# Algebraic Expressions and Identities Class 8 Extra Questions Maths Chapter 9

**Extra Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities**

### Algebraic Expressions and Identities Class 8 Extra Questions Very Short Answer Type

Question 1.

Write two examples of each of

(i) Monomials

(ii) Binomials

(iii) Trinomials

Solution:

(i) Monomials:

(a) 3x

(b) 5xy^{2}

(ii) Binomials:

(a) p + q

(b) -5a + 2b

(iii) Trinomials:

(a) a + b + c

(b) x^{2} + x + 2

Question 2.

Identify the like expressions.

5x, -14x, 3x^{2} + 1, x^{2}, -9x^{2}, xy, -3xy

Solution:

Like terms: 5x and -14x, x^{2} and -9x^{2}, xy and -3xy

Question 3.

Identify the terms and their coefficients for each of the following expressions:

(i) 3x^{2}y – 5x

(ii) xyz – 2y

(iii) -x – x^{2}

Solution:

Question 4.

Add: -3a^{2}b^{2}, –\(\frac { 5 }{ 2 }\) a^{2}b^{2}, 4a^{2}b^{2}, \(\frac { 2 }{ 3 }\) a^{2}b^{2}

Solution:

Question 5.

Add: 8x^{2} + 7xy – 6y^{2}, 4x^{2} – 3xy + 2y^{2} and -4x^{2} + xy – y^{2}

Solution:

Question 6.

Subtract: (4x + 5) from (-3x + 7)

Solution:

(-3x + 7) – (4x + 5) = -3x + 7 – 4x – 5 = -3x – 4x + 7 – 5 = -7x + 2

Question 7.

Subtract: 3x^{2} – 5x + 7 from 5x^{2} – 7x + 9

Solution:

(5x^{2} – 7x + 9) – (3x^{2} – 5x + 7)

= 5x^{2} – 7x + 9 – 3x^{2} + 5x – 7

= 5x^{2} – 3x^{2} + 5x – 7x + 9 – 7

= 2x^{2} – 2x + 2

Question 8.

Multiply the following expressions:

(a) 3xy^{2} × (-5x^{2}y)

(b) \(\frac { 1 }{ 2 }\) x^{2}yz × \(\frac { 2 }{ 3 }\) xy^{2}z × \(\frac { 1 }{ 5 }\) x^{2}yz

Solution:

Question 9.

Find the area of the rectangle whose length and breadths are 3x^{2}y m and 5xy^{2} m respectively.

Solution:

Length = 3x^{2}y m, breadth = 5xy^{2} m

Area of rectangle = Length × Breadth = (3x^{2}y × 5xy^{2}) sq m = (3 × 5) × x^{2}y × xy^{2} sq m = 15x^{3}y^{3} sq m

Question 10.

Multiply x^{2} + 7x – 8 by -2y.

Solution:

### Algebraic Expressions and Identities Class 8 Extra Questions Short Answer Type

Question 11.

Simplify the following:

(i) a^{2} (b^{2} – c^{2}) + b^{2} (c^{2} – a^{2}) + c^{2} (a^{2} – b^{2})

(ii) x^{2}(x – 3y^{2}) – xy(y^{2} – 2xy) – x(y^{3} – 5x^{2})

Solution:

(i) a^{2} (b^{2} – c^{2}) + b^{2} (c^{2} – a^{2}) + c^{2} (a^{2} – b^{2})

= a^{2}b^{2} – a^{2}c^{2}) + b^{2}c^{2} – b^{2}a^{2}) + c^{2}a^{2} – c^{2}b^{2})

= 0

(ii) x^{2}(x – 3y^{2}) – xy(y^{2} – 2xy) – x(y^{3} – 5x^{2})

= x^{3} – 3x^{2}y^{2} – xy^{3} + 2x^{2}y^{2} – xy^{3} + 5x^{3}

= x^{3} + 5x^{3} – 3x^{2}y^{2} + 2x^{2}y^{2} – xy^{3} – xy^{3}

= 6x^{3} – x^{2}y^{2} – 2xy^{3}

Question 12.

Multiply (3x^{2} + 5y^{2}) by (5x^{2} – 3y^{2})

Solution:

(3x^{2} + 5y^{2}) × (5x^{2} – 3y^{2})

= 3x^{2}(5x^{2} – 3y^{2}) + 5y^{2}(5x^{2} – 3y^{2})

= 15x^{4} – 9x^{2}y^{2} + 25x^{2}y^{2} – 15y^{4}

= 15x^{4} + 16x^{2}y^{2} – 15y^{4}

Question 13.

Multiply (6x^{2} – 5x + 3) by (3x^{2} + 7x – 3)

Solution:

(6x^{2} – 5x + 3) × (3x^{2} + 7x – 3)

= 6x^{2}(3x^{2} + 7x – 3) – 5x(3x^{2} + 7x – 3) + 3(3x^{2} + 7x – 3)

= 18x^{4} + 42x^{3} – 18x^{2} – 15x^{3} – 35x^{2} + 15x + 9x^{2} + 21x – 9

= 18x^{4} + 42x^{3} – 15x^{3} – 18x^{2} – 35x^{2} + 9x^{2} + 15x + 21x – 9

= 18x^{4} + 27x^{3} – 44x^{2} + 36x – 9

Question 14.

Simplify:

2x^{2}(x + 2) – 3x (x^{2} – 3) – 5x(x + 5)

Solution:

2x^{2}(x + 2) – 3x(x^{2} – 3) – 5x(x + 5)

= 2x^{3} + 4x^{2} – 3x^{3} + 9x – 5x^{2} – 25x

= 2x^{3} – 3x^{3} – 5x^{2} + 4x^{2} + 9x – 25x

= -x^{3} – x^{2} – 16x

Question 15.

Multiply x^{2} + 2y by x^{3} – 2xy + y^{3} and find the value of the product for x = 1 and y = -1.

Solution:

(x^{2} + 2y) × (x^{3} – 2xy + y^{3})

= x^{2}(x^{3} – 2xy + y^{3}) + 2y(x^{3} – 2xy + y^{3})

= x^{5} – 2x^{3}y + x^{2}y^{3} + 2x^{3}y – 4xy^{2} + 2y^{4}

= x^{5} + x^{2}y^{3} – 4xy^{2} + 2y^{4}

Put x = 1 and y = -1

= (1)^{5} + (1)^{2} (-1)^{3} – 4(1)(-1)^{2} + 2(-1)^{4}

= 1 + (1) (-1) – 4(1)(1) + 2(1)

= 1 – 1 – 4 + 2

= -2

Question 16.

Using suitable identity find:

(i) 48^{2} (NCERT Exemplar)

(ii) 96^{2}

(iii) 231^{2} – 131^{2}

(iv) 97 × 103

(v) 181^{2} – 19^{2} = 162 × 200 (NCERT Exemplar)

Solution:

Question 17.

Solution:

Question 18.

Verify that (11pq + 4q)^{2} – (11pq – 4q)^{2} = 176pq^{2} (NCERT Exemplar)

Solution:

LHS = (11pq + 4q)^{2} – (11pq – 4q)^{2} = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q)

[using a^{2} -b^{2} = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q]

= (22pq) (8q)

= 176 pq^{2}

= RHS.

Hence Verified.

Question 19.

Find the value of \(\frac { { 38 }^{ 2 }-{ 22 }^{ 2 } }{ 16 }\), using a suitable identity. (NCERT Exemplar)

Solution:

Question 20.

Find the value of x, if 10000x = (9982)^{2} – (18)^{2} (NCERT Exemplar)

Solution:

RHS = (9982)^{2} – (18)^{2} = (9982 + 18)(9982 – 18)

[Since a^{2} -b^{2} = (a + b) (a – b)]

= (10000) × (9964)

LHS = (10000) × x

Comparing L.H.S. and RHS, we get

10000x = 10000 × 9964

x = 9964