5}+ 15y^{4}+ 16y^{3}+ 4y^{2}+ 10y – 35 by 3y^{2}+ 5, we getRemainder = 0

Therefore, 3y

^{2 }+ 5 is a factor of 6y^{5}+ 15y^{4}+ 16y^{3}+ 4y^{2}+ 10y – 35

### (v) z^{2} + 3 is a factor of z^{5}– 9z

**Solution:**

On dividing z

^{5 }– 9z by z^{2 }+ 3, we get-3z

^{3 }– 9zRemainder = 0

Therefore, z

^{2}+ 3 is a factor of z^{5}– 9z

### (vi) 2x^{2} – x + 3 is a factor of 6x^{5}– x^{4} + 4x^{3} – 5x^{2} – x – 15

**Solution:-**

On dividing 6x

^{5 }– x^{4 }+ 4x^{3}– 5x^{2}– x – 15 by 2x^{2}– x + 3-10x

^{2}+ 5x – 15Remainder = 0

Therefore, 2x

^{2}– x + 3 is a factor of 60x^{5 }– x^{4}+ 4x^{3}– 5x^{2}– x – 15

### Question 24. Find the value of ‘a’, if x + 2 is a factor of 4x^{4 }+ 2x^{3} – 3x^{2} + 8x + 5a.

**Solution:**

Given that, x + 2 is a factor of 4x

^{4}+ 2x^{3 }– 3x^{2}+ 8x + 5a,On dividing 4x

^{4}+ 2x^{3}– 3x^{2}+ 8x + 5a by x + 2, we getRemainder = 5a + 20

5a + 20 = 0

a = -4

### Question 25. What must be added to x^{4 }+ 2x^{3} – 2x^{2 }+ x – 1 so that the resulting polynomial is exactly divisible by x^{2} + 2x – 3.

**Solution:**

On dividing x

^{4 }+ 2x^{3 }– 2x^{2}+ x – 1 by x^{2}+ 2x – 3, we getRemainder = 0

The No. added to given polynomial to get remainder 0 will be :

x + 2 = 0