Students can refer to RD Sharma Solutions for Class 8 Maths Chapter 8 Division of Algebraic Expressions, which is the best study material for those students who want to top their board exams. Subject experts at BYJUâ€™S have formulated RD Sharma Class 8 Maths solutionsÂ which includes all the questions providing a clear understanding of important concepts by giving detailed explanations for the problems. The PDFs of this chapter are available here, learners can download for free from the links given below. Chapter 8- Division of Algebraic Expressions contains six exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

- Polynomials and degree of a polynomial in two variables.
- Division of a monomial by a monomial.
- Division of polynomials in one variable.
- Division of a polynomial by a monomial.
- Division of a polynomial by a binomial by using long division method.
- Division of polynomials by using factorization.

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### EXERCISE 8.1 PAGE NO: 8.2

**1. Write the degree of each of the following polynomials:**

**(i) 2x ^{3}Â + 5x^{2}Â – 7**

**(ii) 5x ^{2}Â â€“ 3xÂ + 2**

**(iii) 2x + x ^{2} – 8**

**(iv) 1/2y ^{7} â€“ 12y^{6} + 48y^{5} – 10**

**(v) 3x ^{3} + 1**

**(vi) 5 **

**(vii) 20x ^{3} + 12x^{2}y^{2} â€“ 10y^{2} + 20**

**Solution:**

**(i)** 2x^{3}Â + 5x^{2}Â – 7

We know that in a polynomial, degree is the highest power of the variable.

The degree of the polynomial, 2x^{3}Â + 5x^{2}Â – 7 is 3.

**(ii)** 5x^{2}Â â€“ 3xÂ + 2

The degree of the polynomial, 5x^{2}Â â€“ 3xÂ + 2 is 2.

**(iii)** 2x + x^{2} â€“ 8

The degree of the polynomial, 2x + x^{2} â€“ 8 is 2.

**(iv)** 1/2y^{7} â€“ 12y^{6} + 48y^{5} â€“ 10

The degree of the polynomial, 1/2y^{7} â€“ 12y^{6} + 48y^{5} â€“ 10 is 7.

**(v)** 3x^{3} + 1

The degree of the polynomial, 3x^{3} + 1 is 3

**(vi)** 5

The degree of the polynomial, 5 is 0 (since 5 is a constant number).

**(vii)** 20x^{3} + 12x^{2}y^{2} â€“ 10y^{2} + 20

The degree of the polynomial, 20x^{3} + 12x^{2}y^{2} â€“ 10y^{2} + 20 is 4.

**2. Which of the following expressions are not polynomials?**

**(i) x ^{2} + 2x^{-2}**

**(ii) âˆš(ax) + x ^{2} â€“ x^{3}**

**(iii) 3y ^{3} – âˆš5y + 9**

**(iv) ax ^{1/2} + ax + 9x^{2} + 4**

**(v) 3x ^{-3} + 2x^{-1} + 4x + 5**

**Solution:**

**(i)** x^{2} + 2x^{-2}

The given expression is not a polynomial.

Because a polynomial does not contain any negative powers or fractions.

**(ii)** âˆš(ax) + x^{2} â€“ x^{3}

The given expression is a polynomial.

Because the polynomial has positive powers.

**(iii)** 3y^{3} – âˆš5y + 9

The given expression is a polynomial.

Because the polynomial has positive powers.

**(iv)** ax^{1/2} + ax + 9x^{2} + 4

The given expression is not a polynomial.

Because a polynomial does not contain any negative powers or fractions.

**(v)** 3x^{-3} + 2x^{-1} + 4x + 5

The given expression is not a polynomial.

Because a polynomial does not contain any negative powers or fractions.

**3. Write each of the following polynomials in the standard from. Also, write their degree:**

**(i) x ^{2} + 3 + 6x + 5x^{4} **

**(ii) a ^{2} + 4 + 5a^{6} **

**(iii) (x ^{3} – 1) (x^{3} – 4)**

**(iv) (y ^{3} – 2) (y^{3} + 11)**

**(v) (a ^{3} â€“ 3/8) (a^{3} + 16/17)**

**(vi) (a + 3/4) (a + 4/3)**

**Solution:**

** (i)** x^{2} + 3 + 6x + 5x^{4}

The standard form of the polynomial is written in either increasing or decreasing order of their powers.

3 + 6x + x^{2} + 5x^{4} or 5x^{4} + x^{2} + 6x + 3

The degree of the given polynomial is 4.

**(ii)** a^{2} + 4 + 5a^{6}

The standard form of the polynomial is written in either increasing or decreasing order of their powers.

4 + a^{2} + 5a^{6} or 5a^{6} + a^{2} + 4

The degree of the given polynomial is 6.

**(iii)** (x^{3} – 1) (x^{3} – 4)

x^{6} â€“ 4x^{3} â€“ x^{3} + 4

x^{6} â€“ 5x^{3} + 4

The standard form of the polynomial is written in either increasing or decreasing order of their powers.

x^{6} â€“ 5x^{3} + 4 or 4 â€“ 5x^{3} + x^{6}

The degree of the given polynomial is 6.

**(iv)** (y^{3} – 2) (y^{3} + 11)

y^{6} + 11y^{3} â€“ 2y^{3} â€“ 22

y^{6} + 9y^{3} â€“ 22

y^{6} + 9y^{3} â€“ 22 or -22 + 9y^{3} + y^{6}

The degree of the given polynomial is 6.

**(v)** (a^{3} â€“ 3/8) (a^{3} + 16/17)

a^{6} + 16a^{3}/17 â€“ 3a^{3}/8 â€“ 6/17

a^{6} + 77/136a^{3} â€“ 48/136

a^{6} + 77/136a^{3} â€“ 48/136 or -48/136 + 77/136a^{3} + a^{6}

The degree of the given polynomial is 6.

**(vi)** (a + 3/4) (a + 4/3)

a^{2} + 4a/3 + 3a/4 + 1

a^{2} + 25a/12 + 1

a^{2} + 25a/12 + 1 or 1 + 25a/12 + a^{2}

The degree of the given polynomial is 2.

### EXERCISE 8.2 PAGE NO: 8.4

**Divide:**

**1. 6x ^{3}y^{2}z^{2} by 3x^{2}yz**

**Solution:**

We have,

6x^{3}y^{2}z^{2} / 3x^{2}yz

By using the formula a^{n} / a^{m} = a^{n-m}

6/3 x^{3-2} y^{2-1} z^{2-1}

2xyz

**2. 15m ^{2}n^{3} by 5m^{2}n^{2}**

**Solution:**

We have,

15m^{2}n^{3} / 5m^{2}n^{2}

By using the formula a^{n} / a^{m} = a^{n-m}

15/5 m^{2-2} n^{3-2}

3n

**3. 24a ^{3}b^{3} by -8ab**

**Solution:**

We have,

24a^{3}b^{3} / -8ab

By using the formula a^{n} / a^{m} = a^{n-m}

24/-8 a^{3-1} b^{3-1}

-3a^{2}b^{2}

**4. -21abc ^{2} by 7abc**

**Solution:**

We have,

-21abc^{2} / 7abc

By using the formula a^{n} / a^{m} = a^{n-m}

-21/7 a^{1-1} b^{1-1} c^{2-1}

-3c

**5. 72xyz ^{2} by -9xz**

**Solution:**

We have,

72xyz^{2} / -9xz

By using the formula a^{n} / a^{m} = a^{n-m}

72/-9 x^{1-1} y z^{2-1}

-8yz

**6. -72a ^{4}b^{5}c^{8}Â by -9a^{2}b^{2}c^{3}**

**Solution:**

We have,

-72a^{4}b^{5}c^{8}Â / -9a^{2}b^{2}c^{3}

By using the formula a^{n} / a^{m} = a^{n-m}

-72/-9 a^{4-2} b^{5-2} c^{8-3}

8a^{2}b^{3}c^{5}

**Simplify:**

**7. 16m ^{3}y^{2} / 4m^{2}y**

**Solution:**

We have,

16m^{3}y^{2} / 4m^{2}y

By using the formula a^{n} / a^{m} = a^{n-m}

16/4 m^{3-2} y^{2-1}

4my

**8. 32m ^{2}n^{3}p^{2} / 4mnp**

**Solution:**

We have,

32m^{2}n^{3}p^{2} / 4mnp

By using the formula a^{n} / a^{m} = a^{n-m}

32/4 m^{2-1} n^{3-1} p^{2-1}

8mn^{2}p

### EXERCISE 8.3 PAGE NO: 8.6

**Divide:**

**1. x + 2x ^{2} + 3x^{4} â€“ x^{5} by 2x**

**Solution:**

We have,

(x + 2x^{2} + 3x^{4} â€“ x^{5}) / 2x

x/2x + 2x^{2}/2x + 3x^{4}/2x â€“ x^{5}/2x

By using the formula a^{n} / a^{m} = a^{n-m}

1/2 x^{1-1} + x^{2-1} + 3/2 x^{4-1} â€“ 1/2 x^{5-1}

1/2 + x + 3/2 x^{3} â€“ 1/2 x^{4}

**2. y ^{4} â€“ 3y^{3} + 1/2y^{2} by 3y**

**Solution:**

We have,

(y^{4} â€“ 3y^{3} + 1/2y^{2})/ 3y

y^{4}/3y â€“ 3y^{3}/3y + (Â½)y^{2}/3y

By using the formula a^{n} / a^{m} = a^{n-m}

1/3 y^{4-1} â€“ y^{3-1} + 1/6 y^{2-1}

1/3y^{3} â€“ y^{2} + 1/6y

**3. -4a ^{3} + 4a^{2} + a by 2a**

**Solution:**

We have,

(-4a^{3} + 4a^{2} + a) / 2a

-4a^{3}/2a + 4a^{2}/2a + a/2a

By using the formula a^{n} / a^{m} = a^{n-m}

-2a^{3-1} + 2a^{2-1} + 1/2 a^{1-1}

-2a^{2} + 2a + Â½

**4. â€“x ^{6} + 2x^{4} + 4x^{3} + 2x^{2} by âˆš2x^{2}**

**Solution:**

We have,

(â€“x^{6} + 2x^{4} + 4x^{3} + 2x^{2}) / âˆš2x^{2}

-x^{6}/âˆš2x^{2} + 2x^{4}/âˆš2x^{2} + 4x^{3}/âˆš2x^{2} + 2x^{2}/âˆš2x^{2}

By using the formula a^{n} / a^{m} = a^{n-m}

-1/âˆš2 x^{6-2} + 2/âˆš2 x^{4-2} + 4/âˆš2 x^{3-2} + 2/âˆš2 x^{2-2}

-1/âˆš2 x^{4} + âˆš2x^{2} + 2âˆš2x + âˆš2

**5. -4a ^{3} + 4a^{2} + a by 2a**

**Solution:**

We have,

(-4a^{3} + 4a^{2} + a) / 2a

-4a^{3}/2a + 4a^{2}/2a + a/2a

By using the formula a^{n} / a^{m} = a^{n-m}

-2a^{3-1} + 2a^{2-1} + 1/2a^{1-1}

-2a^{2} + 2a + Â½

**6. âˆš3a ^{4} + 2âˆš3a^{3} + 3a^{2} â€“ 6a by 3a**

**Solution:**

We have,

(âˆš3a^{4} + 2âˆš3a^{3} + 3a^{2} â€“ 6a) / 3a

âˆš3a^{4}/3a + 2âˆš3a^{3}/3a + 3a^{2}/3a â€“ 6a/3a

By using the formula a^{n} / a^{m} = a^{n-m}

âˆš3/3 a^{4-1} + 2âˆš3/3 a^{3-1} + a^{2-1} â€“ 2a^{1-1}

1/âˆš3 a^{3} + 2/âˆš3 a^{2} + a â€“ 2

### EXERCISE 8.4 PAGE NO: 8.11

**Divide:**

**1. 5x ^{3} â€“ 15x^{2} + 25x by 5x**

**Solution:**

We have,

(5x^{3} â€“ 15x^{2} + 25x) / 5x

5x^{3}/5x â€“ 15x^{2}/5x + 25x/5x

By using the formula a^{n} / a^{m} = a^{n-m}

5/5 x^{3-1} â€“ 15/5 x^{2-1} + 25/5 x^{1-1}

x^{2} â€“ 3x + 5

**2. 4z ^{3} + 6z^{2} â€“ z by -1/2z**

**Solution:**

We have,

(4z^{3} + 6z^{2} â€“ z) / -1/2z

4z^{3}/(-1/2z) + 6z^{2}/(-1/2z) â€“ z/(-1/2z)

By using the formula a^{n} / a^{m} = a^{n-m}

-8 z^{3-1} â€“ 12z^{2-1} + 2 z^{1-1}

-8z^{2} â€“ 12z + 2

**3. 9x ^{2}y â€“ 6xy + 12xy^{2} by -3/2xy**

**Solution:**

We have,

(9x^{2}y â€“ 6xy + 12xy^{2}) / -3/2xy

9x^{2}y/(-3/2xy) â€“ 6xy/(-3/2xy) + 12xy^{2}/(-3/2xy)

By using the formula a^{n} / a^{m} = a^{n-m}

(-9Ã—2)/3 x^{2-1}y^{1-1} â€“ (-6Ã—2)/3 x^{1-1}y^{1-1} + (-12Ã—2)/3 x^{1-1}y^{2-1}

-6x + 4 â€“ 8y

**4. 3x ^{3}y^{2} + 2x^{2}y + 15xy by 3xy**

**Solution:**

We have,

(3x^{3}y^{2} + 2x^{2}y + 15xy) / 3xy

3x^{3}y^{2}/3xy + 2x^{2}y/3xy + 15xy/3xy

By using the formula a^{n} / a^{m} = a^{n-m}

3/3 x^{3-1}y^{2-1} + 2/3 x^{2-1}y^{1-1} + 15/3 x^{1-1}y^{1-1}

x^{2}y + 2/3x + 5

**5. x ^{2 }+ 7x + 12 by x + 4**

**Solution:**

We have,

(x^{2} + 7x + 12) / (x + 4)

By using long division method

âˆ´ (x^{2} + 7x + 12) / (x + 4) = x + 3

**6. 4y ^{2} + 3y + 1/2 by 2y + 1**

**Solution:**

We have,

4y^{2} + 3y + 1/2 by (2y + 1)

By using long division method

âˆ´ (4y^{2} + 3y + 1/2) / (2y + 1) = 2y + 1/2

**7. 3x ^{3} + 4x^{2} + 5x + 18 by x + 2**

**Solution:**

We have,

(3x^{3} + 4x^{2} + 5x + 18) / (x + 2)

By using long division method

âˆ´ (3x^{3} + 4x^{2} + 5x + 18) / (x + 2) = 3x^{2} â€“ 2x + 9

**8. 14x ^{2} â€“ 53x + 45 by 7x – 9**

**Solution:**

We have,

(14x^{2} â€“ 53x + 45) / (7x â€“ 9)

By using long division method

âˆ´ (14x^{2} â€“ 53x + 45) / (7x â€“ 9) = 2x â€“ 5

**9. -21 + 71x â€“ 31x ^{2} â€“ 24x^{3} by 3 â€“ 8x**

**Solution:**

We have,

-21 + 71x â€“ 31x^{2} â€“ 24x^{3} by 3 â€“ 8x

(-24x^{3} â€“ 31x^{2} + 71x â€“ 21) / (3 â€“ 8x)

By using long division method

âˆ´ (-24x^{3} â€“ 31x^{2} + 71x â€“ 21) / (3 â€“ 8x) = 3x^{2} + 5x â€“ 7

**10. 3y ^{4} â€“ 3y^{3} â€“ 4y^{2} â€“ 4y by y^{2} – 2y**

**Solution:**

We have,

(3y^{4} â€“ 3y^{3} â€“ 4y^{2} â€“ 4y) / (y^{2} – 2y)

By using long division method

âˆ´ (3y^{4} â€“ 3y^{3} â€“ 4y^{2} â€“ 4y) / (y^{2} – 2y) = 3y^{2} + 3y + 2

**11. 2y ^{5} + 10y^{4} + 6y^{3} + y^{2} + 5y + 3 by 2y^{3} + 1**

**Solution:**

We have,

(2y^{5} + 10y^{4} + 6y^{3} + y^{2} + 5y + 3) / (2y^{3} + 1)

By using long division method

âˆ´ (2y^{5} + 10y^{4} + 6y^{3} + y^{2} + 5y + 3) / (2y^{3} + 1) = y^{2} + 5y + 3

**12. x ^{4} â€“ 2x^{3} + 2x^{2} + x + 4 by x^{2} + x + 1**

**Solution:**

We have,

(x^{4} â€“ 2x^{3} + 2x^{2} + x + 4) / (x^{2} + x + 1)

By using long division method

âˆ´ (x^{4} â€“ 2x^{3} + 2x^{2} + x + 4) / (x^{2} + x + 1) = x^{2} â€“ 3x + 4

**13. m ^{3} â€“ 14m^{2} + 37m â€“ 26 by m^{2} â€“ 12m + 13**

**Solution:**

We have,

(m^{3} â€“ 14m^{2} + 37m â€“ 26) / (m^{2} â€“ 12m + 13)

By using long division method

âˆ´ (m^{3} â€“ 14m^{2} + 37m â€“ 26) / (m^{2} â€“ 12m + 13) = m â€“ 2

**14. x ^{4} + x^{2} + 1 by x^{2} + x + 1**

**Solution:**

We have,

(x^{4} + x^{2} + 1) / (x^{2} + x + 1)

By using long division method

âˆ´ (x^{4} + x^{2} + 1) / (x^{2} + x + 1) = x^{2} â€“ x + 1

**15. x ^{5} + x^{4} + x^{3} + x^{2} + x + 1 by x^{3} + 1**

**Solution:**

We have,

(x^{5} + x^{4} + x^{3} + x^{2} + x + 1) / (x^{3} + 1)

By using long division method

âˆ´ (x^{5} + x^{4} + x^{3} + x^{2} + x + 1) / (x^{3} + 1) = x^{2} + x + 1

**Divide each of the following and find the quotient and remainder:**

**16. 14x ^{3} â€“ 5x^{2} + 9x â€“ 1 by 2x â€“ 1**

**Solution:**

We have,

(14x^{3} â€“ 5x^{2} + 9x â€“ 1) / (2x â€“ 1)

By using long division method

âˆ´ Quotient is 7x^{2} + x + 5 and the Remainder is 4.

**17. 6x ^{3} â€“ x^{2} â€“ 10x â€“ 3 by 2x â€“ 3 **

**Solution:**

We have,

(6x^{3} â€“ x^{2} â€“ 10x â€“ 3) / (2x â€“ 3)

By using long division method

âˆ´ Quotient is 3x^{2} + 4x + 1 and the Remainder is 0.

**18. 6x ^{3} + 11x^{2} â€“ 39x â€“ 65 by 3x^{2} + 13x + 13 **

**Solution:**

We have,

(6x^{3} + 11x^{2} â€“ 39x â€“ 65) / (3x^{2} + 13x + 13)

By using long division method

âˆ´ Quotient is 2x – 5 and the Remainder is 0.

**19. 30x ^{4} + 11x^{3} â€“ 82x^{2} â€“ 12x + 48 by 3x^{2} + 2x â€“ 4**

**Solution:**

We have,

(30x^{4} + 11x^{3} â€“ 82x^{2} â€“ 12x + 48) / (3x^{2} + 2x â€“ 4)

By using long division method

âˆ´ Quotient is 10x^{2} â€“ 3x – 12 and the Remainder is 0.

**20. 9x ^{4} â€“ 4x^{2} + 4 by 3x^{2} â€“ 4x + 2**

**Solution:**

We have,

(9x^{4} â€“ 4x^{2} + 4) / (3x^{2} â€“ 4x + 2)

By using long division method

âˆ´ Quotient is 3x^{2} + 4x + 2 and the Remainder is 0.

**21. Verify division algorithm i.e. Dividend = Divisor Ã—Â Quotient + Remainder, in each of the following. Also, write the quotient and remainder:**

**Dividend divisor **

**(i) 14x ^{2} + 13x â€“ 15 7x â€“ 4 **

**(ii) 15z ^{3} â€“ 20z^{2} + 13z â€“ 12 3z â€“ 6 **

**(iii) 6y ^{5} â€“ 28y^{3} + 3y^{2} + 30y â€“ 9 2x^{2} â€“ 6 **

**(iv) 34x â€“ 22x ^{3} â€“ 12x^{4} â€“ 10x^{2} â€“ 75 3x + 7**

**(v) 15y ^{4} â€“ 16y^{3} + 9y^{2} â€“ 10/3y + 6 3y â€“ 2 **

**(vi) 4y ^{3} + 8y + 8y^{2} + 7 2y^{2} â€“ y + 1**

**(vii) 6y ^{4} + 4y^{4} + 4y^{3} + 7y^{2} + 27y + 6 2y^{3} + 1**

**Solution:**

**(i)** Dividend divisor

14x^{2} + 13x â€“ 15 7x â€“ 4

By using long division method

Let us verify, Dividend = Divisor Ã—Â Quotient + Remainder

14x^{2} + 13x â€“ 15 = (7x – 4) Ã— (2x + 3) + (-3)

= 14x^{2} + 21x â€“ 8x -12 -3

= 14x^{2} + 13x â€“ 15

Hence, verified.

âˆ´ Quotient is 2x + 3 and the Remainder is -3.

**(ii)** Dividend divisor

15z^{3} â€“ 20z^{2} + 13z â€“ 12 3z â€“ 6

By using long division method

Let us verify, Dividend = Divisor Ã—Â Quotient + Remainder

15z^{3} â€“ 20z^{2} + 13z â€“ 12 = (3z – 6) Ã— (5z^{2} + 10z/3 + 11) + 54

= 15z^{3} + 10z^{2} + 33z â€“ 30z^{2} â€“ 20z + 54

= 15z^{2} â€“ 20z^{2} + 13z â€“ 12

Hence, verified.

âˆ´ Quotient is 5z^{2} + 10z/3 + 11 and the Remainder is 54.

**(iii)** Dividend divisor

6y^{5} â€“ 28y^{3} + 3y^{2} + 30y â€“ 9 2x^{2} â€“ 6

By using long division method

Let us verify, Dividend = Divisor Ã—Â Quotient + Remainder

6y^{5} â€“ 28y^{3} + 3y^{2} + 30y â€“ 9 = (2x^{2} â€“ 6) Ã— (3y^{3} â€“ 5y + 3/2) + 0

= 6y^{5} â€“ 10y^{3} + 3y^{2} â€“ 18y^{3} + 30y â€“ 9

= 6y^{5} â€“ 28y^{3} + 3y^{2} + 30y â€“ 9

Hence, verified.

âˆ´ Quotient is 3y^{3} â€“ 5y + 3/2 and the Remainder is 0.

**(iv)** Dividend divisor

34x â€“ 22x^{3} â€“ 12x^{4} â€“ 10x^{2} â€“ 75 3x + 7

-12x^{4} â€“ 22x^{3} â€“ 10x^{2} + 34x – 75

By using long division method

Let us verify, Dividend = Divisor Ã—Â Quotient + Remainder

-12x^{4} â€“ 22x^{3} â€“ 10x^{2} + 34x â€“ 75 = (3x + 7) Ã— (-4x^{3} + 2x^{2} â€“ 8x + 30) â€“ 285

= -12x^{4} + 6x^{3} â€“ 24x^{2} â€“ 28x^{3} + 14x^{2} + 90x â€“ 56x + 210 -285

= -12x^{4} â€“ 22x^{3} â€“ 10x^{2} + 34x â€“ 75

Hence, verified.

âˆ´ Quotient is -4x^{3} + 2x^{2} – 8x + 30 and the Remainder is -285.

**(v)** Dividend divisor

15y^{4} â€“ 16y^{3} + 9y^{2} â€“ 10/3y + 6 3y â€“ 2

By using long division method

Let us verify, Dividend = Divisor Ã—Â Quotient + Remainder

15y^{4} â€“ 16y^{3} + 9y^{2} â€“ 10/3y + 6 = (3y â€“ 2) Ã— (5y^{3} â€“ 2y^{2} + 5y/3) + 6

= 15y^{4} â€“ 6y^{3} + 5y^{2} â€“ 10y^{3} + 4y^{2} â€“ 10y/3 + 6

= 15y^{4} â€“ 16y^{3} + 9y^{2} â€“ 10/3y + 6

Hence, verified.

âˆ´ Quotient is 5y^{3} â€“ 2y^{2} + 5y/3 and the Remainder is 6.

**(vi)** Dividend divisor

4y^{3} + 8y + 8y^{2} + 7 2y^{2} â€“ y + 1

4y^{3} + 8y^{2} + 8y + 7

By using long division method

Let us verify, Dividend = Divisor Ã—Â Quotient + Remainder

4y^{3} + 8y^{2} + 8y + 7 = (2y^{2} â€“ y + 1) Ã— (2y + 5) + 11y + 2

= 4y^{3} + 10y^{2} â€“ 2y^{2} â€“ 5y + 2y + 5 + 11y + 2

= 4y^{3} + 8y^{2} + 8y + 7

Hence, verified.

âˆ´ Quotient is 2y + 5 and the Remainder is 11y + 2.

**(vii)** Dividend divisor

6y^{5} + 4y^{4} + 4y^{3} + 7y^{2} + 27y + 6 2y^{3} + 1

By using long division method

Let us verify, Dividend = Divisor Ã—Â Quotient + Remainder

6y^{5} + 4y^{4} + 4y^{3} + 7y^{2} + 27y + 6 = (2y^{3} + 1) Ã— (3y^{2} + 2y + 2) + 4y^{2} + 25y + 4

= 6y^{5} + 4y^{4} + 4y^{3} + 3y^{2} + 2y + 2 + 4y^{2} + 25y + 4

= 6y^{5} + 4y^{4} + 4y^{3} + 7y^{2} + 27y + 6

Hence, verified.

âˆ´ Quotient is 3y^{2} + 2y + 2 and the Remainder is 4y^{2} + 25y + 4.

**22. Divide 15y ^{4}+ 16y^{3} + 10/3y â€“ 9y^{2} â€“ 6 by 3y – 2Â Write down the coefficients of the terms in the quotient.**

**Solution:**

We have,

(15y^{4}+ 16y^{3} + 10/3y â€“ 9y^{2} â€“ 6) / (3y â€“ 2)Â

By using long division method

âˆ´ Quotient is 5y^{3} + 26y^{2}/3 + 25y/9 + 80/27

So the coefficients of the terms in the quotient are:

Coefficient of y^{3}Â = 5

Coefficient of y^{2}Â =Â 26/3

Coefficient of y =Â 25/9

Constant term =Â 80/27

**23. Using division of polynomials state whether(i)Â x + 6Â is a factor of x ^{2} â€“ x – 42(ii) 4x – 1 is a factor of 4x^{2} â€“ 13x – 12Â (iii) 2y – 5 is a factor of 4y^{4} â€“ 10y^{3} â€“ 10y^{2} + 30y â€“ 15(iv) 3y^{2} + 5 is a factor of 6y^{5} + 15y^{4} + 16y^{3} + 4y^{2} + 10y – 35Â (v)Â z^{2} + 3 is a factor of z^{5} â€“ 9z(vi) 2x^{2} â€“ x + 3Â is a factor of 6x^{5 }â€“ x^{4} + 4x^{3} â€“ 5x^{2} â€“ x â€“ 15**

**Solution:**

**(i)Â **x + 6Â is a factor of x^{2} â€“ x – 42

Firstly let us perform long division method

Since the remainder is 0, we can say that x + 6Â is a factor of x^{2} â€“ x â€“ 42

**(ii) **4x – 1 is a factor of 4x^{2} â€“ 13x – 12Â

Firstly let us perform long division method

Since the remainder is -15, 4x – 1 is not a factor of 4x^{2} â€“ 13x – 12Â

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

Firstly let us perform long division method

Since the remainder is 5y^{3} â€“ 45y^{2}/2 + 30y â€“ 15, 2y – 5 is not a factor of 4y^{4} â€“ 10y^{3} â€“ 10y^{2} + 30y â€“ 15

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

Firstly let us perform long division method

Since the remainder is 0, 3y^{2} + 5 is a factor of 6y^{5} + 15y^{4} + 16y^{3} + 4y^{2} + 10y – 35**Â **

**(v)Â **z^{2} + 3 is a factor of z^{5} â€“ 9z

Firstly let us perform long division method

Since the remainder is 0, z^{2} + 3 is a factor of z^{5} â€“ 9z

**(vi) **2x^{2} â€“ x + 3Â is a factor of 6x^{5 }â€“ x^{4} + 4x^{3} â€“ 5x^{2} â€“ x â€“ 15

Firstly let us perform long division method

Since the remainder is 0, 2x^{2} â€“ x + 3Â is a factor of 6x^{5 }â€“ x^{4} + 4x^{3} â€“ 5x^{2} â€“ x â€“ 15

**24. Find the value of a, if x + 2 is a factor of 4x ^{4} + 2x^{3} â€“ 3x^{2} + 8x + 5a**

**Solution:**

We know that x + 2 is a factor of 4x^{4} + 2x^{3} â€“ 3x^{2} + 8x + 5a

Let us equate x + 2 = 0

x = -2

Now let us substitute x = -2 in the equation 4x^{4} + 2x^{3} â€“ 3x^{2} + 8x + 5a

4(-2)^{4} + 2(-2)^{3} â€“ 3(-2)^{2} + 8(-2) + 5a = 0

64 â€“ 16 â€“ 12 â€“ 16 + 5a = 0

20 + 5a = 0

5a = -20

a = -20/5

= -4

**25. What must be added to x ^{4} + 2x^{3} â€“ 2x^{2} + x – 1Â so that the resulting polynomial is exactly divisible by x^{2} + 2x -3.**

**Solution:**

Firstly let us perform long division method

By long division method we got remainder as â€“x + 2,

âˆ´ x â€“ 2 has to be added to x^{4} + 2x^{3} â€“ 2x^{2} + x – 1Â so that the resulting polynomial is exactly divisible by x^{2} + 2x -3.

### EXERCISE 8.5 PAGE NO: 8.15

**1. Divide the first polynomial by the second polynomial in each of the following. Also, write the quotient and remainder:**

**(i) 3x ^{2} + 4x + 5, x – 2**

**(ii) 10x ^{2} â€“ 7x + 8, 5x – 3**

**(iii) 5y ^{3} â€“ 6y^{2} + 6y â€“ 1, 5y â€“ 1**

**(iv) x ^{4} â€“ x^{3} + 5x, x – 1**

**(v) y ^{4} + y^{2}, y^{2} â€“ 2**

**Solution:**

**(i) **3x^{2} + 4x + 5, x – 2

By using long division method

âˆ´ the Quotient is 3x + 10 and the Remainder is 25.

**(ii) **10x^{2} â€“ 7x + 8, 5x – 3

By using long division method

âˆ´ the Quotient is 2x â€“ 1/5 and the Remainder is 37/5.

**(iii) **5y^{3} â€“ 6y^{2} + 6y â€“ 1, 5y â€“ 1

By using long division method

âˆ´ the Quotient is y^{2} â€“ y + 1 and the Remainder is 0.

**(iv) **x^{4} â€“ x^{3} + 5x, x – 1

By using long division method

âˆ´ the Quotient is x^{3} + 5 and the Remainder is 5.

**(v)** y^{4} + y^{2}, y^{2} â€“ 2

By using long division method

âˆ´ the Quotient is y^{2} + 3 and the Remainder is 6.

**2. Find Whether or not the first polynomial is a factor of the second:**

**(i) x + 1, 2x ^{2} + 5x + 4**

**(ii) y â€“ 2, 3y ^{3} + 5y^{2} + 5y + 2**

**(iii) 4x ^{2} â€“ 5, 4x^{4} + 7x^{2} + 15**

**(iv) 4 â€“ z, 3z ^{2} â€“ 13z + 4**

**(v) 2a â€“ 3, 10a ^{2} â€“ 9a – 5**

**(vi) 4y + 1, 8y ^{2} â€“ 2y + 1**

**Solution:**

**(i) **x + 1, 2x^{2} + 5x + 4

Let us perform long division method,

Since remainder is 1 thereforeÂ the first polynomial isÂ notÂ a factor of the second polynomial.

**(ii) **y â€“ 2, 3y^{3} + 5y^{2} + 5y + 2

Let us perform long division method,

Since remainder is 56 thereforeÂ the first polynomial isÂ notÂ a factor of the second polynomial.

**(iii) **4x^{2} â€“ 5, 4x^{4} + 7x^{2} + 15

Let us perform long division method,

Since remainder is 30 thereforeÂ the first polynomial isÂ notÂ a factor of the second polynomial.

**(iv) **4 â€“ z, 3z^{2} â€“ 13z + 4

Let us perform long division method,

Since remainder is 0 thereforeÂ the first polynomial isÂ a factor of the second polynomial.

**(v) **2a â€“ 3, 10a^{2} â€“ 9a – 5

Let us perform long division method,

Since remainder is 4 thereforeÂ the first polynomial isÂ notÂ a factor of the second polynomial.

**(vi) **4y + 1, 8y^{2} â€“ 2y + 1

Let us perform long division method,

Since remainder is 2 thereforeÂ the first polynomial isÂ notÂ a factor of the second polynomial.

### EXERCISE 8.6 PAGE NO: 8.17

**Divide:**

**1. x ^{2} â€“ 5x + 6 by x â€“ 3**

**Solution:**

We have,

(x^{2} â€“ 5x + 6) / (x â€“ 3)

Let us perform long division method,

âˆ´ the Quotient is x â€“ 2

**2. ax ^{2} â€“ ay^{2} by ax+ay**

**Solution:**

We have,

(ax^{2} â€“ ay^{2})/ (ax+ay)

(ax^{2} â€“ ay^{2})/ (ax+ay) = (x – y) + 0/(ax+ay)

= (x – y)

âˆ´ the answer is (x â€“ y)

**3. x ^{4} â€“ y^{4} by x^{2} â€“ y^{2}**

**Solution:**

We have,

(x^{4} â€“ y^{4})/ (x^{2} â€“ y^{2})

(x^{4} â€“ y^{4})/ (x^{2} â€“ y^{2}) = x^{2} + y^{2} + 0/(x^{2} â€“ y^{2})

= x^{2} + y^{2}

âˆ´ the answer is (x^{2} + y^{2})

**4. acx ^{2} + (bc + ad)x + bd by (ax + b)**

**Solution:**

We have,

(acx^{2} + (bc + ad) x + bd) / (ax + b)

(acx^{2} + (bc + ad) x + bd) / (ax + b) = cx + d + 0/ (ax + b)

= cx + d

âˆ´ the answer is (cx + d)

**5. (a ^{2} + 2ab + b^{2}) â€“ (a^{2} + 2ac + c^{2}) by 2a + b + c**

**Solution:**

We have,

[(a^{2}+ 2ab + b

^{2}) â€“ (a

^{2}+ 2ac + c

^{2})] / (2a + b + c) [(a

^{2}+ 2ab + b

^{2}) â€“ (a

^{2}+ 2ac + c

^{2})] / (2a + b + c) = b â€“ c + 0/(2a + b + c)

= b â€“ c

âˆ´ the answer is (b â€“ c)

**6. 1/4x ^{2} â€“ 1/2x â€“ 12 by 1/2x â€“ 4**

**Solution**:

We have,

(1/4x^{2} â€“ 1/2x â€“ 12) / (1/2x â€“ 4)

Let us perform long division method,

âˆ´ the Quotient is x/2 + 3