# RD Sharma Solutions for Class 8 Maths Chapter 8 Division of Algebraic Expressions

Students can refer to RD Sharma Solutions for Class 8 Maths Chapter 8 Division of Algebraic Expressions, which is the best study material for them to excel in the annual exams. Subject experts at BYJU’S have formulated RD Sharma Class 8 Maths Solutions, which include 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, and students can download them for free from the links given below. Chapter 8 – Division of Algebraic Expressions contains six exercises, and the RD Sharma Solutions available on this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts covered 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 the long division method.
• Division of polynomials by using factorization.

## RD Sharma Solutions for Class 8 Maths Chapter 8 Division of Algebraic Expressions

### EXERCISE 8.1 PAGE NO: 8.2

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

(i) 2x3 + 5x2 – 7

(ii) 5x2 – 3x + 2

(iii) 2x + x2 – 8

(iv) 1/2y7 – 12y6 + 48y5 – 10

(v) 3x3 + 1

(vi) 5

(vii) 20x3 + 12x2y2 – 10y2 + 20

Solution:

(i) 2x3 + 5x2 – 7

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

The degree of the polynomial, 2x3 + 5x2 – 7, is 3.

(ii) 5x2 – 3x + 2

The degree of the polynomial, 5x2 – 3x + 2, is 2.

(iii) 2x + x2 – 8

The degree of the polynomial, 2x + x2 – 8, is 2.

(iv) 1/2y7 – 12y6 + 48y5 – 10

The degree of the polynomial, 1/2y7 – 12y6 + 48y5 – 10, is 7.

(v) 3x3 + 1

The degree of the polynomial, 3x3 + 1, is 3

(vi) 5

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

(vii) 20x3 + 12x2y2 – 10y2 + 20

The degree of the polynomial, 20x3 + 12x2y2 – 10y2 + 20, is 4.

2. Which of the following expressions are not polynomials?

(i) x2 + 2x-2

(ii) √(ax) + x2 – x3

(iii) 3y3 – √5y + 9

(iv) ax1/2 + ax + 9x2 + 4

(v) 3x-3 + 2x-1 + 4x + 5

Solution:

(i) x2 + 2x-2

The given expression is not a polynomial.

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

(ii) √(ax) + x2 – x3

The given expression is not a polynomial.

Because the polynomial has 1/2 as the power of variable x, which is not a non-negative integer.

(iii) 3y3 – √5y + 9

The given expression is a polynomial.

Because the polynomial has positive powers.

(iv) ax1/2 + ax + 9x2 + 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 form. Also, write their degree.

(i) x2 + 3 + 6x + 5x4

(ii) a2 + 4 + 5a6

(iii) (x3 – 1) (x3 – 4)

(iv) (y3 – 2) (y3 + 11)

(v) (a3 – 3/8) (a3 + 16/17)

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

Solution:

(i) x2 + 3 + 6x + 5x4

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

3 + 6x + x2 + 5x4 or 5x4 + x2 + 6x + 3

The degree of the given polynomial is 4.

(ii) a2 + 4 + 5a6

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

4 + a2 + 5a6 or 5a6 + a2 + 4

The degree of the given polynomial is 6.

(iii) (x3 – 1) (x3 – 4)

x6 – 4x3 – x3 + 4

x6 – 5x3 + 4

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

x6 – 5x3 + 4 or 4 – 5x3 + x6

The degree of the given polynomial is 6.

(iv) (y3 – 2) (y3 + 11)

y6 + 11y3 – 2y3 – 22

y6 + 9y3 – 22

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

y6 + 9y3 – 22 or -22 + 9y3 + y6

The degree of the given polynomial is 6.

(v) (a3 – 3/8) (a3 + 16/17)

a6 + 16a3/17 – 3a3/8 – 6/17

a6 + 77/136a3 – 48/136

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

a6 + 77/136a3 – 48/136 or -48/136 + 77/136a3 + a6

The degree of the given polynomial is 6.

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

a2 + 4a/3 + 3a/4 + 1

a2 + 25a/12 + 1

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

a2 + 25a/12 + 1 or 1 + 25a/12 + a2

The degree of the given polynomial is 2.

### EXERCISE 8.2 PAGE NO: 8.4

Divide:

1. 6x3y2z2 by 3x2yz

Solution:

We have,

6x3y2z2 / 3x2yz

By using the formula an / am = an-m

6/3 x3-2 y2-1 z2-1

2xyz

2. 15m2n3 by 5m2n2

Solution:

We have,

15m2n3 / 5m2n2

By using the formula an / am = an-m

15/5 m2-2 n3-2

3n

3. 24a3b3 by -8ab

Solution:

We have,

24a3b3 / -8ab

By using the formula an / am = an-m

24/-8 a3-1 b3-1

-3a2b2

4. -21abc2 by 7abc

Solution:

We have,

-21abc2 / 7abc

By using the formula an / am = an-m

-21/7 a1-1 b1-1 c2-1

-3c

5. 72xyz2 by -9xz

Solution:

We have,

72xyz2 / -9xz

By using the formula an / am = an-m

72/-9 x1-1 y z2-1

-8yz

6. -72a4b5c8 by -9a2b2c3

Solution:

We have,

-72a4b5c8 / -9a2b2c3

By using the formula an / am = an-m

-72/-9 a4-2 b5-2 c8-3

8a2b3c5

Simplify:

7. 16m3y2 / 4m2y

Solution:

We have,

16m3y2 / 4m2y

By using the formula an / am = an-m

16/4 m3-2 y2-1

4my

8. 32m2n3p2 / 4mnp

Solution:

We have,

32m2n3p2 / 4mnp

By using the formula an / am = an-m

32/4 m2-1 n3-1 p2-1

8mn2p

### EXERCISE 8.3 PAGE NO: 8.6

Divide:

1. x + 2x2 + 3x4 – x5 by 2x

Solution:

We have,

(x + 2x2 + 3x4 – x5) / 2x

x/2x + 2x2/2x + 3x4/2x – x5/2x

By using the formula an / am = an-m

1/2 x1-1 + x2-1 + 3/2 x4-1 – 1/2 x5-1

1/2 + x + 3/2 x3 – 1/2 x4

2. y4 – 3y3 + 1/2y2 by 3y

Solution:

We have,

(y4 – 3y3 + 1/2y2)/ 3y

y4/3y – 3y3/3y + (½)y2/3y

By using the formula an / am = an-m

1/3 y4-1 – y3-1 + 1/6 y2-1

1/3y3 – y2 + 1/6y

3. -4a3 + 4a2 + a by 2a

Solution:

We have,

(-4a3 + 4a2 + a) / 2a

-4a3/2a + 4a2/2a + a/2a

By using the formula an / am = an-m

-2a3-1 + 2a2-1 + 1/2 a1-1

-2a2 + 2a + ½

4. –x6 + 2x4 + 4x3 + 2x2 by √2x2

Solution:

We have,

(–x6 + 2x4 + 4x3 + 2x2) / √2x2

-x6/√2x2 + 2x4/√2x2 + 4x3/√2x2 + 2x2/√2x2

By using the formula an / am = an-m

-1/√2 x6-2 + 2/√2 x4-2 + 4/√2 x3-2 + 2/√2 x2-2

-1/√2 x4 + √2x2 + 2√2x + √2

5. -4a3 + 4a2 + a by 2a

Solution:

We have,

(-4a3 + 4a2 + a) / 2a

-4a3/2a + 4a2/2a + a/2a

By using the formula an / am = an-m

-2a3-1 + 2a2-1 + 1/2a1-1

-2a2 + 2a + ½

6. √3a4 + 2√3a3 + 3a2 – 6a by 3a

Solution:

We have,

(√3a4 + 2√3a3 + 3a2 – 6a) / 3a

√3a4/3a + 2√3a3/3a + 3a2/3a – 6a/3a

By using the formula an / am = an-m

√3/3 a4-1 + 2√3/3 a3-1 + a2-1 – 2a1-1

1/√3 a3 + 2/√3 a2 + a – 2

### EXERCISE 8.4 PAGE NO: 8.11

Divide:

1. 5x3 – 15x2 + 25x by 5x

Solution:

We have,

(5x3 – 15x2 + 25x) / 5x

5x3/5x – 15x2/5x + 25x/5x

By using the formula an / am = an-m

5/5 x3-1 – 15/5 x2-1 + 25/5 x1-1

x2 – 3x + 5

2. 4z3 + 6z2 – z by -1/2z

Solution:

We have,

(4z3 + 6z2 – z) / -1/2z

4z3/(-1/2z) + 6z2/(-1/2z) – z/(-1/2z)

By using the formula an / am = an-m

-8 z3-1 – 12z2-1 + 2 z1-1

-8z2 – 12z + 2

3. 9x2y – 6xy + 12xy2 by -3/2xy

Solution:

We have,

(9x2y – 6xy + 12xy2) / -3/2xy

9x2y/(-3/2xy) – 6xy/(-3/2xy) + 12xy2/(-3/2xy)

By using the formula an / am = an-m

(-9×2)/3 x2-1y1-1 – (-6×2)/3 x1-1y1-1 + (-12×2)/3 x1-1y2-1

-6x + 4 – 8y

4. 3x3y2 + 2x2y + 15xy by 3xy

Solution:

We have,

(3x3y2 + 2x2y + 15xy) / 3xy

3x3y2/3xy + 2x2y/3xy + 15xy/3xy

By using the formula an / am = an-m

3/3 x3-1y2-1 + 2/3 x2-1y1-1 + 15/3 x1-1y1-1

x2y + 2/3x + 5

5. x2 + 7x + 12 by x + 4

Solution:

We have,

(x2 + 7x + 12) / (x + 4)

By using the long division method

∴ (x2 + 7x + 12) / (x + 4) = x + 3

6. 4y2 + 3y + 1/2 by 2y + 1

Solution:

We have,

4y2 + 3y + 1/2 by (2y + 1)

By using the long division method

∴ (4y2 + 3y + 1/2) / (2y + 1) = 2y + 1/2

7. 3x3 + 4x2 + 5x + 18 by x + 2

Solution:

We have,

(3x3 + 4x2 + 5x + 18) / (x + 2)

By using the long division method

∴ (3x3 + 4x2 + 5x + 18) / (x + 2) = 3x2 – 2x + 9

8. 14x2 – 53x + 45 by 7x – 9

Solution:

We have,

(14x2 – 53x + 45) / (7x – 9)

By using the long division method

∴ (14x2 – 53x + 45) / (7x – 9) = 2x – 5

9. -21 + 71x – 31x2 – 24x3 by 3 – 8x

Solution:

We have,

-21 + 71x – 31x2 – 24x3 by 3 – 8x

(-24x3 – 31x2 + 71x – 21) / (3 – 8x)

By using the long division method

∴ (-24x3 – 31x2 + 71x – 21) / (3 – 8x) = 3x2 + 5x – 7

10. 3y4 – 3y3 – 4y2 – 4y by y2 – 2y

Solution:

We have,

(3y4 – 3y3 – 4y2 – 4y) / (y2 – 2y)

By using the long division method

∴ (3y4 – 3y3 – 4y2 – 4y) / (y2 – 2y) = 3y2 + 3y + 2

11. 2y5 + 10y4 + 6y3 + y2 + 5y + 3 by 2y3 + 1

Solution:

We have,

(2y5 + 10y4 + 6y3 + y2 + 5y + 3) / (2y3 + 1)

By using the long division method

∴ (2y5 + 10y4 + 6y3 + y2 + 5y + 3) / (2y3 + 1) = y2 + 5y + 3

12. x4 – 2x3 + 2x2 + x + 4 by x2 + x + 1

Solution:

We have,

(x4 – 2x3 + 2x2 + x + 4) / (x2 + x + 1)

By using the long division method

∴ (x4 – 2x3 + 2x2 + x + 4) / (x2 + x + 1) = x2 – 3x + 4

13. m3 – 14m2 + 37m – 26 by m2 – 12m + 13

Solution:

We have,

(m3 – 14m2 + 37m – 26) / (m2 – 12m + 13)

By using the long division method

∴ (m3 – 14m2 + 37m – 26) / (m2 – 12m + 13) = m – 2

14. x4 + x2 + 1 by x2 + x + 1

Solution:

We have,

(x4 + x2 + 1) / (x2 + x + 1)

By using the long division method

∴ (x4 + x2 + 1) / (x2 + x + 1) = x2 – x + 1

15. x5 + x4 + x3 + x2 + x + 1 by x3 + 1

Solution:

We have,

(x5 + x4 + x3 + x2 + x + 1) / (x3 + 1)

By using the long division method

∴ (x5 + x4 + x3 + x2 + x + 1) / (x3 + 1) = x2 + x + 1

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

16. 14x3 – 5x2 + 9x – 1 by 2x – 1

Solution:

We have,

(14x3 – 5x2 + 9x – 1) / (2x – 1)

By using the long division method

∴ Quotient is 7x2 + x + 5 and the Remainder is 4.

17. 6x3 – x2 – 10x – 3 by 2x – 3

Solution:

We have,

(6x3 – x2 – 10x – 3) / (2x – 3)

By using the long division method

∴ Quotient is 3x2 + 4x + 1 and the Remainder is 0.

18. 6x3 + 11x2 – 39x – 65 by 3x2 + 13x + 13

Solution:

We have,

(6x3 + 11x2 – 39x – 65) / (3x2 + 13x + 13)

By using the long division method

∴ Quotient is 2x – 5 and the Remainder is 0.

19. 30x4 + 11x3 – 82x2 – 12x + 48 by 3x2 + 2x – 4

Solution:

We have,

(30x4 + 11x3 – 82x2 – 12x + 48) / (3x2 + 2x – 4)

By using the long division method

∴ Quotient is 10x2 – 3x – 12 and the Remainder is 0.

20. 9x4 – 4x2 + 4 by 3x2 – 4x + 2

Solution:

We have,

(9x4 – 4x2 + 4) / (3x2 – 4x + 2)

By using the long division method

∴ Quotient is 3x2 + 4x + 2 and the Remainder is 0.

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

Dividend divisor

(i) 14x2 + 13x – 15 7x – 4

(ii) 15z3 – 20z2 + 13z – 12 3z – 6

(iii) 6y5 – 28y3 + 3y2 + 30y – 9 2x2 – 6

(iv) 34x – 22x3 – 12x4 – 10x2 – 75 3x + 7

(v) 15y4 – 16y3 + 9y2 – 10/3y + 6 3y – 2

(vi) 4y3 + 8y + 8y2 + 7 2y2 – y + 1

(vii) 6y4 + 4y4 + 4y3 + 7y2 + 27y + 6 2y3 + 1

Solution:

(i) Dividend divisor

14x2 + 13x – 15 7x – 4

By using the long division method

Let us verify, Dividend = Divisor × Quotient + Remainder

14x2 + 13x – 15 = (7x – 4) × (2x + 3) + (-3)

= 14x2 + 21x – 8x -12 -3

= 14x2 + 13x – 15

Hence, verified.

∴ The quotient is 2x + 3 and the Remainder is -3.

(ii) Dividend divisor

15z3 – 20z2 + 13z – 12 3z – 6

By using the long division method

Let us verify, Dividend = Divisor × Quotient + Remainder

15z3 – 20z2 + 13z – 12 = (3z – 6) × (5z2 + 10z/3 + 11) + 54

= 15z3 + 10z2 + 33z – 30z2 – 20z + 54

= 15z2 – 20z2 + 13z – 12

Hence, verified.

∴ The quotient is 5z2 + 10z/3 + 11 and the Remainder is 54.

(iii) Dividend divisor

6y5 – 28y3 + 3y2 + 30y – 9 2x2 – 6

By using the long division method

Let us verify, Dividend = Divisor × Quotient + Remainder

6y5 – 28y3 + 3y2 + 30y – 9 = (2x2 – 6) × (3y3 – 5y + 3/2) + 0

= 6y5 – 10y3 + 3y2 – 18y3 + 30y – 9

= 6y5 – 28y3 + 3y2 + 30y – 9

Hence, verified.

∴ The quotient is 3y3 – 5y + 3/2 and the Remainder is 0.

(iv) Dividend divisor

34x – 22x3 – 12x4 – 10x2 – 75 3x + 7

-12x4 – 22x3 – 10x2 + 34x – 75

By using the long division method

Let us verify, Dividend = Divisor × Quotient + Remainder

-12x4 – 22x3 – 10x2 + 34x – 75 = (3x + 7) × (-4x3 + 2x2 – 8x + 30) – 285

= -12x4 + 6x3 – 24x2 – 28x3 + 14x2 + 90x – 56x + 210 -285

= -12x4 – 22x3 – 10x2 + 34x – 75

Hence, verified.

∴ The quotient is -4x3 + 2x2 – 8x + 30 and the Remainder is -285.

(v) Dividend divisor

15y4 – 16y3 + 9y2 – 10/3y + 6 3y – 2

By using the long division method

Let us verify, Dividend = Divisor × Quotient + Remainder

15y4 – 16y3 + 9y2 – 10/3y + 6 = (3y – 2) × (5y3 – 2y2 + 5y/3) + 6

= 15y4 – 6y3 + 5y2 – 10y3 + 4y2 – 10y/3 + 6

= 15y4 – 16y3 + 9y2 – 10/3y + 6

Hence, verified.

∴ The quotient is 5y3 – 2y2 + 5y/3 and the Remainder is 6.

(vi) Dividend divisor

4y3 + 8y + 8y2 + 7 2y2 – y + 1

4y3 + 8y2 + 8y + 7

By using the long division method

Let us verify, Dividend = Divisor × Quotient + Remainder

4y3 + 8y2 + 8y + 7 = (2y2 – y + 1) × (2y + 5) + 11y + 2

= 4y3 + 10y2 – 2y2 – 5y + 2y + 5 + 11y + 2

= 4y3 + 8y2 + 8y + 7

Hence, verified.

∴ The quotient is 2y + 5 and the Remainder is 11y + 2.

(vii) Dividend divisor

6y5 + 4y4 + 4y3 + 7y2 + 27y + 6 2y3 + 1

By using the long division method

Let us verify, Dividend = Divisor × Quotient + Remainder

6y5 + 4y4 + 4y3 + 7y2 + 27y + 6 = (2y3 + 1) × (3y2 + 2y + 2) + 4y2 + 25y + 4

= 6y5 + 4y4 + 4y3 + 3y2 + 2y + 2 + 4y2 + 25y + 4

= 6y5 + 4y4 + 4y3 + 7y2 + 27y + 6

Hence, verified.

∴ The quotient is 3y2 + 2y + 2 and the Remainder is 4y2 + 25y + 4.

22. Divide 15y4+ 16y3 + 10/3y – 9y2 – 6 by 3y – 2 Write down the coefficients of the terms in the quotient.

Solution:

We have,

(15y4+ 16y3 + 10/3y – 9y2 – 6) / (3y – 2)

By using the long division method

∴ The quotient is 5y3 + 26y2/3 + 25y/9 + 80/27

So the coefficients of the terms in the quotient are:

Coefficient of y3 = 5

Coefficient of y2 = 26/3

Coefficient of y = 25/9

Constant term = 80/27

23. Using the division of polynomials, state whether
(i) x + 6 is a factor of x2 – x – 42
(ii) 4x – 1 is a factor of 4x2 – 13x – 12
(iii) 2y – 5 is a factor of 4y4 – 10y3 – 10y2 + 30y – 15
(iv) 3y2 + 5 is a factor of 6y5 + 15y4 + 16y3 + 4y2 + 10y – 35
(v) z2 + 3 is a factor of z5 – 9z
(vi) 2x2 – x + 3 is a factor of 6x5 – x4 + 4x3 – 5x2 – x – 15

Solution:

(i) x + 6 is a factor of x2 – x – 42

Firstly let us perform long division method

Since the remainder is 0, we can say that x + 6 is a factor of x2 – x – 42

(ii) 4x – 1 is a factor of 4x2 – 13x – 12

Firstly let us perform long division method

Since the remainder is -15, 4x – 1 is not a factor of 4x2 – 13x – 12

(iii) 2y – 5 is a factor of 4y4 – 10y3 – 10y2 + 30y – 15

Firstly let us perform long division method

Since the remainder is 5y3 – 45y2/2 + 30y – 15, 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

Firstly let us perform long division method

Since the remainder is 0, 3y2 + 5 is a factor of 6y5 + 15y4 + 16y3 + 4y2 + 10y – 35

(v) z2 + 3 is a factor of z5 – 9z

Firstly let us perform long division method

Since the remainder is 0, z2 + 3 is a factor of z5 – 9z

(vi) 2x2 – x + 3 is a factor of 6x5 – x4 + 4x3 – 5x2 – x – 15

Firstly let us perform long division method

Since the remainder is 0, 2x2 – x + 3 is a factor of 6x5 – x4 + 4x3 – 5x2 – x – 15

24. Find the value of a, if x + 2 is a factor of 4x4 + 2x3 – 3x2 + 8x + 5a

Solution:

We know that x + 2 is a factor of 4x4 + 2x3 – 3x2 + 8x + 5a

Let us equate x + 2 = 0

x = -2

Now let us substitute x = -2 in the equation 4x4 + 2x3 – 3x2 + 8x + 5a

4(-2)4 + 2(-2)3 – 3(-2)2 + 8(-2) + 5a = 0

64 – 16 – 12 – 16 + 5a = 0

20 + 5a = 0

5a = -20

a = -20/5

= -4

25. What must be added to x4 + 2x3 – 2x2 + x – 1 so that the resulting polynomial is exactly divisible by x2 + 2x -3.

Solution:

Firstly let us perform long division method

By long division method, we got the remainder as –x + 2,

∴ x – 2 has to be added to x4 + 2x3 – 2x2 + x – 1 so that the resulting polynomial is exactly divisible by x2 + 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) 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

By using the long division method

∴ the Quotient is 3x + 10 and the Remainder is 25.

(ii) 10x2 – 7x + 8, 5x – 3

By using the long division method

∴ the Quotient is 2x – 1/5 and the Remainder is 37/5.

(iii) 5y3 – 6y2 + 6y – 1, 5y – 1

By using the long division method

∴ the Quotient is y2 – y + 1 and the Remainder is 0.

(iv) x4 – x3 + 5x, x – 1

By using the long division method

∴ the Quotient is x3 + 5 and the Remainder is 5.

(v) y4 + y2, y2 – 2

By using the long division method

∴ the Quotient is y2 + 3 and the Remainder is 6.

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, 4x4 + 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

Let us perform long division method,

Since the remainder is 1, the first polynomial is not a factor of the second polynomial.

(ii) y – 2, 3y3 + 5y2 + 5y + 2

Let us perform long division method,

Since the remainder is 56, the first polynomial is not a factor of the second polynomial.

(iii) 4x2 – 5, 4x4 + 7x2 + 15

Let us perform long division method,

Since the remainder is 30, the first polynomial is not a factor of the second polynomial.

(iv) 4 – z, 3z2 – 13z + 4

Let us perform long division method,

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

(v) 2a – 3, 10a2 – 9a – 5

Let us perform long division method,

Since the remainder is 4, the first polynomial is not a factor of the second polynomial.

(vi) 4y + 1, 8y2 – 2y + 1

Let us perform long division method,

Since the remainder is 2, the first polynomial is not a factor of the second polynomial.

### EXERCISE 8.6 PAGE NO: 8.17

Divide:

1. x2 – 5x + 6 by x – 3

Solution:

We have,

(x2 – 5x + 6) / (x – 3)

Let us perform long division method,

∴ the Quotient is x – 2

2. ax2 – ay2 by ax+ay

Solution:

We have,

(ax2 – ay2)/ (ax+ay)

(ax2 – ay2)/ (ax+ay) = (x – y) + 0/(ax+ay)

= (x – y)

∴ the answer is (x – y)

3. x4 – y4 by x2 – y2

Solution:

We have,

(x4 – y4)/ (x2 – y2)

(x4 – y4)/ (x2 – y2) = x2 + y2 + 0/(x2 – y2)

= x2 + y2

∴ the answer is (x2 + y2)

4. acx2 + (bc + ad)x + bd by (ax + b)

Solution:

We have,

(acx2 + (bc + ad) x + bd) / (ax + b)

(acx2 + (bc + ad) x + bd) / (ax + b) = cx + d + 0/ (ax + b)

= cx + d

∴ the answer is (cx + d)

5. (a2 + 2ab + b2) – (a2 + 2ac + c2) by 2a + b + c

Solution:

We have,

[(a2 + 2ab + b2) – (a2 + 2ac + c2)] / (2a + b + c) [(a2 + 2ab + b2) – (a2 + 2ac + c2)] / (2a + b + c) = b – c + 0/(2a + b + c)

= b – c

∴ the answer is (b – c)

6. 1/4x2 – 1/2x – 12 by 1/2x – 4

Solution:

We have,

(1/4x2 – 1/2x – 12) / (1/2x – 4)

Let us perform long division method,

∴ the Quotient is x/2 + 3