RBSE Maths Chapter 9 – Algebraic Expressions Mathematics Class 8 Important questions and solutions are available here. The additional, exercise important questions and solutions of Chapter 9, available at BYJU’S, contain detailed explanations. All these important questions are based on the new pattern designed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 8 solutions.

Chapter 9 of the RBSE Class 8 Maths will help the students to solve problems related to power of expression, like and unlike terms, addition and subtraction of algebraic expressions, multiplication of algebraic expressions, multiplying a monomial by a monomial, multiplying a binomial or a trinomial by a monomial, multiplying a binomial by a binomial, multiplying a binomial by trinomial, standard identities, application of identities.

### RBSE Maths Chapter 9: Exercise 9.1 Textbook Important Questions and Solutions

**Question 1: Find the product of the following pairs of monomials.**

**(i) 3, 5x **

**(ii) -5p, -2q **

**(iii) 7t ^{2}, -3n^{2}**

**(iv) 6m, 3n **

**(v) -5x ^{2}, -2x **

**Solution: **

= 3 * 5 * (x)

= 15x

[ii] -5p, -2q= (-5) x p x (-2) x q

= (-5) x (-2) x pq

= 10pq

[iii] 7t^{2}, -3n

^{2}

= 7 x t^{2} x (-3) x n^{2}

= 7 x (-3) x t^{2} x n^{2}

= -21t^{2}n^{2}

= 6 x m x 3 x n

= 6 x 3 x m x n

= 18 mn

[v] -5x^{2}, -2x

= (-5) * (x^{2}) * (-2) * (-x)

= (-5) * (-2) * (x^{2}) * (-x)

= 10x^{3}

**Question 2: Complete the table below.**

**Solution: **

* |
7 |
x |
y |
2z |
a |
-5b |
c |

7 |
49 | 7x | 7y | 14z | 7a | -35b | 7c |

x |
7x | x^{2} |
xy | 2zx | ax | -5bx | cx |

2y |
14y | 2xy | 2y^{2} |
4yz | 2ay | -10by | 2cy |

-3a |
-21a | -3ax | -3ay | -6az | -3a^{2} |
15ab | -3ac |

b |
7b | bx | by | 2bz | ab | -5b^{2} |
bc |

y |
7y | xy | y^{2} |
2yz | ay | -5by | cy |

2x^{3} |
14x^{3} |
2x^{4} |
2x^{3}y |
4x^{3}z |
2x^{3}a |
-10x^{3}b |
2x^{3}c |

a^{4} |
7a^{4} |
a^{4}x |
a^{4}y |
2a^{4}z |
a^{5} |
-5a^{4}b |
a^{4}c |

z^{2} |
7z^{2} |
xz^{2} |
yz^{2} |
2z^{3} |
az^{2} |
-5bz^{2} |
cz^{2} |

**Question 3: Multiply the following monomial.**

**[i] xy, x ^{2}y, xy, x**

**[ii] m, n, mn, m ^{3}n, mn^{3}**

**[iii] kl, lm, km, klm**

**Solution: **

^{2}y, xy, x

= (x) * (y) * (x) * (x) * (y) * (x) * (y) * (x)

= (x) * (x) * (x) * (x) * (x) * (y) * (y) * (y)

= x^{5}y^{3}

^{3}n, mn

^{3}

= m x n x m x n x m x m x m x n x m x n x n x n

= m x m x m x m x m x m x n x n x n x n x n x n

= m^{6}n^{6}

= k x l x l x m x k x m x k x l x m

= k x k x k x l x l x l x m x m x m

= k^{3}l^{3}m^{3}

**Question 4: Find the simple interest = [PTR] / [100], if principal P = 4x ^{2}, time (T) = 5x and rate of interest R = 5y.**

**Solution:**

Simple interest = [PTR] / [100]

= [(4x^{2}) * (5x) * (5y)] / [100]

= [(4 * (x) * (x) * 5 * (x) * 5 * (y)] / [100]

= [4 * 5 * 5 * (x) * (x) * (x) * (y)] / [100]

= [100x^{3}y] / [100]

= x^{3}y

### RBSE Maths Chapter 9: Exercise 9.2 Textbook Important Questions and Solutions

**Question 1: Multiply the binomials given below.**

**[i] (2x + 5) and (3x – 7)**

**[ii] (x – 8) and (3y + 5)**

**[iii] (1.5p – 0.5q) and (1.5p + 0.5q)**

**[iv] (a + 3b) and (x + 5)**

**[v] (2lm + 3l ^{2}) and (3lm – 5l^{2})**

**[vi] [(¾)a ^{2} + 3b^{2}] and [4a^{2} – (5/3)b^{2}]**

**Solution:**

= 2x (3x – 7) + 5 (3x – 7)

= (2x) * (3x) – (2x) * 7 + 5 * (3x) – (5 * 7)

= 2 * (x) * 3 * (x) – 2 * (x) * 7 + 5 * 3 * (x) – 35

= 2 * 3 * (x) * (x) – 2 * 7 * (x) + 15 * (x) – 35

= 6 * (x^{2}) – 14 * (x) + 15x – 35

= 6x^{2} – 14x + 15x – 35

= 6x^{2} + x – 35

= x (3y + 5) – 8 (3y + 5)

= (x) * 3y + (x) * 5 – 8 * 3 * y – 8 * 5

= (x) * 3y + 5x – 8 * 3 * y – 40

= 3 * (x) * y + 5x – 24 * y – 40

= (3x) * y + 5x – 24y – 40

= 3xy + 5x – 24y – 40

[iii] (1.5p – 0.5q) and (1.5p + 0.5q)= 1.5p (1.5p + 0.5q) – 0.5q (1.5p + 0.5q)

= 1.5p x 1.5p + 1.5p x 0.5q – 0.5q x 1.5p – 0.5q x 0.5q

= 2.25p^{2} x + 0.75 x pq – 0.75 x qp – 0.25 q^{2}

= 2.25p^{2} + 0.75pq – 0.75pq – 0.25q^{2}

= 2.25p^{2} – 0.25q^{2}

= a (x + 5) + 3b (x + 5)

= a x (x) + a x 5 + 3b x (x) + 3b x 5

= ax + 5a + 3bx + 3 x 5 x b

= ax + 5a+ 3bx + 15b

[v] (2lm + 3l^{2}) and (3lm – 5l

^{2})

= 2lm (3lm – 5l^{2}) + 3l^{2} (3lm – 5l^{2})

= [2lm x 3lm] – [2lm x 5l^{2}] + 3l^{2} x 3lm – [3l^{2} x 5l^{2}]

= 6l^{2}m^{2} – 10l^{3}m + 9l^{3}m – 15l^{4}

= 6l^{2}m^{2 }– l^{3}m – 15l^{4}

^{2}+ 3b

^{2}] and [4a

^{2}– (5/3)b

^{2}]

**Question 2: Find the product of the following.**

**[i] (3x + 8) (5 – 2x)**

**[ii] (x + 3y) (3x – y)**

**[iii] (a ^{2} + b) (a + b^{2}) **

**[iv] (p ^{2} – q^{2}) (2p + q)**

**Solution:**

**Question 3: Simplify the following.**

**[i] (x + 5) (x – 7) + 35**

**[ii] (a ^{2} – 3) (b^{2 }+ 3) + 5**

**[iii] (t + s ^{2}) (t^{2} – s)**

**[iv] (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)**

**[v] (a + b) (a ^{2} – ab + b^{2})**

**[vi] (a + b + c) (a + b – c)**

**[vii] (a + b) (a – b) – a ^{2} + b^{2}**

**Solution:**

### RBSE Maths Chapter 9: Exercise 9.3 Textbook Important Questions and Solutions

**Question 1: Find the products using a suitable identity.**

**[i] (x + 5) (x + 5)**

**[ii] (3x + 2) (3x + 2)**

**[iii] (5a – 7) (5a – 7)**

**[iv] (3p – [½]) (3p – [½])**

**[v] (1.2m – 0.3) (1.2m + 0.3)**

**[vi] (x ^{2} + y^{2}) (x^{2} – y^{2})**

**[vii] (7a – 9b) (7a – 9b)**

**Solution:**

**Question 2: Use the identity (x + a) (x + b) = x ^{2} + (a + b)x + ab to find the following products.**

**[i] (x + 1) (x + 2)**

**[ii] (3x + 5) (3x + 1)**

**[iii] (4x – 5) (4x – 1)**

**Solution:**

Using the identity **(x + a) (x + b) = x ^{2} + (a + b)x + ab**

a = 1; b = 2

= x^{2} + (1 + 2)x + 2 x 1

= x^{2} + 3x + 2

Using the identity **(x + a) (x + b) = x ^{2} + (a + b)x + ab**

a = 5; b = 1

= [3x]^{2} + (5 + 1)* 3x + 5 x 1

= 9x^{2} + 6 * 3x + 5

= 9x^{2} + 18x + 5

Using the identity **(x + a) (x + b) = x ^{2} + (a + b)x + ab**

a = (-5); b = (-1)

= [4x]^{2} + ([-5] + [-1]) * 4x + (-5 * -1)

= 16x^{2} + (-6) * 4x + [5]

= 16x^{2} – 24x + 5

**Question 3: Find the following squares by using identities.**

**[i] (b – 7) ^{2}**

**[ii] (xy + 3z) ^{2}**

**Solution:**

**Question 4: Simplify the following problems.**

**[i] (a ^{2} – b^{2})^{2}**

**[ii] (2n + 5) ^{2} – (2n – 5)^{2}**

**Solution:**

**Question 5: Show that **

**[i] (2a + 3b) ^{2} – (2a – 3b)^{2} = 24ab**

**[ii] (4x + 5) ^{2} – 80x = (4x – 5)^{2}**

**Solution:**

**Question 6: Using the identities, evaluate the following.**

**[i] 99 ^{2}**

**[ii] 103 ^{2}**

**Solution:**

^{2}

= (100 – 1)^{2}

(a – b)^{2} = a^{2} – 2ab + b^{2}

a = 100; b = 1

= (100)^{2} – 2 * (100) * (1) + 1^{2}

= 10000 – 200 + 1

= 9801

[ii] 103^{2}

= (100 + 3)^{2}

(a + b)^{2} = a^{2} + 2ab + b^{2}

a = 100; b = 3

= (100)^{2} + 2 * (100) * (3) + 3^{2}

= 10000 + 600 + 9

= 10609

**Question 7: Using a ^{2} – b^{2} = (a + b) (a – b), find the values of the following.**

**[i] 10 ^{2} – 99^{2}**

**[ii] (10.3) ^{2} – (9.7)^{2}**

**Solution:**

^{2}– 99

^{2}

= (10 + 99) (10 – 99)

= (109) (-89)

= -9701

[ii] (10.3)^{2}– (9.7)

^{2}

= (10.3 + 9.7) (10.3 – 9.7)

= (20) (0.6)

= 12

**Question 8: Using a ^{2} – b^{2} = (a + b) (a – b), find the values of the following.**

**[i] 103 x 102**

**[ii] 7.1 x 7.3**

**[iii] 102 x 99**

**[iv] 9.8 x 9.6**

**Solution: **

= (100 + 3) (100 + 2)

Using the identity **(x + a) (x + b) = x ^{2} + (a + b)x + ab**

a = 3; b = 2

= 100^{2} + (3 + 2) * 100 + [3 x 2]

= 10000 + 5 * 100 + 6

= 10000 + 500 + 6

= 10506

[ii] 7.1 x 7.3= (7 + 0.1) (7 + 0.3)

Using the identity **(x + a) (x + b) = x ^{2} + (a + b)x + ab**

a = 0.1; b = 0.3

= 7^{2} + [0.1 + 0.3] * 7 + [0.1 x 0.3]

= 49 + [0.4] * 7 + 0.03

= 49 + 2.8 + 0.03

= 51.83

[iii] 102 x 99= (100 + 2) (100 – 1)

Using the identity **(x + a) (x + b) = x ^{2} + (a + b)x + ab**

a = 2; b = (-1)

= 100^{2} + (2 + [-1]) * 100 + (2) * (-1)

= 10000 + 100 – 2

= 10098

[iv] 9.8 x 9.6= (9 + 0.8) (9 + 0.6)

Using the identity **(x + a) (x + b) = x ^{2} + (a + b)x + ab**

a = 0.8; b = 0.6

= 9^{2} + [0.8 + 0.6] * 9 + [0.8 x 0.6]

= 81 + [1.4] * 9 + 0.48

= 81 + 12.6 + 0.48

= 94.08