Chapter 11: Algebraic Expressions

Exercise 11(A)

Question 1(i)

The sum of a - b + ab, b - c + bc and c - a + ca is:

0
2(a + b + c)
ab + bc + ca
none of these

Answer

The sum of a - b + ab, b - c + bc and c - a + ca is,

= a - b + ab + b - c + bc + c - a + ca

= (a - a) + (- b + b) + ab + bc + ca.

= ab + bc + ca

Hence, option 3 is the correct option.

Question 1(ii)

(x³ - 5x² + 7)+ (3x² + 5x - 2)+ (2x³ - x + 7) is equal to:

3x³ - 2x² + 4x + 12
3x³ + 2x² - 4x + 12
3x³ - 2x² - 4x - 12
3x³ + 2x² + 4x + 12

Answer

(x³ - 5x² + 7) + (3x² + 5x - 2) + (2x³ - x + 7)

= (x³ + 2x³) + (-5x² + 3x²) + (5x - x) + (7 - 2 + 7)

= 3x³ - 2x² + 4x + 12

Hence, option 1 is the correct option.

Question 1(iii)

(x³ - 5x² + 3x + 2) - (6x² - 4x³ + 3x + 5) is equal to:

5x³ + 11x² - 2
5x³ - 11x² + 3
5x³ - 11x² - 3
5x³ + 11x² + 3

Answer

(x³ - 5x² + 3x + 2) - (6x² - 4x³ + 3x + 5)

= x³ - 5x² + 3x + 2 - 6x² + 4x³ - 3x - 5

= (x³ + 4x³) + (- 5x² - 6x²) (+ 3x - 3x) + (2 - 5)

= 5x³ - 11x² - 3

Hence, option 3 is the correct option.

Question 1(iv)

p - (p - q) - q - (q - p) is equal to:

q - p
p - q
p + q
2p - q

Answer

p - (p - q) - q - (q - p)

= p - p + q - q - q + p

= (p - p + p) + (q - q - q)

= p - q

Hence, option 2 is the correct option.

Question 1(v)

(ab - bc) - (ca - bc) + (ca - ab) is:

bc - ab
ab - ca
2(ab - bc - ca)
0

Answer

(ab - bc) - (ca - bc) + (ca - ab)

= ab - bc - ca + bc + ca - ab

= (ab - ab) + (bc - bc) + (- ca + ca)

= 0

Hence, option 4 is the correct option.

Question 2

Separate the constants and variables from the following:

- 7, 7 + x, 7x + yz, $\sqrt{5}$ , $\sqrt{xy}$ , $\dfrac{3yz}{8}$ , 4.5y - 3x, 8 - 5, 8 - 5x, 8x - 5y x p and 3y²z ÷ 4x

Answer

Constants = -7, $\sqrt{5}$ and 8 - 5

Variables = 7 + x, 7x + yz, $\sqrt{xy}$ , $\dfrac{3yz}{8}$ , 4.5y - 3x, 8 - 5x, 8x - 5y x p and 3y²z ÷ 4x

Question 3

Write the number of terms in each of the following polynomials:

(i) 5x² + 3 x ax

(ii) ax ÷ 4 - 7

(iii) ax - by + y x z

(iv) 23 + a x b ÷ 2

Answer

(i) Terms are 5x² and 3 x ax

Hence, number of terms = 2.

(ii) Terms are ax ÷ 4 and 7

Hence, number of terms = 2.

(iii) Terms are ax , by and y x z

Hence, number of terms = 3.

(iv) Terms are 23 and a x b ÷ 2

Hence, number of terms = 2.

Question 4

Separate monomials, binomials, trinomials and multinomial from the following algebraic expressions:

8 - 3x, xy², 3y² - 5y + 8, 9x - 3x² + 15x³ - 7, 3x x 5y, 3x ÷ 5y, 2y ÷ 7 + 3x - 7 and 4 - ax² + bx + y

Answer

Monomials are those algebraic expression which consist of one term.

xy², 3x x 5y and 3x ÷ 5y are monomials.

Binomials are those algebraic expression which consist of two terms.

8 - 3x is binomial.

Trinomials are those algebraic expression which consist of three terms.

3y² - 5y + 8 and 2y ÷ 7 + 3x - 7 are trinomials

Multinominal are those algebraic expression which consist two or more terms.

8 - 3x, 3y² - 5y + 8, 9x - 3x² + 15x³ - 7, 2y ÷ 7 + 3x - 7 and 4 - ax² + bx + y are multinomials.

Question 5

Write the degree of each polynomial given below:

(i) xy + 7z

(ii) x² - 6x³ + 8

(iii) y - 6y² + 5y⁸

(iv) xyz - 3

(v) xy + yz² - zx³

(vi) x⁵y⁷ - 8x³y⁸ + 10x⁴y⁴z⁴

Answer

(i) Degree = 2
Reason — As the polynomial contains 3 variables, we will find the sum of powers of each term.
The sum of powers of the term xy = 1 + 1 = 2
The sum of powers of the term 7z = 1
∵ Highest sum of powers = 2
∴ Degree of polynomial = 2

(ii) Degree = 3
Reason — As the polynomial contains 1 variable, degree of a polynomial is the highest power of the variable in a polynomial expression.
Highest power of polynomial = 3
∴ Degree of polynomial = 3

(iii) Degree = 8
Reason — As the polynomial contains 1 variable, degree of a polynomial is the highest power of the variable in a polynomial expression.
Highest power of polynomial = 8
∴ Degree of polynomial = 8

(iv) Degree = 3
Reason — As the polynomial contains 3 variables, we will find the sum of powers of each term.
The sum of powers of the term xyz = 1 + 1 + 1 = 3
The powers of the term 3 = 0
∵ Highest sum of powers = 3
∴ Degree of polynomial = 3

(v) Degree = 4
Reason — As the polynomial contains 3 variables, we will find the sum of powers of each term.
The sum of powers of the term xy = 1 + 1 = 2
The sum of powers of the term yz² = 1 + 2 = 3
The sum of powers of the term zx³ = 1 + 3 = 4
∵ Highest sum of powers = 4
∴ Degree of polynomial = 4

(vi) Degree = 12
Reason — As the polynomial contains 3 variables, we will find the sum of powers of each term.
The sum of powers of the term x⁵y⁷ = 5 + 7 = 12
The sum of powers of the term - 8x³y⁸ = 3 + 8 = 11
The sum of powers of the term 10x⁴y⁴z⁴ = 4 + 4 + 4 = 12
∵ Highest sum of powers = 12
∴ Degree of polynomial = 12

Question 6(i)

Write the coefficient of:

ab in 7abx

Answer

Coefficient of ab = 7x.

Question 6(ii)

Write the coefficient of:

7a in 7abx

Answer

Coefficient of 7a = bx.

Question 6(iii)

Write the coefficient of:

5x² in 5x² - 5x

Answer

Coefficient of 5x² = 1.

Question 6(iv)

Write the coefficient of:

8 in a² - 8ax + a

Answer

Coefficient of 8 = -ax.

Question 6(v)

Write the coefficient of:

4xy in x² - 4xy + y²

Answer

Coefficient of 4xy = -1.

Question 7(i)

Evaluate:

-7x² + 18x² + 3x² - 5x²

Answer

-7x² + 18x² + 3x² - 5x²

Arrange the polynomial with like terms together and combine them.

= (-7x² + 18x²) + 3x² - 5x²

= (11x² + 3x²) - 5x²

= (14x² - 5x²)

= 9x²

Hence, -7x² + 18x² + 3x² - 5x²= 9x²

Question 7(ii)

Evluate:

b²y - 9b²y + 2b²y - 5b²y

Answer

b²y - 9b²y + 2b²y - 5b²y

Arrange the polynomial with like terms together and combine them.

= (b²y - 9b²y) + 2b²y - 5b²y

= (-8b²y + 2b²y) - 5b²y

= (-6b²y - 5b²y)

= -11b²y

Hence, b²y - 9b²y + 2b²y - 5b²y = -11b²y

Question 7(iii)

Evaluate:

abx - 15abx - 10abx + 32abx

Answer

abx - 15abx - 10abx + 32abx

Arrange the polynomial with like terms together and combine them.

= (abx - 15abx) - 10abx + 32abx

= (-14abx - 10abx) + 32abx

= (-24abx + 32abx)

= 8abx

Hence, abx - 15abx - 10abx + 32abx = 8abx

Question 7(iv)

Evaluate:

7x - 9y + 3 - 3x - 5y + 8

Answer

7x - 9y + 3 - 3x - 5y + 8

Arrange the polynomial with like terms together and combine them.

(7x - 3x) - 9y + 3 - 5y + 8

= 4x + (- 9y - 5y) + 3 + 8

= 4x - 14y + (3 + 8)

= 4x - 14y + 11

Hence, 7x - 9y + 3 - 3x - 5y + 8 = 4x - 14y + 11

Question 7(v)

Evaluate:

3x² + 5xy - 4y² + x² - 8xy - 5y²

Answer

3x² + 5xy - 4y² + x² - 8xy - 5y²

Arrange the polynomial with like terms together and combine them.

= (3x² + x²) + 5xy - 4y² - 8xy - 5y²

= 4x² + (5xy - 8xy) - 4y² - 5y²

= 4x² - 3xy + (- 4y² - 5y²)

= 4x² - 3xy - 9y²

Hence, 3x² + 5xy - 4y² + x² - 8xy - 5y² = 4x² - 3xy - 9y²

Question 8(i)

Add :

5a + 3b, a - 2b, 3a + 5b

Answer

5a + 3b + a - 2b + 3a + 5b

Arrange the polynomial with like terms together and combine them.

= (5a + a + 3a) + (3b - 2b + 5b)

= 9a + 6b

Hence, addition of 5a + 3b, a - 2b, 3a + 5b is 9a + 6b.

Question 8(ii)

Add :

8x - 3y + 7z, - 4x + 5y - 4z, - x - y - 2z

Answer

8x - 3y + 7z + (- 4x + 5y - 4z )+ (- x - y - 2z)

= 8x - 3y + 7z - 4x + 5y - 4z - x - y - 2z

Arrange the polynomial with like terms together and combine them.

= (8x - 4x - x) + (- 3y + 5y - y) + (7z - 4z - 2z)

= 3x + y + z

Hence, the addition of 8x - 3y + 7z, - 4x + 5y - 4z, - x - y - 2z is 3x + y + z.

Question 8(iii)

Add :

3b - 7c + 10, 5c - 2b - 15, 15 + 12c + b

Answer

3b - 7c + 10 + 5c - 2b - 15 + 15 + 12c + b

Arrange the polynomial with like terms together and combine them.

= (3b - 2b + b) + (-7c + 5c + 12c) + (10 - 15 + 15)

= 2b + 10c + 10

Hence, addition of 3b - 7c + 10, 5c - 2b - 15, 15 + 12c + b is 2b + 10c + 10

Question 8(iv)

Add :

a - 3b + 3, 2a + 5 - 3c, 6c - 15 + 6b

Answer

a - 3b + 3 + 2a + 5 - 3c + 6c - 15 + 6b

Arrange the polynomial with like terms together and combine them.

= (a + 2a) + (- 3b + 6b) + (3 + 5 - 15) + (- 3c + 6c)

= 3a + 3b - 7 + 3c

Hence, the addition of a - 3b + 3, 2a + 5 - 3c, 6c - 15 + 6b is 3a + 3b + 3c - 7

Question 8(v)

Add :

13ab - 9cd - xy, 5xy, 15cd - 7ab, 6xy - 3cd

Answer

13ab - 9cd - xy + 5xy + 15cd - 7ab + 6xy - 3cd

Arrange the polynomial with like terms together and combine them.

= (13ab - 7ab) + (-9cd + 15cd - 3cd) + (-xy + 5xy + 6xy)

= 6ab + 3cd + 10xy

Hence, the addition of 13ab - 9cd - xy, 5xy, 15cd - 7ab, 6xy - 3cd is 6ab + 3cd + 10xy.

Question 8(vi)

Add :

x³ - x²y + 5xy² + y³, -x³ - 9xy² + y³, 3x²y + 9xy²

Answer

x³ - x²y + 5xy² + y³ + (-x³ - 9xy² + y³) + 3x²y + 9xy²

Arrange the polynomial with like terms together and combine them.

= (x³ - x³) + (- x²y + 3x²y) + (5xy² - 9xy² + 9xy²) + (y³ + y³)

= 0 + 2x²y + 5xy² + 2y³

Hence, the addition of x³ - x²y + 5xy² + y³, -x³ - 9xy² + y³, 3x²y + 9xy² is 2x²y + 5xy² + 2y³.

Question 9

Find the total savings of a boy who saves ₹ (4x - 6y), ₹ (6x + 2y), ₹ (4y - x) and ₹ (y - 2x) in four consecutive weeks.

Answer

The total savings = ₹ (4x - 6y) + ₹ (6x + 2y) + ₹ (4y - x) + ₹ (y - 2x)

= 4x - 6y + 6x + 2y + 4y - x + y - 2x

Arrange the polynomial with like terms together and combine them.

= (4x + 6x - x - 2x) + (- 6y + 2y + 4y + y)

= 7x + y

Hence, the total saving is ₹ (7x + y).

Question 10(i)

Subtract:

4xy² from 3xy²

Answer

3xy² - 4xy²

= -xy²

Hence, 3xy² - 4xy² = - 1xy²

Question 10(ii)

Subtract:

- 2x²y + 3xy² from 8x²y

Answer

8x²y - (-2x²y + 3xy²)

= 8x²y + 2x²y - 3xy²

Arrange the polynomial with like terms together and combine them.

= (8x²y + 2x²y) - 3xy²

= 10x²y - 3xy²

Hence, 8x²y - (- 2x²y + 3xy²) = 10x²y - 3xy²

Question 10(iii)

Subtract:

3a - 5b + c + 2d from 7a - 3b + c - 2d

Answer

7a - 3b + c - 2d - (3a - 5b + c + 2d)

= 7a - 3b + c - 2d - 3a + 5b - c - 2d

Arrange the polynomial with like terms together and combine them.

= (7a - 3a) + (- 3b + 5b) + (c - c) + (- 2d - 2d)

= 4a + 2b - 4d

Hence, 7a - 3b + c - 2d - (3a - 5b + c + 2d) = 4a + 2b - 4d.

Question 10(iv)

Subtract:

x³ - 4x - 1 from 3x³ - x² + 6

Answer

3x³ - x² + 6 - (x³ - 4x - 1)

= 3x³ - x² + 6 - x³ + 4x + 1

Arrange the polynomial with like terms together and combine them.

= (3x³ - x³) - x² + 4x + (6 + 1)

= 2x³ - x² + 4x + 7

Hence, 3x³ - x² + 6 - (x³ - 4x - 1) = 2x³ - x² + 4x + 7

Question 10(v)

Subtract:

6a + 3 from a³ - 3a² + 4a + 1

Answer

a³ - 3a² + 4a + 1 - (6a + 3)

= a³ - 3a² + 4a + 1 - 6a - 3

Arrange the polynomial with like terms together and combine them.

= a³ - 3a² + (4a - 6a) + (1 - 3)

= a³ - 3a² - 2a - 2

Hence, a³ - 3a² + 4a + 1 - (6a + 3) = a³ - 3a² - 2a - 2.

Question 10(vi)

Subtract:

cab - 4cad - cbd from 3abc + 5bcd - cda

Answer

3abc + 5bcd - cda - (cab - 4cad - cbd)

= 3abc + 5bcd - cda - cab + 4cad + cbd

Arrange the polynomial with like terms together and combine them.

= (3abc - abc) + (5bcd + bcd) + (- cda + 4cda)

= 2abc + 6bcd + 3cda

Hence, 3abc + 5bcd - cda - (cab - 4cad - cbd) = 2abc + 6bcd + 3cda.

Question 10(vii)

Subtract:

a² + ab + b² from 4a² - 3ab + 2b²

Answer

4a² - 3ab + 2b² - (a² + ab + b²)

= 4a² - 3ab + 2b² - a² - ab - b²

Arrange the polynomial with like terms together and combine them.

= (4a² - a²) + (- 3ab - ab) + (2b² - b²)

= 3a² - 4ab + b²

Hence, 4a² - 3ab + 2b² - (a² + ab + b²) = 3a² - 4ab + b².

Question 11(i)

Take away - 3x³ + 4x² - 5x + 6 from 3x³ - 4x² + 5x - 6.

Answer

3x³ - 4x² + 5x - 6 - (- 3x³ + 4x² - 5x + 6)

= 3x³ - 4x² + 5x - 6 + 3x³ - 4x² + 5x - 6

Arrange the polynomial with like terms together and combine them.

= (3x³ + 3x³) + (- 4x² - 4x²) + (5x + 5x) + (- 6 - 6)

= 6x³ - 8x² + 10x - 12

Hence, 3x³ - 4x² + 5x - 6 - (- 3x³ + 4x² - 5x + 6) = 6x³ - 8x² + 10x - 12.

Question 11(ii)

Take m² + m + 4 from -m² + 3m + 6 and the result from m² + m + 1.

Answer

Difference between m² + m + 4 and - m² + 3m + 6

- m² + 3m + 6 - (m² + m + 4)

= - m² + 3m + 6 - m² - m - 4

Arrange the polynomial with like terms together and combine them.

= (- m² - m²) + (3m - m) + (6 - 4)

= - 2m² + 2m + 2

And, the difference between - 2m² + 2m + 2 and m² + m + 1

(m² + m + 1) - (- 2m² + 2m + 2)

= m² + m + 1 + 2m² - 2m - 2

Arrange the polynomial with like terms together and combine them.

= ( m² + 2m²) + (m - 2m) + (1 - 2)

= 3m² - m - 1

Hence, the result is 3m² - m - 1.

Question 12

Subtract the sum of 5y² + y - 3 and y² - 3y + 7 from 6y² + y - 2.

Answer

The sum of 5y² + y - 3 and y² - 3y + 7

5y² + y - 3 + (y² - 3y + 7)

= 5y² + y - 3 + y² - 3y + 7

Arrange the polynomial with like terms together and combine them.

= (5y² + y²) + (y - 3y) + (- 3 + 7)

= 6y² - 2y + 4

The difference of 6y² - 2y + 4 from 6y² + y - 2

= 6y² + y - 2 - (6y² - 2y + 4)

= 6y² + y - 2 - 6y² + 2y - 4

= (6y² - 6y²) + (y + 2y) + (- 2 - 4)

= 0 + 3y - 6

Hence, the result is 3y - 6.

Question 13

What must be added to x⁴ - x³ + x² + x + 3 to obtain x⁴ + x² - 1 ?

Answer

Let the number Z must be added to x⁴ - x³ + x² + x + 3 to obtain x⁴ + x² - 1

⇒ Z + (x⁴ - x³ + x² + x + 3) = (x⁴ + x² - 1)

⇒ Z = (x⁴ + x² - 1) - (x⁴ - x³ + x² + x + 3)

⇒ Z = x⁴ + x² - 1 - x⁴ + x³ - x² - x - 3

Arrange the polynomial with like terms together and combine them.

⇒ Z = (x⁴ - x⁴) + x³ + (x² - x²) - x + (- 1 - 3)

⇒ Z = 0 + x³ + 0 - x - 4

Hence, the number is x³ - x - 4.

Question 14(i)

How much more than 2x² + 4xy + 2y² is 5x² + 10xy - y²?

Answer

Let 5x² + 10xy - y² be more than 2x² + 4xy + 2y² by Z.

2x² + 4xy + 2y² + Z = 5x² + 10xy - y²

⇒ Z = 5x² + 10xy - y² - (2x² + 4xy + 2y²)

⇒ Z = 5x² + 10xy - y² - 2x² - 4xy - 2y²

Arrange the polynomial with like terms together and combine them.

⇒ Z = (5x² - 2x²) + (10xy - 4xy) + (- y² - 2y²)

⇒ Z = 3x² + 6xy - 3y²

Hence, the number is 3x² + 6xy - 3y².

Question 14(ii)

How much less 2a² + 1 is than 3a² - 6 ?

Answer

Let 2a² + 1 be less than 3a² - 6 by Z.

3a² - 6 - Z = 2a² + 1

⇒ Z = 3a² - 6 - (2a² + 1)

⇒ Z = 3a² - 6 - 2a² - 1

Arrange the polynomial with like terms together and combine them.

⇒ Z = (3a² - 2a²) + (- 6 - 1)

⇒ Z = a² - 7

Hence, the number is a² - 7.

Question 15

If x = 6a + 8b + 9c; y = 2b - 3a - 6c and z = c - b + 3a; find :

(i) x + y + z

(ii) x - y + z

(iii) 2x - y - 3z

(iv) 3y - 2z - 5x

Answer

(i) If x = 6a + 8b + 9c; y = 2b - 3a - 6c and z = c - b + 3a, then

x + y + z = (6a + 8b + 9c) + (2b - 3a - 6c) + (c - b + 3a)

= 6a + 8b + 9c + 2b - 3a - 6c + c - b + 3a

Arrange the polynomial with like terms together and combine them.

= (6a - 3a + 3a) + (8b + 2b - b) + (9c - 6c + c)

= 6a + 9b + 4c

Hence, x + y + z = 6a + 9b + 4c

(ii) If x = 6a + 8b + 9c; y = 2b - 3a - 6c and z = c - b + 3a, then

x - y + z = (6a + 8b + 9c) - (2b - 3a - 6c) + (c - b + 3a)

= 6a + 8b + 9c - 2b + 3a + 6c + c - b + 3a

Arrange the polynomial with like terms together and combine them.

= (6a + 3a + 3a) + (8b - 2b - b) + (9c + 6c + c)

= 12a + 5b + 16c

Hence, x - y + z = 12a + 5b + 16c

(iii) If x = 6a + 8b + 9c; y = 2b - 3a - 6c and z = c - b + 3a, then

2x - y - 3z = 2(6a + 8b + 9c) - (2b - 3a - 6c) - 3(c - b + 3a)

= 12a + 16b + 18c - 2b + 3a + 6c - 3c + 3b - 9a

Arrange the polynomial with like terms together and combine them.

= (12a + 3a - 9a) + (16b - 2b + 3b) + (18c + 6c - 3c)

= 6a + 17b + 21c

Hence, 2x - y - 3z = 6a + 17b + 21c

(iv) If x = 6a + 8b + 9c; y = 2b - 3a - 6c and z = c - b + 3a, then

3y - 2z - 5x = 3(2b - 3a - 6c) - 2(c - b + 3a) - 5(6a + 8b + 9c)

= 6b - 9a - 18c - 2c + 2b - 6a - 30a - 40b - 45c

Arrange the polynomial with like terms together and combine them.

= (-9a - 6a - 30a) + (6b + 2b - 40b) + (-18c - 2c - 45c)

= -45a - 32b - 65c

Hence, 3y - 2z - 5x = -45a - 32b - 65c

Exercise 11(B)

Question 1(i)

(9x⁴ - 8x³ - 12x) x (3x) is equal to:

27x⁵ - 24x⁴ + 36x²
27x⁵ - 24x⁴ - 36x²
27x⁵ + 24x⁴ - 36x²
27x⁵ + 24x⁴ + 36x²

Answer

(3x) $\times$ (9x⁴ - 8x³ - 12x)

= 3x $\times$ 9x⁴ - 3x $\times$ 8x³ - 3x $\times$ 12x

= 27x⁽⁴⁺¹⁾ - 24x⁽³⁺¹⁾ - 36x⁽¹⁺¹⁾

= 27x^5 - 24x^4 - 36x^2

Hence, option 2 is the correct option.

Question 1(ii)

(9x⁴ - 12x³ - 18x) ÷ (3x) is equal to:

3x³ + 4x² + 6
3x³ + 4x² - 6
3x³ - 4x² - 6
3x⁴ - 4x² + 6

Answer

(9x⁴ - 12x³ - 18x) ÷ (3x)

= (9x⁴ ÷ 3x) - (12x³ ÷ 3x) - (18x ÷ 3x)

= 3x^(4-1) - 4x^(3-1) - 6x^(1-1)

= 3x³ - 4x² - 6

Hence, option 3 is the correct option.

Question 1(iii)

$\Big(\dfrac{10}{3}xy^2z\Big)$ x $\Big(-\dfrac{9}{5}x^2z\Big)$ is equal io:

- 6x³y²z²
6x³y²z²
$-\dfrac{2}{3}x^2y^2z^2$
$\dfrac{9}{5}xy^3z^2$

Answer

$\Big(\dfrac{10}{3}xy^2z\Big) \times \Big(-\dfrac{9}{5}x^2z\Big)\\[1em] = \Big(-\dfrac{10 \times 9}{3 \times 5}x^{1+2}y^2z^{1+1}\Big)\\[1em] = \Big(-\dfrac{90}{15}x^{3}y^2z^{2}\Big)\\[1em] = (-6x^3y^2z^2)$

Hence, option 1 is the correct option.

Question 1(iv)

(x³ + y²) x 10x² is equal to :

10x⁵ + 10xy
10x⁵ + 10x²y²
10x⁵ - 10x²y²
-10x⁵ - 10x²y²

Answer

(x³+ y²) $\times$ 10x²

= 10x² $\times$ x³ + 10x² $\times$ y²

= 10x⁽²⁺³⁾ + 10x²y²

= 10x⁵ + 10x²y²

Hence, option 2 is the correct option.

Question 1(v)

(a³ - b³) ÷ (a - b) is equal to :

a² + b² + ab
a² + b² - ab
a² - b² + ab
a² - b² - ab

Answer

(a³ - b³) ÷ (a - b)

[∵ As we know, (x³ - y³) = (x - y)(x² + y² + xy)]

∴ (a³ - b³) ÷ (a - b) = $\dfrac{(a - b)(a^2 + b^2 + ab)}{(a - b)}$

= $\dfrac{\cancel{(a - b)}(a^2 + b^2 + ab)}{\cancel{(a - b)}}$

= (a² + b² + ab)

Hence, option 1 is the correct option.

Question 2(i)

Multiply :

8ab² by - 4a³b⁴

Answer

8ab² x (- 4a³b⁴)

= 8 x (-4) a⁽¹⁺³⁾ b⁽²⁺⁴⁾

= -32 a⁴ b⁶

Hence, 8ab² x (- 4a³b⁴) = -32a⁴ b⁶

Question 2(ii)

Multiply :

$\dfrac{2}{3}$ ab by - $\dfrac{1}{4}$ a²b

Answer

$\dfrac{2}{3}ab \times \Big(-\dfrac{1}{4}a^2b\Big)\\[1em] = -\dfrac{2 \times 1}{3 \times 4}a^{1+2}b^{1+1}\\[1em] = -\dfrac{2}{12}a^3b^2\\[1em] = -\dfrac{1}{6}a^3b^2$

Hence, $\dfrac{2}{3}$ ab x (- $\dfrac{1}{4}$ a²b) = $-\dfrac{1}{6}$ a³b².

Question 2(iii)

Multiply :

-5cd² by - 5cd²

Answer

(-5cd²) x (- 5cd²)

= (-5) x (-5) c⁽¹⁺¹⁾d⁽²⁺²⁾

= 25c²d⁴

Hence, (-5cd²) x (-5cd²) = 25c²d⁴

Question 2(iv)

Multiply :

4a and 6a + 7

Answer

4a x (6a + 7)

= 4a x 6a + 4a x 7

= 24a⁽¹⁺¹⁾ + 28a

= 24a² + 28a

Hence, 4a x (6a + 7) = 24a² + 28a

Question 2(v)

Multiply :

-8x and 4 - 2x - x²

Answer

-8x $\times$ (4 - 2x - x²)

= (-8x) $\times$ 4 - (-8x) $\times$ 2x - (-8x) $\times$ x²

= -32x + 16x⁽¹⁺¹⁾ + 8x⁽¹⁺²⁾

= -32x + 16x² + 8x³

Hence, (-8x) x (4 - 2x - x²) = -32x + 16x² + 8x³

Question 2(vi)

Multiply :

2a² - 5a - 4 and -3a

Answer

-3a x (2a² - 5a - 4)

= -3a x 2a² - (-3a) x 5a - (-3a) x 4

= -6a⁽²⁺¹⁾ + 15a⁽¹⁺¹⁾ + 12a

= -6a³ + 15a² + 12a

Hence, (2a² - 5a - 4) x (-3a) = -6a³ + 15a² + 12a

Question 2(vii)

Multiply :

x + 4 by x - 5

Answer

(x + 4) $\times$ (x - 5)

= x $\times$ (x - 5) + 4 $\times$ (x - 5)

= x $\times$ x - x $\times$ 5 + 4 $\times$ x - 4 $\times$ 5

= x² - 5x + 4x - 20

= x² - x - 20

Hence, (x + 4) x (x - 5) = x² - x - 20

Question 2(viii)

Multiply :

5a - 1 by 7a - 3

Answer

(5a - 1) x (7a - 3)

= 5a x (7a - 3) - 1 x (7a - 3)

= 5a x 7a - 5a x 3 - 1 x 7a - 1 x (-3)

= 35a⁽¹⁺¹⁾ - 15a - 7a + 3

= 35a² - 22a + 3

Hence, (5a - 1) x (7a - 3) = 35a² - 22a + 3

Question 2(ix)

Multiply :

12a + 5b by 7a - b

Answer

(12a + 5b) x (7a - b)

= 12a x 7a - 12a x b + 5b x 7a - 5b x b

= 84a⁽¹⁺¹⁾ - 12ab + 35ab - 5b⁽¹⁺¹⁾

= 84a² + 23ab - 5b²

Hence, (12a + 5b) x (7a - b) = 84a² + 23ab - 5b²

Question 2(x)

Multiply :

x² + x + 1 by 1 - x

Answer

(x² + x + 1) $\times$ (1 - x)

= 1 $\times$ (x² + x + 1) - x $\times$ (x² + x + 1)

= (1 $\times$ x² + 1 $\times$ x + 1 $\times$ 1) - (x $\times$ x² + x $\times$ x + x $\times$ 1)

= x² + x + 1 - (x³ + x² + x)

= x² + x + 1 - x³ - x² - x

= (x² - x²) + (x - x) + 1 - x³

= 1 - x³

Hence, (x² + x + 1) $\times$ (1 - x) = 1 - x³

Question 2(xi)

Multiply :

2m² - 3m - 1 and 4m² - m - 1

Answer

(2m² - 3m - 1) x (4m² - m - 1)

= 2m² x 4m² - 2m² x m - 2m² x 1 - 3m x 4m² - 3m x (-m) - 3m x (-1) - 1 x 4m² - 1 x (- m) - 1 x (-1)

= 8m⁽²⁺²⁾ - 2m⁽²⁺¹⁾ - 2m² - 12m⁽²⁺¹⁾ + 3m⁽¹⁺¹⁾ + 3m - 4m² + m + 1

= 8m⁴ - 2m³ - 2m² - 12m³ + 3m² + 3m - 4m² + m + 1

= 8m⁴ - 2m³ - 12m³ - 2m² + 3m² - 4m² + 3m + m + 1

= 8m⁴ - 14m³ - 3m² + 4m + 1

Hence, (2m² - 3m - 1) x (4m² - m - 1) = 8m⁴ - 14m³ - 3m² + 4m + 1

Question 2(xii)

Multiply :

a², ab and b²

Answer

(a² x ab) x b²

= a⁽²⁺¹⁾b x b²

= a³b x b²

= a³b⁽²⁺¹⁾

= a³b³

Hence, a² x ab x b² = a³b³

Question 2(xiii)

Multiply :

abx, - 3a²x and 7b²x³

Answer

abx $\times$ (- 3a²x) $\times$ 7b²x³

= (-3 $\times$ 7) a⁽¹⁺²⁾b⁽¹⁺²⁾x^(1+1+3)

= -21 a³b³x⁵

Hence, (abx) $\times$ (- 3a²x) $\times$ (7b²x³) = -21a³b³x⁵

Question 2(xiv)

Multiply :

- 3bx, - 5xy and - 7b³y²

Answer

-3bx $\times$ (-5xy) $\times$ (-7b³y²)

= -3 $\times$ (-5) $\times$ (-7) b⁽¹⁺³⁾x⁽¹⁺¹⁾y⁽¹⁺²⁾

= -105 b⁴x²y³

Hence, (- 3bx) $\times$ (- 5xy) $\times$ (- 7b³y²) = -105b⁴x²y³

Question 2(xv)

Multiply :

- $\dfrac{3}{2}$ x⁵y³ and $\dfrac{4}{9}$ a²x³y

Answer

$-\dfrac{3}{2}x^5y^3 \times \dfrac{4}{9}a^2x^3y\\[1em] = -\dfrac{3 \times 4}{2 \times 9}a^2x^{5+3}y^{3+1}\\[1em] = -\dfrac{12}{18}a^2x^8y^4\\[1em] = -\dfrac{2}{3}a^2x^8y^4\\[1em]$

Hence, - $\dfrac{3}{2}$ x⁵ y³ x $\dfrac{4}{9}$ a²x³y = - $\dfrac{2}{3}$ a²x⁸y⁴

Question 2(xvi)

Multiply :

$-\dfrac{2}{3}$ a⁷b² and $-\dfrac{9}{4}$ ab⁵

Answer

$-\dfrac{2}{3}a^7b^2 \times \Big(-\dfrac{9}{4}ab^5\Big)\\[1em] = \dfrac{2 \times 9}{3 \times 4}a^{7 + 1}b^{2+5}\\[1em] = \dfrac{18}{12}a^8b^7\\[1em] = \dfrac{3}{2}a^8b^7\\[1em]$

Hence, - $\dfrac{2}{3}$ a⁷b² x - $\dfrac{9}{4}$ ab⁵ = $\dfrac{3}{2}$ a⁸b⁷

Question 2(xvii)

Multiply :

2a³ - 3a²b and $-\dfrac{1}{2}$ ab²

Answer

$(2a^3 - 3a^2b) \times \Big(-\dfrac{1}{2}ab^2\Big)\\[1em] = \Big[2a^3 \times (- \dfrac{1}{2}ab^2) - 3a^2b \times (- \dfrac{1}{2}ab^2)\Big]\\[1em] = - \dfrac{1 \times 2}{2}a^{3+1}b^2 + \dfrac{1 \times 3}{2}a^{2+1}b^{1+2}\\[1em] = - \dfrac{2}{2}a^4b^2 + \dfrac{3}{2}a^3b^3\\[1em] = - a^4b^2 + \dfrac{3}{2}a^3b^3\\[1em]$

Hence, 2a³ - 3a²b x $-\dfrac{1}{2}$ ab² = -a⁴b² + $\dfrac{3}{2}$ a³b³

Question 2(xviii)

Multiply :

2x + $\dfrac{1}{2}$ y and 2x - $\dfrac{1}{2}$ y

Answer

$\Big(2x + \dfrac{1}{2}y\Big) \times \Big(2x - \dfrac{1}{2}y\Big)\\[1em] = 2x \times \Big(2x - \dfrac{1}{2}y\Big) + \dfrac{1}{2}y \times \Big(2x - \dfrac{1}{2}y\Big)\\[1em] =\Big(2x \times 2x - 2x \times \dfrac{1}{2}y\Big) + \Big(\dfrac{1}{2}y \times 2x - \dfrac{1}{2}y \times \dfrac{1}{2}y\Big)\\[1em] =\Big(4x^{1+1} - \dfrac{2xy}{2}\Big) + \Big(\dfrac{2xy}{2} - \dfrac{1 \times 1}{2 \times 2}y^{1+1}\Big)\\[1em] = (4x^2 - xy) + \Big(xy - \dfrac{1}{4}y^{2}\Big)\\[1em] = 4x^2 - xy + xy - \dfrac{1}{4}y^{2}\\[1em] = 4x^2 - \cancel{xy} + \cancel{xy} - \dfrac{1}{4}y^{2}\\[1em] = 4x^2 - \dfrac{1}{4}y^2$

Hence, 2x + $\dfrac{1}{2}$ y x 2x - $\dfrac{1}{2}$ y = 4x² - $\dfrac{1}{4}$ y²

Question 3(i)

Multiply :

5x² - 8xy + 6y² - 3 by - 3xy

Answer

(5x² - 8xy + 6y² - 3) $\times$ (- 3xy)

= 5x² $\times$ (- 3xy) - 8xy $\times$ (- 3xy) + 6y² $\times$ (- 3xy) - 3 $\times$ (- 3xy)

= -15x⁽²⁺¹⁾y + 24x⁽¹⁺¹⁾y⁽¹⁺¹⁾ - 18xy⁽²⁺¹⁾ + 9xy

= -15x³y + 24x²y² - 18xy³ + 9xy

Hence, (5x² - 8xy + 6y² - 3) x (- 3xy) = -15x³y + 24x²y² - 18xy³ + 9xy

Question 3(ii)

Multiply :

3 - $\dfrac{2}{3}$ xy + $\dfrac{5}{7}$ xy² - $\dfrac{16}{21}$ x²y by - 21x²y²

Answer

$\Big(3 - \dfrac{2}{3}xy + \dfrac{5}{7}xy^2 - \dfrac{16}{21}x^2y\Big) \times (-21x^2y^2)\\[1em] = \Big(3 \times (-21x^2y^2) - \dfrac{2}{3}xy \times (-21x^2y^2) + \dfrac{5}{7}xy^2 \times (-21x^2y^2) - \dfrac{16}{21}x^2y \times (-21x^2y^2)\Big)\\[1em] = \Big(-63x^2y^2 - \dfrac{2 \times (-21)}{3}x^{1+2}y^{1+2} + \dfrac{5\times (-21)}{7}x^{1+2}y^{2+2} - \dfrac{16 \times (-21)}{21}x^{2+2}y^{1+2}\Big)\\[1em] = -63x^2y^2 - \dfrac{-42}{3}x^3y^3 + \dfrac{-105}{7}x^3y^4 + \dfrac{16 \times \cancel{(21)}}{\cancel{21}}x^4y^3\\[1em] = -63x^2y^2 + 14x^3y^3 - 15x^3y^4 + 16x^4y^3$

Hence, (3 - $\dfrac{2}{3} xy + \dfrac{5}{7} xy^2 - \dfrac{16}{21}$ x²y) x (- 21x²y²) = -63x²y² + 14x³y³ - 15x³y⁴ + 16x⁴y³

Question 3(iii)

Multiply :

6x³ - 5x + 10 by 4 - 3x²

Answer

(6x³ - 5x + 10) $\times$ (4 - 3x²)

= 4 $\times$ (6x³ - 5x + 10) - 3x² $\times$ (6x³ - 5x + 10)

= 24x³ - 20x + 40 - 18x⁽³⁺²⁾ + 15x⁽¹⁺²⁾ - 30x²

= 24x³ - 20x + 40 - 18x⁵ + 15x³ - 30x²

= - 18x⁵ + 24x³ + 15x³ - 30x² - 20x + 40

= - 18x⁵ + 39x³ - 30x² - 20x + 40

Hence, (6x³ - 5x + 10) x (4 - 3x²) = - 18x⁵ + 39x³ - 30x² - 20x + 40

Question 3(iv)

Multiply :

2y - 4y³ + 6y⁵ by y² + y - 3

Answer

(2y - 4y³ + 6y⁵) x (y² + y - 3)

= 2y x (y² + y - 3) - 4y³ x (y² + y - 3) + 6y⁵ x (y² + y - 3)

= 2y⁽¹⁺²⁾ + 2y⁽¹⁺¹⁾ - 6y - 4y⁽³⁺²⁾ - 4y⁽³⁺¹⁾ + 12y³ + 6y⁽⁵⁺²⁾ + 6y⁽⁵⁺¹⁾ - 18y⁵

= 2y³ + 2y² - 6y - 4y⁵ - 4y⁴ + 12y³ + 6y⁷ + 6y⁶ - 18y⁵

= 6y⁷ + 6y⁶ - 18y⁵ - 4y⁵ - 4y⁴ + 12y³ + 2y³ + 2y² - 6y

= 6y⁷ + 6y⁶ - 22y⁵ - 4y⁴ + 14y³ + 2y² - 6y

Hence, (2y - 4y³ + 6y⁵) x (y² + y - 3) = 6y⁷ + 6y⁶ - 22y⁵ - 4y⁴ + 14y³ + 2y² - 6y

Question 3(v)

Multiply :

5p² + 25pq + 4q² by 2p² - 2pq + 3q²

Answer

(5p² + 25pq + 4q²) x (2p² - 2pq + 3q²)

= 5p² x (2p² - 2pq + 3q²) + 25pq x (2p² - 2pq + 3q²) + 4q² x (2p² - 2pq + 3q²)

= 10p⁽²⁺²⁾ - 10p⁽²⁺¹⁾q + 15p²q² + 50p⁽¹⁺²⁾q - 50p⁽¹⁺¹⁾q⁽¹⁺¹⁾ + 75pq⁽¹⁺²⁾ + 8p²q² - 8pq⁽²⁺¹⁾ + 12q⁽²⁺²⁾

= 10p⁴ - 10p³q + 15p²q² + 50p³q - 50p²q² + 75pq³ + 8p²q² - 8pq³ + 12q⁴

= 10p⁴ - 10p³q + 50p³q + 15p²q² - 50p²q² + 8p²q² + 75pq³ - 8pq³ + 12q⁴

= 10p⁴ + 40p³q - 27p²q² + 67pq³ + 12q⁴

Hence, (5p² + 25pq + 4q²) x (2p² - 2pq + 3q²) = 10p⁴ + 40p³q - 27p²q² + 67pq³ + 12q⁴

Question 4(i)

Simplify:

(7x - 8) (3x + 2)

Answer

(7x - 8) (3x + 2)

= 7x (3x + 2) - 8 (3x + 2)

= 21x⁽¹⁺¹⁾ + 14x - 24x - 16

= 21x² + 14x - 24x - 16

= 21x² - 10x - 16

Hence, (7x - 8) (3x + 2) = 21x² - 10x - 16

Question 4(ii)

Simplify:

(px - q) (px + q)

Answer

(px - q) (px + q)

= px (px + q) - q (px + q)

= p⁽¹⁺¹⁾x⁽¹⁺¹⁾ + pqx - pqx - q⁽¹⁺¹⁾

= p²x² + pqx - pqx - q²

= p²x² - q²

Hence, (px - q) (px + q) = p²x² - q²

Question 4(iii)

Simplify:

(5a + 5b - c) (2b - 3c)

Answer

(5a + 5b - c) (2b - 3c)

= 5a (2b - 3c) + 5b (2b - 3c) - c (2b - 3c)

= 10ab - 15ac + 10b⁽¹⁺¹⁾ - 15bc - 2bc + 3c⁽¹⁺¹⁾

= 10ab - 15ac + 10b² - 17bc + 3c²

Hence, (5a + 5b - c) (2b - 3c) = 10ab - 15ac + 10b² - 17bc + 3c²

Question 4(iv)

Simplify:

(4x - 5y) (5x - 4y)

Answer

(4x - 5y) (5x - 4y)

= 4x (5x - 4y) - 5y (5x - 4y)

= 20x⁽¹⁺¹⁾ - 16xy - 25xy + 20y⁽¹⁺¹⁾

= 20x² - 41xy + 20y²

Hence, (4x - 5y) (5x - 4y) = 20x² - 41xy + 20y²

Question 4(v)

Simplify:

(3y + 4z) (3y - 4z) + (2y + 7z) (y + z)

Answer

(3y + 4z) (3y - 4z) + (2y + 7z) (y + z)

= [3y(3y - 4z) + 4z(3y - 4z)] + [2y (y + z) + 7z (y + z)]

= [9y² - 12yz + 12yz - 16z²] + [2y² + 2yz + 7zy + 7z²]

= 9y² - 16z² + 2y² + 9zy + 7z²

= 9y² + 2y² + 9zy - 16z² + 7z²

= 11y² + 9yz - 9z²

Hence, (3y + 4z) (3y - 4z) + (2y + 7z) (y + z) = 11y² + 9yz - 9z²

Question 5

The adjacent sides of a rectangle are x² - 4xy + 7y² and x³ - 5xy². Find its area.

Answer

Given:

Dimensions of the rectangle = x² - 4xy + 7y² and x³ - 5xy²

As we know, area of rectangle = l x b

So, putting the value

(x² - 4xy + 7y²) $\times$ (x³ - 5xy²)

= x² $\times$ (x³ - 5xy²) - 4xy $\times$ (x³ - 5xy²) + 7y² $\times$ (x³ - 5xy²)

= x⁽²⁺³⁾ - 5x⁽²⁺¹⁾y² - 4x⁽¹⁺³⁾y + 20x⁽¹⁺¹⁾y⁽¹⁺²⁾ + 7x³y² - 35xy⁽²⁺²⁾

= x⁵ - 5x³y² + 7x³y² - 4x⁴y + 20x²y³ - 35xy⁴

= x⁵ + 2x³y² - 4x⁴y + 20x²y³ - 35xy⁴

Hence, the area of the rectangle = x⁵ + 2x³y² - 4x⁴y + 20x²y³ - 35xy⁴

Question 6

The base and the altitude of a triangle are (3x - 4y) and (6x + 5y) respectively. Find its area.

Answer

Given:

Base of a triangle = (3x - 4y)

Altitude of a triangle = (6x + 5y)

As we know that the area of triangle = $\dfrac{1}{2}$ x base x altitude

$\dfrac{1}{2} \times (3x - 4y) \times (6x + 5y)\\[1em] = \dfrac{1}{2} \times (3x (6x + 5y) - 4y (6x + 5y))\\[1em] = \dfrac{1}{2} \times (18x^2 + 15xy - 24xy - 20y^2)\\[1em] = \dfrac{1}{2} \times (18x^2 - 9xy - 20y^2)$

Hence, the area of tirangle = $\dfrac{1}{2}$ (18x² - 9xy - 20y²) sq. unit

Question 7

Multiply - 4xy³ and 6x²y and verify your result for x = 2 and y = 1.

Answer

(- 4xy³) $\times$ (6x²y)

= - 24x⁽¹⁺²⁾y⁽³⁺¹⁾

= - 24x³y⁴

Hence, (- 4xy³) $\times$ (6x²y) = - 24x³y⁴

Putting x = 2 and y = 1

Taking LHS:

(- 4xy³) x (6x²y)

= (- 4 x 2 x 1³) x (6 x 2² x 1)

= (- 4 x 2 x 1) x (6 x 4 x 1)

= (- 8) x (24)

= - 192

Taking RHS:

- 24x³y⁴

= - 24 x 2³ x 1⁴

= - 24 x 8 x 1

= - 192

Hence, LHS = RHS

So, (- 4xy³) $\times$ (6x²y) = - 24x³y⁴

Question 8

Find the value of (3x³) x (-5xy²) x (2x²yz³) for x = 1, y = 2 and z = 3.

Answer

(3x³) $\times$ (-5xy²) $\times$ (2x²yz³)

= (- 15x⁽³⁺¹⁾y²) $\times$ (2x²yz³)

= (- 15x⁴y²) $\times$ (2x²yz³)

= - 30x⁽⁴⁺²⁾y⁽²⁺¹⁾z³

= - 30x⁶y³z³

Putting the value x = 1, y = 2 and z = 3, we get

- 30 x 1⁶ x 2³ x 3³

= - 30 x 1 x 8 x 27

= - 6480

Hence, the value of (3x³) x (-5xy²) x (2x²yz³) for x = 1, y = 2 and z = 3 is - 6480

Question 9

Evaluate (3x⁴y²) (2x²y³) for x = 1 and y = 2.

Answer

(3x⁴y²) $\times$ (2x²y³)

= 6x⁽⁴⁺²⁾y⁽²⁺³⁾

= 6x⁶y⁵

Putting x = 1 and y = 2, we get

6 x 1⁶ x 2⁵

= 6 x 1 x 32

= 192

Hence, (3x⁴y²) (2x²y³) for x = 1 and y = 2 is 192

Question 10

Evaluate (x⁵) x (3x²) x (-2x) for x = 1.

Answer

Given:

(x⁵) $\times$ (3x²) $\times$ (-2x)

= 3x⁽⁵⁺²⁾ $\times$ (-2x)

= 3x⁷ $\times$ (-2x)

= - 6x⁽⁷⁺¹⁾

= - 6x⁸

For x = 1

- 6 x 1⁸

= - 6 x 1

= - 6

Hence, the value of (x⁵) $\times$ (3x²) x (-2x) for x = 1 is -6

Question 11(i)

Divide:

- 70a³ by 14a²

Answer

- 70a³ ÷ 14a²

= - $\dfrac{70}{14}$ a^(3-2)

= - 5 a¹

Hence, - 70a³ ÷ 14a² = - 5 a

Question 11(ii)

Divide:

24x³y³ by - 8y²

Answer

(24x³y³) ÷ (- 8y²)

= - $\dfrac{24}{8}$ x³y^(3-2)

= - 3 x³y¹

Hence, (24x³y³) ÷ (- 8y²) = - 3 x³y

Question 11(iii)

Divide:

15a⁴b by - 5a³b

Answer

(15a⁴b) ÷ (- 5a³b)

= - $\dfrac{15}{5}$ a^(4-3)b^(1-1)

= - 3a¹b⁰

Hence, (15a⁴b) ÷ (- 5a³b) = - 3a

Question 11(iv)

Divide:

- 24x⁴d³ by - 2x²d⁵

Answer

(- 24x⁴d³) ÷ (- 2x²d⁵)

= $\dfrac{24}{2}$ x^(4-2)d^(3-5)

= 12 x²d^(-2)

= $\dfrac{12x^2}{d^2}$

Hence, (- 24x⁴d³) ÷ (- 2x²d⁵) = $\dfrac{12x^2}{d^2}$

Question 11(v)

Divide:

63a⁴b⁵c⁶ by - 9a²b⁴c³

Answer

(63a⁴b⁵c⁶) ÷ (- 9a²b⁴c³)

= - $\dfrac{63}{9}$ a^(4-2)b^(5-4)c^(6-3)

= - 7a²b¹c³

Hence, (63a⁴b⁵c⁶) ÷ (- 9a²b⁴c³) = - 7a²bc³

Question 11(vi)

Divide:

8x - 10y + 6c by 2

Answer

(8x - 10y + 6c) ÷ 2

= (8x ÷ 2) - (10y ÷ 2) + (6c ÷ 2)

= 4x - 5y + 3c

Hence, (8x - 10y + 6c) ÷ 2 = 4x - 5y + 3c

Question 11(vii)

Divide:

15a³b⁴ - 10a⁴b³ - 25a³b⁶ by -5a³b²

Answer

(15a³b⁴ - 10a⁴b³ - 25a³b⁶) ÷ (-5a³b²)

= (15a³b⁴ ÷ (- 5a³b²)) - (10a⁴b³ ÷ (- 5a³b²)) - (25a³b⁶ ÷ (- 5a³b²))

= $\Big($ - $\dfrac{15}{5}$ a^(3-3)b^(4-2) $\Big)$ - $\Big($ - $\dfrac{10}{5}$ a^(4-3)b^(3-2) $\Big)$ - $\Big($ - $\dfrac{25}{5}$ a^(3-3)b^(6-2) $\Big)$

= - 3 a⁰b² + 2 a¹b¹ + 5 a⁰b⁴

= - 3b² + 2 ab + 5b⁴

Hence, (15a³b⁴ - 10a⁴b³ - 25a³b⁶) ÷ (- 5a³b²) = - 3b² + 2 ab + 5b⁴

Question 11(viii)

Divide:

- 14x⁶y³ - 21x⁴y⁵ + 7x⁵y⁴ by 7x²y²

Answer

(- 14x⁶y³ - 21x⁴y⁵ + 7x⁵y⁴) ÷ 7x²y²

= (- 14x⁶y³ ÷ 7x²y²) - (21x⁴y⁵ ÷ 7x²y²) + (7x⁵y⁴ ÷ 7x²y²)

= $\Big($ - $\dfrac{14}{7}$ x^(6-2)y^(3-2) $\Big)$ - $\Big($ $\dfrac{21}{7}$ x^(4-2)y^(5-2) $\Big)$ - $\Big($ - $\dfrac{7}{7}$ x^(5-2)y^(4-2) $\Big)$

= - 2 x⁴y¹ - 3 x²y³ + 1 x³y²

Hence, (- 14x⁶y³ - 21x⁴y⁵ + 7x⁵y⁴) ÷ 7x²y² = - 2 x⁴y - 3 x²y³ + x³y²

Question 11(ix)

Divide:

a² + 7a + 12 by a + 4

Answer

(a² + 7a + 12) ÷ (a + 4)

= (a² + 3a + 4a + 12) ÷ (a + 4)

= [a(a + 3) + 4(a + 3)] ÷ (a + 4)

= [(a + 3)(a + 4)] ÷ (a + 4)

= $\dfrac{(a + 3)\cancel{(a + 4)}}{\cancel{(a + 4)}}$

= (a + 3)

Hence, (a² + 7a + 12) ÷ (a + 4) = a + 3

Question 11(x)

Divide:

x² + 3x - 54 by x - 6

Answer

(x² + 3x - 54) ÷ (x - 6)

= (x² + 9x - 6x - 54) ÷ (x - 6)

= [x(x + 9) - 6(x + 9)] ÷ (x - 6)

= [(x + 9)(x - 6)] ÷ (x - 6)

= $\dfrac{(x + 9)\cancel{(x - 6)}}{\cancel{(x - 6)}}$

= (x + 9)

Hence, (x² + 3x - 54) ÷ (x - 6) = x + 9

Question 11(xi)

Divide:

12x² + 7xy - 12y² by 3x + 4y

Answer

(12x² + 7xy - 12y²) ÷ (3x + 4y)

= (12x² + 16xy - 9xy - 12y²) ÷ (3x + 4y)

= [4x(3x + 4y) - 3y(3x + 4y)] ÷ (3x + 4y)

= [(3x + 4y)(4x - 3y)] ÷ (3x + 4y)

= $\dfrac{[\cancel{(3x + 4y)}(4x - 3y)]}{\cancel{(3x + 4y)}}$

= 4x - 3y

Hence, (12x² + 7xy - 12y²) ÷ (3x + 4y) = (4x - 3y)

Question 11(xii)

Divide:

x⁶ - 8 by x² - 2

Answer

(x⁶ - 8) ÷ (x² - 2)

= (x^2x3 - 2³) ÷ (x² - 2)

[Using the formula (a³ - b³) = (a - b)(a² + ab + b²)]

= [(x² - 2)(x⁴ + 2x² + 4)] ÷ (x² - 2)

= $\dfrac{\cancel{(x^2 - 2)}(x^4 + 2x^2 + 4)}{\cancel{(x^2 - 2)}}$

= (x⁴ + 2x² + 4)

Hence,(x⁶ - 8) ÷ (x² - 2) = (x⁴ + 2x² + 4)

Question 11(xiii)

Divide:

6x³ - 13x² - 13x + 30 by 2x² - x - 6

Answer

Dividing 6x³ - 13x² - 13x + 30 by 2x² - x - 6

$\begin{array}{l} \phantom{2x^2 - x - 6)}{3x - 5} \\ 2x^2 - x - 6\overline{\smash{\big)}6x^3 - 13x^2 - 13x + 30} \\ \phantom{2x^2 - x - 6}\underline{\underset{-}{}6x^3 \underset{+}{-}\phantom{0}3x^2 \underset{+}{-}18x} \\ \phantom{{2x^2 - x - 6}2x^3+}-10x^2 + 5x + 30 \\ \phantom{{2x^2 - x - 6}2x^3+}\underline{\underset{+}{-}10x^2 \underset{-}{+} 5x \underset{-}{+} 30} \\ \phantom{{2x^2 - x - 6}{2x^3+}{+5x^2222}}\times \end{array}$

Hence, (6x³ - 13x² - 13x + 30) ÷ (2x² - x - 6) = 3x - 5

Question 11(xiv)

Divide:

4a² + 12ab + 9b² - 25c² by 2a + 3b + 5c

Answer

Dividing 4a² + 12ab + 9b² - 25c² by 2a + 3b + 5c

$\begin{array}{l} \phantom{2a + 3b + 5c)}{2a + 3b - 5c} \\ 2a + 3b + 5c\overline{\smash{\big)}4a^2 + 12ab + 9b^2 - 25c^2} \\ \phantom{2a + 3b + 5c}\underline{\underset{-}{}4a^2 \underset{-}{+}6ab \phantom{+ 9b^2 - 25c^2} \underset{-}{+}10ac} \\ \phantom{{2a + 3b + 5c}2x^3+}6ab - 10ac + 9b^2 - 25c^2 \\ \phantom{{2a + 3b + 5c}2x+}\underline{\underset{-}{}6ab \phantom{- 10ac} \underset{-}{+} 9b^2 \phantom{- 25c^2} \underset{-}{+} 15bc} \\ \phantom{{2a + 3b + 5c}{2x^3+}{+5x^2+}}-10ac - 25c^2 -15bc \\ \phantom{{2a + 3b + 5c}{2x^3+}{+5x^2+\enspace}}\underline{\underset{+}{-}10ac \underset{+}{-} 25c^2 \underset{+}{-} 15bc} \\ \phantom{{2a + 3b + 5c}{2x^3+}{+5x^2+\qquad}{-23x}}\times \end{array}$

Hence, (4a² + 12ab + 9b² - 25c²) ÷ (2a + 3b + 5c) = 2a + 3b - 5c

Question 11(xv)

Divide:

16 + 8x + x⁶ - 8x³ - 2x⁴ + x² by x + 4 - x³

Answer

Dividing 16 + 8x + x⁶ - 8x³ - 2x⁴ + x² by x + 4 - x³

⇒ Dividing x⁶ - 2x⁴ - 8x³ + x² + 8x + 16 by -x³ + x + 4

$\begin{array}{l} \phantom{-x^3 + x + 4)}{-x^3 + x + 4} \\ -x^3 + x + 4\overline{\smash{\big)}x^6 - 2x^4 - 8x^3 + x^2 + 8x + 16} \\ \phantom{-x^3 + x + 4}\underline{\underset{-}{}x^6 \underset{+}{-}\phantom{2}x^4 \underset{+}{-}4x^3} \\ \phantom{{-x^3 + x + 4}2x^3+}-x^4 - 4x^3 + x^2 + 8x + 16 \\ \phantom{{-x^3 + x + 4}2x+2}\underline{\underset{+}{-}x^4 \phantom{- 4x^3} \underset{-}{+} x^2 \underset{-}{+} 4x} \\ \phantom{{-x^3 + x + 4}{2x^3+}{+x+}}-4x^3 \phantom{+ x^2} + 4x + 16 \\ \phantom{{-x^3 + x + 4}{2x^3+}{+541)}}\underline{\underset{+}{-}4x^3 \phantom{+ x^2} \underset{+}{-} 4x \underset{-}{+} 16} \\ \phantom{{-x^3 + x + 4}{2x^3+}{+5x^2+\qquad}{-23x}}\times \end{array}$

Hence, (16 + 8x + x⁶ - 8x³ - 2x⁴ + x²) ÷ (x + 4 - x³) = -x³ + x + 4

Question 12(i)

Find the quotient and the remainder (if any), when:

a³ - 5a² + 8a + 15 is divided by a + 1.

Answer

Dividing a³ - 5a² + 8a + 15 by a + 1

$\begin{array}{l} \phantom{a + 1)}{a^2 - 6a + 14} \\ a + 1\overline{\smash{\big)}a^3 - 5a^2 + 8a + 15} \\ \phantom{a + 1}\underline{\underset{-}{}a^3 \underset{-}{+}a^2} \\ \phantom{{a + 1}2x}-6a^2 +8a + 15 \\ \phantom{{a + 1}21x}\underline{\underset{+}{-}6a^2 \underset{+}{-} 6a} \\ \phantom{{a + 1}{2x^3+}{+5x^2+}}14a + 15 \\ \phantom{{a + 1}{2x^3+}{+5x^2}}\underline{\underset{+}{-}14a \underset{-}{+} 14} \\ \phantom{{a + 1}{a^3 - 5a^2 + 8a + 152}} 1 \end{array}$

Quotient = a² - 6a + 14

Remainder = 1

Verification:

Quotient x Divisor + Remainder

= (a² - 6a + 14) $\times$ (a + 1) + 1

= a $\times$ (a² - 6a + 14) + 1 $\times$ (a² - 6a + 14) + 1

= a⁽¹⁺²⁾ - 6a⁽¹⁺¹⁾ + 14a + a² - 6a + 14 + 1

= a³ - 6a² + 14a + a² - 6a + 14 + 1

= a³ + (- 6a² + a²) + (14a - 6a) + (14 + 1)

= a³ - 5a² + 8a + 15

= Dividend

Question 12(ii)

Find the quotient and the remainder (if any), when:

3x⁴ + 6x³ - 6x² + 2x - 7 is divided by x - 3.

Answer

Dividing 3x⁴ + 6x³ - 6x² + 2x - 7 by x - 3

$\begin{array}{l} \phantom{x - 3)}{3x^3 + 15x^2 + 39x + 119} \\ x - 3\overline{\smash{\big)}3x^4 + 6x^3 - 6x^2 + 2x - 7} \\ \phantom{x - 3}\underline{\underset{-}{}3x^4 \underset{+}{-}9x^3} \\ \phantom{{x - 3)}{3x^4567}}15x^3 - 6x^2 + 2x - 7 \\ \phantom{{x - 3}{3x^456}}\underline{\underset{-}{+}15x^3 \underset{+}{-} 45x^2} \\ \phantom{{x - 3}{3x^456 + 15x^3 + }}39x^2 + 2x - 7 \\ \phantom{{x - 3}{3x^456 + 15x^3}}\underline{\underset{-}{+}39x^2 \underset{+}{-} 117x} \\ \phantom{{x - 3}{3x^456 + 15x^3 + 39x^2 +}}119x - 7 \\ \phantom{{x - 3}{3x^456 + 15x^3 + 39x^2}}\underline{\underset{-}{+}119x \underset{+}{-} 357} \\ \phantom{{x - 3}{3x^456 + 15x^3 + 39x^2 + 117x +}}350 \end{array}$

Quotient = 3x³ + 15x² + 39x + 119

Remainder = 350

Verification:

Quotient x Divisor + Remainder

= (3x³ + 15x² + 39x + 119) $\times$ (x - 3) + 350

= x $\times$ (3x³ + 15x² + 39x + 119) - 3 $\times$ (3x³ + 15x² + 39x + 119) + 350

= 3x⁽³⁺¹⁾ + 15x⁽²⁺¹⁾ + 39x⁽¹⁺¹⁾ + 119x - 9x³ - 45x² - 117x - 357 + 350

= 3x⁴ + 15x³ + 39x² + 119x - 9x³ - 45x² - 117x - 357 + 350

= 3x⁴ + (15x³ - 9x³) + (39x² - 45x²) + (119x - 117x) + (- 357 + 350)

= 3x⁴ + 6x³ - 6x²+ 2x - 7

= Dividend

Question 12(iii)

Find the quotient and the remainder (if any), when:

6x² + x - 15 is divided by 3x + 5.

Answer

(6x² + x - 15) ÷ (3x + 5)

Dividing 6x² + x - 15 by 3x + 5

$\begin{array}{l} \phantom{3x + 5)}{2x - 3} \\ 3x + 5\overline{\smash{\big)}6x^2 + x - 15} \\ \phantom{3x + 5}\underline{\underset{-}{}6x^2 \underset{-}{+}10x} \\ \phantom{{3x + 5}6x^2 +}-9x - 15 \\ \phantom{{3x + 5}6x^2 +}\underline{\underset{+}{-}9x \underset{+}{-} 15} \\ \phantom{{3x + 5}6x^2 + 9x -} \times \end{array}$

Quotient = 2x - 3

Remainder = 0

Verification:

Quotient x Divisor + Remainder

= (2x - 3) $\times$ (3x + 5) + 0

= 2x $\times$ (3x + 5) - 3 $\times$ (3x + 5)

= 6x⁽¹⁺¹⁾ + 10x - 9x - 15

= 6x² + (10x - 9x) - 15

= 6x² + x - 15

= Dividend

Question 13

The area of a rectangle is x³ - 8x² + 7 and one of its sides is x - 1. Find the length of the adjacent side.

Answer

Area of the rectangle = x³ - 8x² + 7

One side = (x - 1)

As we know the area of rectangle = One side x Other side

Other side = Area of rectangle ÷ One side

= (x³ - 8x² + 7) ÷ (x - 1)

$\begin{array}{l} \phantom{x - 1)}{x^2 - 7x - 7} \\ x - 1\overline{\smash{\big)}x^3 - 8x^2 + 7} \\ \phantom{x - 1}\underline{\underset{-}{}x^3 \underset{+}{-}x^2} \\ \phantom{{x - 1}12x}-7x^2 + 7 \\ \phantom{{x - 1}221x}\underline{\underset{+}{-}7x^2 \phantom{+ 7} \underset{-}{+} 7x} \\ \phantom{{x - 1}{2x^3+}{+5x}}-7x + 7 \\ \phantom{{x - 1}{2x^3+}{+5x^2}}\underline{\underset{+}{-}7x \underset{-}{+} 7} \\ \phantom{{x - 1}{2x^3+}{+5x^2}{-7x}} \times \end{array}$

Hence, length of adjacent side is x² - 7x - 7.

Exercise 11(C)

Question 1(i)

The value of $x - \overline{x - y}$ is:

2x
- y
y
2x + y

Answer

$x - \overline{x - y}$

= $x - x + y$

= $y$

Hence, option 3 is the correct option.

Question 1(ii)

(5x - 4y) - (5y - 4x) is equal to:

9(x - y)
x - y
(y - x)
0

Answer

(5x - 4y) - (5y - 4x)

= 5x - 4y - 5y + 4x

= (5x + 4x) + (-4x - 5y)

= 9x + (- 9y)

= 9x - 9y

= 9(x - y)

Hence, option 1 is the correct option.

Question 1(iii)

$x (y - x - \overline{x - y})$ is equal to:

2x (x - y)
2x (y - x)
2x (x + y)
-2x (y - x)

Answer

$x (y - x - \overline{x - y})$

= $x (y - x - x + y)$

= $x (2y - 2x)$

= $2x (y - x)$

Hence, option 2 is the correct option.

Question 1(iv)

$(2x - y) + \overline{2x - y}$ is equal to :

0
2y
4x
4x - 2y

Answer

$(2x - y) + \overline{2x - y}$

= $2x - y + 2x - y$

= $4x - 2y$

Hence, option 4 is the correct option.

Question 1(v)

x(y - z) + y(z - x) - z(y - x) is equal to:

0
xy + yz + zx
xy + yz - zx
xy - yz + zx

Answer

x(y - z) + y(z - x) - z(y - x)

= xy - xz + yz - xy - yz + xz

= (xy - xy) + (- xz + xz) + (yz - yz)

= 0 + 0 + 0

Hence, option 1 is the correct option.

Question 2

a² - 2a + {5a² - (3a - 4a²)}

Answer

a² - 2a + {5a² - (3a - 4a²)}

= a² - 2a + {5a² - 3a + 4a²}

= a² - 2a + {(5a² + 4a²) - 3a}

= a² - 2a + {9a² - 3a}

= a² - 2a + 9a² - 3a

= (a² + 9a²) + ( - 2a - 3a)

= 10a² + (- 5a)

= 10a² - 5a

Hence, a² - 2a + {5a² - (3a - 4a²)} = 10a² - 5a

Question 3

$x - y - {x - y - (x + y) - \overline{x - y}}$

Answer

$x - y - {x - y - (x + y) - \overline{x - y}}$

= $x - y - {x - y - (x + y) - x + y}$

= $x - y - {\cancel{x} - \cancel{y} - (x + y) - \cancel{x} + \cancel{y}}$

= $x - y - {-(x + y)}$

= $x - y - {- x - y}$

= $x - \cancel{y} + x + \cancel{y}$

= $2x$

Hence, $x - y - {x - y - (x + y) - \overline{x - y}} = 2x$

Question 4

-3(1 - x²) - 2{x² - (3 - 2x²)}

Answer

-3(1 - x²) - 2{x² - (3 - 2x²)}

= -3(1 - x²) - 2{x² - 3 + 2x²}

= -3(1 - x²) - 2{(x² + 2x²) - 3}

= -3(1 - x²) - 2{3x² - 3}

= -3 + 3x² - 6x² + 6

= (-3 + 6) + (3x² - 6x²)

= 3 + (-3x²)

= 3 - 3x²

Hence, -3(1 - x²) - 2{x² - (3 - 2x²)} = 3 - 3x²

Question 5

$2{m - 3(n + \overline{m - 2n})}$

Answer

$2{m - 3(n + \overline{m - 2n})}$

= $2{m - 3(n + m - 2n)}$

= $2{m - 3((n - 2n) + m)}$

= $2{m - 3(-n + m)}$

= $2{m + 3n - 3m}$

= $2{(m - 3m) + 3n}$

= $2{-2m + 3n}$

= $-4m + 6n$

Hence, $2{m - 3(n + \overline{m - 2n})} = 6n - 4m$

Question 6

$3x - [3x - {3x - (3x - \overline{3x - y})}]$

Answer

$3x - [3x - {3x - (3x - \overline{3x - y})}]$

= $3x - [3x - {3x - (3x - 3x + y)}]$

= $3x - [3x - {3x - (y)}]$

= $3x - [3x - {3x - y}]$

= $3x - [3x - 3x + y]$

= $3x - [y]$

= $3x - y$

Hence, $3x - [3x - {3x - (3x - \overline{3x - y})}] = 3x - y$

Question 7

$p^2x - 2{px - 3x(x^2 - \overline{3a - x^2})}$

Answer

$p^2x - 2{px - 3x(x^2 - \overline{3a - x^2})}$

= $p^2x - 2{px - 3x(x^2 - 3a + x^2)}$

= $p^2x - 2{px - 3x((x^2 + x^2) - 3a)}$

= $p^2x - 2{px - 3x(2x^2 - 3a)}$

= $p^2x - 2{px - 6x^3 + 9ax}$

= $p^2x - 2px + 12x^3 - 18ax$

Hence, $p^2x - 2{px - 3x(x^2 - \overline{3a - x^2})} = p^2x - 2px + 12x^3 - 18ax$

Question 8

$2[6 + 4{m - 6(7 - \overline{n + p}) + q}]$

Answer

$2[6 + 4{m - 6(7 - \overline{n + p}) + q}]$

= $2[6 + 4{m - 6(7 - n - p) + q}]$

= $2[6 + 4{m - 42 + 6n + 6p + q}]$

= $2[6 + 4m - 168 + 24n + 24p + 4q]$

= $2[(6 - 168) + 4m + 24n + 24p + 4q]$

= $2[-162 + 4m + 24n + 24p + 4q]$

= $-324 + 8m + 48n + 48p + 8q$

Hence, $2[6 + 4{m - 6(7 - \overline{n + p}) + q}] = 8m + 48n + 48p + 8q -324$

Question 9

$a - [a - \overline{b + a} - {a - (a - \overline{b - a})}]$

Answer

$a - [a - \overline{b + a} - {a - (a - \overline{b - a})}]$

= $a - [a - \overline{b + a} - {a - (a - b + a)}]$

= $a - [a - \overline{b + a} - {a - (a + a - b)}]$

= $a - [a - \overline{b + a} - {a - (2a - b)}]$

= $a - [a - \overline{b + a} - {a - 2a + b}]$

= $a - [a - \overline{b + a} - {-a + b}]$

= $a - [a - \overline{b + a} + a - b]$

= $a - [a - b - a + a - b]$

= $a - [(a - a + a) + (- b - b)]$

= $a - [ a + (- 2b)]$

= $a - [ a - 2b]$

= $a - a + 2b$

= $2b$

Hence, $a - [a - \overline{b + a} - {a - (a - \overline{b - a})}] = 2b$

Question 10

$3x - [4x - \overline{3x - 5y} - 3{2x - (3x - \overline{2x - 3y})}]$

Answer

$3x - [4x - \overline{3x - 5y} - 3{2x - (3x - \overline{2x - 3y})}]$

= $3x - [4x - \overline{3x - 5y} - 3{2x - (3x - 2x + 3y)}]$

= $3x - [4x - \overline{3x - 5y} - 3{2x - (x + 3y)}]$

= $3x - [4x - \overline{3x - 5y} - 3{2x - x - 3y}]$

= $3x - [4x - \overline{3x - 5y} - 3{x - 3y}]$

= $3x - [4x - \overline{3x - 5y} - 3x + 9y]$

= $3x - [4x - 3x + 5y - 3x + 9y]$

= $3x - [(4x - 3x - 3x) + (5y + 9y)]$

= $3x - [-2x + 14y]$

= $3x + 2x - 14y$

= $5x - 14y$

Hence, $3x - [4x - \overline{3x - 5y} - 3{2x - (3x - \overline{2x - 3y})}] = 5x - 14y$

Question 11

a⁵ ÷ a³ + 3a x 2a

Answer

a⁵ ÷ a³ + 3a x 2a

= a^(5-3) + 3a x 2a

= a² + 3a x 2a

= a² + 6a⁽¹⁺¹⁾

= a² + 6a²

= 7a²

Hence, a⁵ ÷ a³ + 3a x 2a = 7a²

Question 12

x⁵ ÷ (x² x y²) x y³

Answer

$x^5 ÷ (x^2 \times y^2) \times y^3\\[1em] = \dfrac{x^5}{x^2y^2} \times y^3\\[1em] = \dfrac{x^(5-2)}{y^2} \times y^3\\[1em] = \dfrac{x^3}{y^2} \times y^3\\[1em] = \dfrac{x^3y^3}{y^2}\\[1em] = x^3y^{3-2}\\[1em] = x^3y^1\\[1em]$

Hence, x⁵ ÷ (x² $\times$ y²) $\times$ y³ = x³y

Question 13

(x⁵ ÷ x²) x y² x y³

Answer

(x⁵ ÷ x²) $\times$ y² $\times$ y³

= x^(5-2) $\times$ y² $\times$ y³

= x³ $\times$ y² $\times$ y³

= x³ $\times$ y⁽²⁺³⁾

= x³ $\times$ y⁵

Hence, (x⁵ ÷ x²) $\times$ y² $\times$ y³ = x³y⁵

Question 14

(y³ - 5y²) ÷ y x (y - 1)

Answer

(y³ - 5y²) ÷ y x (y - 1)

= (y³ ÷ y - 5y² ÷ y) x (y - 1)

= (y^(3-1) - 5y^(2-1)) x (y - 1)

= (y² - 5y¹) x (y - 1)

= y² x (y - 1) - 5y¹ x (y - 1)

= y⁽²⁺¹⁾ - y² - 5y⁽¹⁺¹⁾ - 5y

= y³ - y² - 5y² - 5y

= y³ - 6y² - 5y

Hence, (y³ - 5y²) ÷ y x (y - 1) = y³ - 6y² - 5y

Test Yourself

Question 1(i)

(-18xy) - (-8xy) is equal to:

10xy
-10xy
26xy
-26xy

Answer

(-18xy) - (-8xy)

= -18xy + 8xy

= - 10xy

Hence, option 2 is the correct option.

Question 1(ii)

(9a + 7b - 6c) - (2a - 3b + 4c) is equal to:

7x + 7b + 10c
7a + 10b - 10c
7a - 10b + 10c
7a - 10b - 10c

Answer

(9a + 7b - 6c) - (2a - 3b + 4c)

= 9a + 7b - 6c - 2a + 3b - 4c

= (9a - 2a) + (7b + 3b) + (- 6c - 4c)

= 7a + 10b + (-10c)

= 7a + 10b -10c

Hence, option 2 is the correct option.

Question 1(iii)

-81a⁵b⁴c³ ÷ (-9a²b²c) is equal to:

-9a³b²c²
3a³b²c²
9a⁴b
9a³b²c²

Answer

-81a⁵b⁴c³ ÷ (-9a²b²c)

= $\dfrac{-81}{-9}$ a^(5-2)b^(4-2)c^(3-1)

= 9 a³b²c²

Hence, option 4 is the correct option.

Question 1(iv)

$-(-pq - \overline{p^2 - pq})$ is equal to:

2pq + p²
-p²
p²
none of these

Answer

$-(-pq - \overline{p^2 - pq})$

= $-(-pq - p^2 + pq)$

= $-((-pq + pq) - p^2)$

= $-(- p^2)$

= $p^2$

Hence, option 3 is the correct option.

Question 1(v)

$x^3 - y^3 - x(y^2 + \overline{x^2 - z^2})$ is equal to:

y³ + xy² + xz²
y³ + xy² - xz²
-y³ + xy² - xz²
-y³ - xy² + xz²

Answer

$x^3 - y^3 - x(y^2 + \overline{x^2 - z^2})$

= $x^3 - y^3 - x(y^2 + x^2 - z^2)$

= $x^3 - y^3 - xy^2 - x^3 + xz^2$

= $(x^3 - x^3) - y^3 - xy^2 + xz^2$

= $- y^3 - xy^2 + xz^2$

Hence, option 4 is the correct option.

Question 1(vi)

Statement 1: The expression 2x⁴ - 3x² + $\dfrac{7}{\text{x}}$ , x ≠ 0 has no constant term.

Statement 2: In an algebraic expression in terms of one variable, the term(s) independent of the variable is called the constant.

Which of the following options is correct?

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Given :

2x⁴ - 3x² + $\dfrac{7}{\text{x}}$

A constant term is a term that does not contain the variable in this case, x.

Thus, the expression contains no constant term.

So, statement 1 is true.

In an algebraic expression in terms of one variable, the term(s) independent of the variable is called the constant.

This is a standard definition in algebra: a constant term is the part of the expression that does not change with the variable i.e., it is independent of the variable.

So, statement 2 is true.

Hence, option 1 is the correct option.

Question 1(vii)

Assertion (A) : 5x + y² - x³, xy + yz + zx, x² - x + 1 are all trinomials.

Reason (R) : An algebraic expression which contains three different terms is called a trinomial.

Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.

Answer

An algebraic expression which contains three different terms is called a trinomial.

So, reason (R) is true.

Given, 5x + y² - x³, xy + yz + zx, x² - x + 1

All three expressions have three distinct terms.

So, assertion (A) is true and, reason (R) is the correct explanation of assertion (A).

Hence, option 1 is the correct option.

Question 1(viii)

Assertion (A) : Refer the following algebraic expression in terms of one variable : 3x + 9, 9 - $\dfrac{3}{\text{x}^2}$ , 100.

Two of them are not polynomials in x.

Reason (R) : An algebraic expression is a polynomial if the power of each term used in it is a whole number.

Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.

Answer

An algebraic expression is a polynomial if the power of each term used in it is a whole number.

So, reason (R) is true

A polynomial in one variable (say, x) must have:

Only non-negative integers (i.e., whole numbers) as exponents of x.
No negative or fractional powers.

So, reason (R) is true.

Given,

Algebraic expressions : 3x + 9, 9 - $\dfrac{3}{\text{x}^2}$ , 100

3x + 9 and 100 (constant polynomial) both are polynomials, while 9 - $\dfrac{3}{\text{x}^2}$ is not as on taking the variable in numerator (9 - 3x^-2) the power becomes negative.

So, assertion (A) is false.

Hence, option 4 is the correct option.

Question 1(ix)

Assertion (A) : 2xyz + 3x² ia a cubic polynomial in three variables.

Reason (R) : An algebraic expression having two or more variables, the highest sum of the powers of all the variables in each term is taken as the degree of the polynomial, if this sum is a whole number.

Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.

Answer

An algebraic expression having two or more variables, the highest sum of the powers of all the variables in each term taken as the degree of the polynomial if this sum is a whole number.

The maximum of these sums gives the degree. These powers must be whole numbers for it to be a valid polynomial.

So, reason (R) is true.

Given; 2xyz + 3x²

Terms; 2xyz, 3x²

For term 2xyz, degree = 1 + 1 + 1 = 3

For term 3x², degree = 2

The degree of a polynomial in multiple variables is the highest total degree among its terms.

Thus, the degree of this polynomial is 3.

Therefore, this is a cubic polynomial in three variables.

So, assertion (A) is true and, reason (R) is the correct explanation of assertion (A).

Hence, option 1 is the correct option.

Question 1(x)

Assertion (A) : 2024x²yz is not a trinomial.

Reason (R) : For combining polynomials, the like terms of the given polynomials are combined together.

Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.

Answer

A trinomial is an algebraic expression with exactly three terms.

Here, 2024x²yz is just a single term — a monomial.

Thus, the statement that it's not a trinomial is correct.

So, assertion (A) is true.

For combining polynomials, the like terms of the given polynomials are combined together.

Like terms are terms with the same variables raised to the same powers.

When combining polynomials, you add or subtract like terms.

So, reason (R) is true but, reason (R) does not explains assertion(A).

Hence, option 2 is the correct option.

Question 2

In $\dfrac{5}{7}$ xy²z³, write the coefficient of:

(i) 5

(ii) $\dfrac{5}{7}$

(iii) 5x

(iv) xy²

(v) z³

(vi) xz³

(vii) 5xy²

(viii) $\dfrac{1}{7}$ yz

(ix) z

(x) yz²

(xi) 5xyz

Answer

(i) Coefficient = $\dfrac{1}{7}$ xy²z³

(ii) Coefficient = xy²z³

(iii) Coefficient = $\dfrac{1}{7}$ y²z³

(iv) Coefficient = $\dfrac{5}{7}$ z³

(v) Coefficient = $\dfrac{5}{7}$ xy²

(vi) Coefficient = $\dfrac{5}{7}$ y²

(vii) Coefficient = $\dfrac{1}{7}$ z³

(viii) Coefficient = 5 xyz³

(ix) Coefficient = $\dfrac{5}{7}$ xy²z²

(x) Coefficient = $\dfrac{5}{7}$ xyz

(xi) Coefficient = $\dfrac{1}{7}$ yz²

Question 3(i)

In each polynomial, given below, separate the like terms:

3xy, - 4yx², 2xy², 2.5x²y, - 8yx, - 3.2y²x and x²y

Answer

Like terms are those terms which have the same variable raised to the same power.

Hence, 3xy,- 8yx ; 2y²x, - 3.2y²x ; -4x²y, 2.5x²y and x²y are like terms.

Question 3(ii)

In each polynomial, given below, separate the like terms:

y²z³, xy²z³, - 5x²yz, - 4y²z³, - 8xz³y², 3x²yz and 2z³y²

Answer

Like terms are those terms which have the same variable raised to the same power.

Hence, y²z³, - 4y²z³ and 2z³y²; xy²z³, - 8xz³y² ; - 5x²yz, 3x²yz are like terms.

Question 4

The sides of a triangle are x² - 3xy + 8, 4x² + 5xy - 3 and 6 - 3x² + 4xy. Find its perimeter.

Answer

The sides of a triangle = (x² - 3xy + 8), (4x² + 5xy - 3) and (6 - 3x² + 4xy)

Perimeter of triangle = sum of all its side.

(x² - 3xy + 8) + (4x² + 5xy - 3) + (6 - 3x² + 4xy)

= x² - 3xy + 8 + 4x² + 5xy - 3 + 6 - 3x² + 4xy

= (x² + 4x² - 3x²) + (- 3xy + 5xy + 4xy) + (- 3 + 6 + 8)

= 2x² + 6xy + 11

Hence, perimeter of triangle = 2x² + 6xy + 11

Question 5

The perimeter of a triangle is 8y² - 9y + 4 and its two sides are 3y² - 5y and 4y² + 12. Find its third side.

Answer

The perimeter of a triangle = 8y² - 9y + 4

Two sides are 3y² - 5y and 4y² + 12

Let the third side be S.

As we know that, perimeter of triangle = sum of all its side.

8y² - 9y + 4 = (3y² - 5y) + (4y² + 12) + S

⇒ 8y² - 9y + 4 = 3y² - 5y + 4y² + 12 + S

⇒ 8y² - 9y + 4 = (3y² + 4y²) - 5y + 12 + S

⇒ 8y² - 9y + 4 = 7y² - 5y + 12 + S

⇒ S = (8y² - 9y + 4) - (7y² - 5y + 12)

⇒ S = 8y² - 9y + 4 - 7y² + 5y - 12

⇒ S = (8y² - 7y²) + (- 9y + 5y) + (4 - 12)

⇒ S = 1y² - 4y - 8

Hence, the third side of triangle = (y² - 4y - 8)

Question 6

The two adjacent sides of a rectangle are 2x² - 5xy + 3z² and 4xy - x² - z². Find its perimeter.

Answer

The two adjacent sides of a rectangle = (2x² - 5xy + 3z²) and (4xy - x² - z²)

As we know the perimeter of rectangle = 2 x (l + b)

= 2 $\times$ [(2x² - 5xy + 3z²) + (4xy - x² - z²)]

= 2 $\times$ [2x² - 5xy + 3z² + 4xy - x² - z²]

= 2 $\times$ [(2x² - x²) + (- 5xy + 4xy) + (3z² - z²)]

= 2 $\times$ [x² - 1xy + 2z²]

= 2 $\times$ x² - 2 $\times$ 1xy + 2 $\times$ 2z²

= 2x² - 2xy + 4z²

Hence, the perimeter of rectangle = 2x² - 2xy + 4z²

Question 7

What must be subtracted from 19x⁴ + 2x³ + 30x - 37 to get 8x⁴ + 22x³ - 7x - 60?

Answer

Let A be the number that should be subtracted from 19x⁴ + 2x³ + 30x - 37 to get 8x⁴ + 22x³ - 7x - 60

(19x⁴ + 2x³ + 30x - 37) - A = (8x⁴ + 22x³ - 7x - 60)

⇒ (19x⁴ + 2x³ + 30x - 37) - (8x⁴ + 22x³ - 7x - 60) = A

⇒ A = 19x⁴ + 2x³ + 30x - 37 - 8x⁴ - 22x³ + 7x + 60

⇒ A = (19x⁴ - 8x⁴) + (2x³ - 22x³) + (30x + 7x) + (- 37 + 60)

⇒ A = 11x⁴ - 20x³ + 37x + 23

Hence, the number is 11x⁴ - 20x³ + 37x + 23.

Question 8

How much smaller is 15x - 18y + 19z than 22x - 20y -13z +26?

Answer

(22x - 20y -13z + 26) - (15x - 18y + 19z)

= 22x - 20y -13z + 26 - 15x + 18y - 19z

= (22x - 15x) + (- 20y + 18y) + (-13z - 19z) + 26

= 7x - 2y -32z + 26

Hence, (22x - 20y -13z +26) is less than (15x - 18y + 19z) by (7x - 2y -32z + 26)

Question 9

How much bigger is 5x²y² - 18xy² - 10x²y than - 5x² + 6x²y - 7xy?

Answer

(5x²y² - 18xy² - 10x²y) - (- 5x² + 6x²y - 7xy)

= 5x²y² - 18xy² - 10x²y + 5x² - 6x²y + 7xy

= 5x²y² - 18xy² + (- 10x²y - 6x²y) + 5x² + 7xy

= 5x²y² - 18xy² - 16x²y + 5x² + 7xy

Hence, (5x²y² - 18xy² - 10x²y) is bigger than (- 5x² + 6x²y - 7xy) by (5x²y² - 18xy² - 16x²y + 5x² + 7xy)

Question 10

If x = 2 and y = 1 ; find the value of (-4x²y³) x (-5x²y⁵).

Answer

(-4x²y³) $\times$ (-5x²y⁵)

= (-4) $\times$ (-5)x⁽²⁺²⁾y⁽³⁺⁵⁾

= 20 x⁴y⁸

When x = 2 and y = 1,

= 20 x 2⁴ x 1⁸

= 20 x 16 x 1

= 320

Hence, if x = 2 and y = 1 ; the value of (-4x²y³) x (-5x²y⁵) = 320.

Question 11(i)

Evaluate:

(3x - 2) (x + 5) for x = 2.

Answer

(3x - 2) (x + 5)

= 3x (x + 5) - 2 (x + 5)

= 3x⁽¹⁺¹⁾ + 15x - 2x - 10

= 3x² + (15x - 2x) - 10

= 3x² + 13x - 10

For x = 2

3 x 2² + 13 x 2 - 10

= 3 x 4 + 26 - 10

= 12 + 26 - 10

= 38 - 10

= 28

Hence, for x = 2, the value of (3x - 2) (x + 5) is 28.

Question 11(ii)

Evaluate:

(2x - 5y) (2x + 3y) for x = 2 and y = 3.

Answer

(2x - 5y) (2x + 3y)

= 2x (2x + 3y) - 5y (2x + 3y)

= 4x⁽¹⁺¹⁾ + 6xy - 10xy - 15y⁽¹⁺¹⁾

= 4x² + (6xy - 10xy) - 15y²

= 4x² - 4xy - 15y²

For x = 2, y = 3

4 x 2² - 4 x 2 x 3 - 15 x 3²

= 4 x 4 - 4 x 2 x 3 - 15 x 9

= 16 - 24 - 135

= - 143

Hence, for x = 2, y = 3, the value (2x - 5y) (2x + 3y) = - 143.

Question 11(iii)

Evaluate:

xz (x² + y²) for x = 2, y = 1 and z = 1.

Answer

xz (x² + y²)

= x⁽²⁺¹⁾z + xzy²

= x³z + xzy²

For x = 2, y = 1 and z = 1

2³ x 1 + 2 x 1 x 1²

= 8 x 1 + 2 x 1 x 1

= 8 + 2

= 10

Hence, for x = 2, y = 1 and z = 1, the value of xz (x² + y²) is 10.

Question 12(i)

Evaluate:

x(x - 5) + 2 for x = 1.

Answer

x(x - 5) + 2

= x² - 5x + 2

For x = 1

= 1² - 5 x 1 + 2

= 1 - 5 + 2

= - 2

Hence, for x = 1, the value of x(x - 5) + 2 is - 2.

Question 12(ii)

Evaluate:

xy²(x - 5y) + 1 for x = 2 and y = 1.

Answer

xy²(x - 5y) + 1

= x⁽¹⁺¹⁾y² - 5xy⁽²⁺¹⁾ + 1

= x²y² - 5xy³ + 1

For x = 2 and y = 1,

2² x 1² - 5 x 2 x 1³ + 1

= 4 x 1 - 5 x 2 x 1 + 1

= 4 - 10 + 1

= - 5

Hence, for x = 2 and y = 1, the value of [xy²(x - 5y) + 1] is - 5.

Question 12(iii)

Evaluate:

2x(3x - 5) - 5(x - 2) - 18 for x = 2.

Answer

2x(3x - 5) - 5(x - 2) - 18

= 6x⁽¹⁺¹⁾ - 10x - 5x + 10 - 18

= 6x² + (- 10x - 5x) + (10 - 18)

= 6x² - 15x - 8

For x = 2,

= 6 x 2² - 15 x 2 - 8

= 6 x 4 - 15 x 2 - 8

= 24 - 30 - 8

= - 14

Hence, for x = 2, the value of [2x(3x - 5) - 5(x - 2) - 18] is - 14.

Question 13

Multiply and then verify:

-3x²y² and (x - 2y) for x = 1 and y = 2.

Answer

(-3x²y²) $\times$ (x - 2y)

= (-3x²y²) $\times$ x - (-3x²y²) $\times$ 2y

= (-3x⁽²⁺¹⁾y²) - (-6x²y⁽²⁺¹⁾)

= -3x³y² + 6x²y³

For x = 1 and y = 2

(-3x²y²) $\times$ (x - 2y) = -3x³y² + 6x²y³

Taking LHS:

(-3x²y²) $\times$ (x - 2y)

= (-3 x 1² x 2²) x (1 - 2 x 2)

= (-3 x 1 x 4) x (1 - 4)

= (-12) x (- 3)

= 36

Now, taking RHS:

-3x³y² + 6x²y³

= -3 x 1³ x 2² + 6 x 1² x 2³

= -3 x 1 x 4 + 6 x 1 x 8

= -12 + 48

= 36

Hence, LHS = RHS.

Question 14(i)

Multiply:

2x² - 4x + 5 by x² + 3x - 7

Answer

(2x² - 4x + 5) $\times$ (x² + 3x - 7)

= 2x² $\times$ (x² + 3x - 7) - 4x $\times$ (x² + 3x - 7) + 5 $\times$ (x² + 3x - 7)

= 2x⁽²⁺²⁾ + 6x⁽²⁺¹⁾ - 14x² - 4x⁽¹⁺²⁾ - 12x⁽¹⁺¹⁾ + 28x + 5x² + 15x - 35

= 2x⁴ + 6x³ - 14x² - 4x³ - 12x² + 28x + 5x² + 15x - 35

= 2x⁴ + (6x³ - 4x³) + (- 14x² - 12x² + 5x²) + (28x + 15x) - 35

= 2x⁴ + 2x³ - 21x² + 43x - 35

Hence, (2x² - 4x + 5) $\times$ (x² + 3x - 7) = 2x⁴ + 2x³ - 21x² + 43x - 35

Question 14(ii)

Multiply:

(ab - 1) (3 - 2ab)

Answer

(ab - 1) (3 - 2ab)

= ab (3 - 2ab) - 1 (3 - 2ab)

= 3ab - 2a⁽¹⁺¹⁾b⁽¹⁺¹⁾ - 3 + 2ab

= 3ab - 2a²b² - 3 + 2ab

= (3ab + 2ab) - 2a²b² - 3

= 5ab - 2a²b² - 3

Hence, (ab - 1) (3 - 2ab) = 5ab - 2a²b² - 3

Question 15

Simplify:

(5 - x) (6 - 5x) (2 - x).

Answer

(5 - x) (6 - 5x) (2 - x)

= [5 (6 - 5x) - x (6 - 5x)](2 - x)

= [30 - 25x - 6x + 5x⁽¹⁺¹⁾](2 - x)

= [30 + (- 25x - 6x) + 5x²](2 - x)

= [30 - 31x + 5x²](2 - x)

= 2 (30 - 31x + 5x²) - x (30 - 31x + 5x²)

= 60 - 62x + 10x² - 30x + 31x⁽¹⁺¹⁾ - 5x⁽²⁺¹⁾

= 60 - 62x + 10x² - 30x + 31x² - 5x³

= 60 + (- 62x - 30x) + (10x² + 31x²) - 5x³

= 60 - 92x + 41x² - 5x³

Hence, (5 - x) (6 - 5x) (2 - x) = 60 - 92x + 41x² - 5x³

Question 16

The product of two numbers is 16x⁴ -1. If one number is 2x - 1, find the other.

Answer

The product = (16x⁴ -1)

One number = (2x - 1)

Let the other number be A.

(2x - 1) $\times$ A = (16x⁴ -1)

⇒ A = $\dfrac{(16x^4 - 1)}{(2x - 1)}$

⇒ A = $\dfrac{(4x^2 - 1)(4x^2 + 1)}{(2x - 1)}$

⇒ A = $\dfrac{(2x - 1)(2x + 1)(4x^2 + 1)}{(2 - 1)}$

⇒ A = $\dfrac{\cancel{(2x - 1)}(2x + 1)(4x^2 + 1)}{\cancel{(2x - 1)}}$

⇒ A = (2x + 1)(4x² + 1)

⇒ A = 2x(4x² + 1) + 1(4x² + 1)

⇒ A = 8x⁽²⁺¹⁾ + 2x + 4x² + 1

⇒ A = 8x³ + 4x² + 2x + 1

Hence, the other number is 8x³ + 4x² + 2x + 1.

Question 17

Divide x⁶ - y⁶ by the product of x² + xy + y² and x - y.

Answer

The product of (x² + xy + y²) and (x - y)

= (x² + xy + y²) $\times$ (x - y)

= x $\times$ (x² + xy + y²) - y $\times$ (x² + xy + y²)

= x⁽¹⁺²⁾ + x⁽¹⁺¹⁾y + xy² - x²y - xy⁽¹⁺¹⁾ - y⁽¹⁺²⁾

= x³ + x²y + xy² - x²y - xy² - y³

= x³ + (x²y - x²y) + (xy² - xy²) - y³

= x³ - y³

Now, (x⁶ - y⁶) ÷ ( x³ - y³)

$\begin{array}{l} \phantom{x^3 - y^3)}{x^3 + y^3} \\ x^3 - y^3\overline{\smash{\big)}x^6 - y^6 \phantom{+ x^3y^3}} \\ \phantom{x^3 - y^3}\underline{\underset{-}{}x^6 \phantom{- y^6} \underset{+}{-}x^3y^3} \\ \phantom{{x^3 - y^3}6x^2 +}x^3y^3 - y^6 \\ \phantom{{x^3 - y^3}6x^21}\underline{\underset{-}{+}x^3y^3 \underset{+}{-} y^6} \\ \phantom{{x^3 - y^3}6x^2 + 9x -} \times \end{array}$

Hence, the answer is (x³ + y³)

Question 18

$3a \times [8b ÷ 4 - 6{a - (5a - \overline{3b - 2a})}]$

Answer

$3a \times [8b ÷ 4 - 6{a - (5a - \overline{3b - 2a})}]$

= $3a \times [8b ÷ 4 - 6{a - (5a - 3b + 2a)}]$

= $3a \times [8b ÷ 4 - 6{a - (5a + 2a - 3b )}]$

= $3a \times [8b ÷ 4 - 6{a - (7a - 3b )}]$

= $3a \times [8b ÷ 4 - 6{a - 7a + 3b }]$

= $3a \times [8b ÷ 4 - 6{-6a + 3b }]$

= $3a \times [8b ÷ 4 + 36a - 18b ]$

= $3a \times [2b + 36a - 18b]$

= $3a \times [2b - 18b + 36a]$

= $3a \times [-16b + 36a]$

= $-48ab + 108a^2$

Hence, $3a \times [8b ÷ 4 - 6{a - (5a - \overline{3b - 2a})}] = 108a^2 - 48ab$

Question 19

7x + 4{x² ÷ (5x ÷ 10)} - 3{2 - x³ ÷ (3x² ÷ x)}

Answer

7x + 4{x² ÷ (5x ÷ 10)} - 3{2 - x³ ÷ (3x² ÷ x)}

= 7x + 4{x² ÷ (5x ÷ 10)} - 3{2 - x³ ÷ 3x}

= 7x + 4{x² ÷ (5x ÷ 10)} - 3{2 - $\dfrac{\text{x}^2}{3}$ }

= 7x + 4{x² ÷ (5x ÷ 10)} - 6 + $\dfrac{3\text{x}^2}{3}$

= 7x + 4{x² ÷ $\dfrac{5\text{x}}{10}$ } - 6 + x²

= 7x + 4{x² ÷ $\dfrac{\text{x}}{2}$ } - 6 + x²

= 7x + 4{ $\dfrac{2\text{x}^2}{\text{x}}$ } - 6 + x²

= 7x + 4 x 2x - 6 + x²

= 7x + 8x - 6 + x²

= x² + 15x - 6

Hence, the answer is x² + 15x - 6

Chapter 11

Algebraic Expressions

Class - 8 Concise Mathematics Selina

Exercise 11(A)

Question 1(i)

Question 1(ii)

Question 1(iii)

Question 1(iv)

Question 1(v)

Question 2

Question 3

Question 4

Question 5

Question 6(i)

Question 6(ii)

Question 6(iii)

Question 6(iv)

Question 6(v)

Question 7(i)

Question 7(ii)

Question 7(iii)

Question 7(iv)

Question 7(v)

Question 8(i)

Question 8(ii)

Question 8(iii)

Question 8(iv)

Question 8(v)

Question 8(vi)

Question 9

Question 10(i)

Question 10(ii)

Question 10(iii)

Question 10(iv)

Question 10(v)

Question 10(vi)

Question 10(vii)

Question 11(i)

Question 11(ii)

Question 12

Question 13

Question 14(i)

Question 14(ii)

Question 15

Exercise 11(B)

Question 1(i)

Question 1(ii)

Question 1(iii)

Question 1(iv)

Question 1(v)

Question 2(i)

Question 2(ii)

Question 2(iii)

Question 2(iv)

Question 2(v)

Question 2(vi)

Question 2(vii)

Question 2(viii)

Question 2(ix)

Question 2(x)

Question 2(xi)

Question 2(xii)

Question 2(xiii)

Question 2(xiv)

Question 2(xv)

Question 2(xvi)

Question 2(xvii)

Question 2(xviii)

Question 3(i)

Question 3(ii)

Question 3(iii)

Question 3(iv)

Question 3(v)

Question 4(i)

Question 4(ii)

Question 4(iii)

Question 4(iv)

Question 4(v)

Question 5

Question 6