Algebraic Expressions and Identities

Algebraic expressions are expressions formed from the variables and the constants. A variable can take any value. The value of an expression changes with the value chosen for variables it contains.

  A number line has infinite number of points. A variable can take position on number line.

  Expressions containing one, two or three terms are called monomial, binomial and trinomial respectively.

  Any expression having one or more terms is called polynomial.

  A monomial is obtained on multiplying any monomial with another monomial.

  The numerical factor of a term is called its coefficient.

  An identity is a standard equality which is true for all the values of the variables in the equality.

  Few commonly used identities are

        I.   (a + b)² = a² + b² + 2ab

       II.   (a – b)² = a² + b² – 2ab

      III.   (a + b)(a – b) =a² – b²

      IV.   (x + a)(x + b) = x² + (a+b)x + ab

Note – Mainly the above formulas are used to solve all the problems of this chapter.

1.  Find the value of: x² – 1/5 at x= -1.

2.  What is the value of x² + y² – 10 at x = 0 and y = 0?

3.  Find the product of 9a, 4ab and -2a.

4.  Simplify (a + b + c)(a + b - c).

5.  Using identities evaluate: 8.56 * 11.60.

6.  Using identities evaluate: (99)².

7.  Simplify x(2x – 1) + 5 and find its value at x = -2.

8.  Evaluate the value of (95)² using identities.

1.  Which of the following is the numerical coefficient of x²y²?

        I.   0

       II.   1

      III.   x²

      IV.   y²

2.  Which of the following is the numerical coefficient of -5xy?

        I.   5

       II.   -x

      III.   -5

      IV.   -y

3.  pqr is what type of polynomial?

        I.   Monomial

       II.   Binomial

      III.   Trimonial

      IV.   None of these

4.  The value of x² - 5 at x= -1 is-

        I.   -2

       II.   -1

      III.   -4

      IV.   -5

5.  a²-b² is a product of

        I.   (a+b)(a-b)

       II.   (a+b)(a+b)

      III.   (a-b)(a-b)

      IV.   None of these

6.  Which of the following is the value of (x+ 1/x)²?

        I.   x² + 1/x²

       II.   x² - 1/x²

      III.   x² + 1/x² + 2

      IV.   x²+ 1/x² + 2x

7.  Which of the following is obtained by subtracting x²-y² from y² - x²?

        I.   -2(x²-y²)

       II.   -2(x² + y²)

      III.   2(x²+ y²)

      IV.   2(x²- y²)

8.  What degree does x³ - x²y² - 8y²+ 2 have?

        I.   2

       II.   3

      III.   4

      IV.   7

9.  What is the value of 5x²⁵ - 3x³² + 2x^-12 at x=1?

        I.   0

       II.   2

      III.   4

      IV.   None of these

10.  What is the product of (x+a) and (x+b)?

        I.   x²+ (a-b)x + ab

       II.   x² + (a+b)x - ab

      III.   x² + (a+b)x - ab

      IV.   x² + (a+b)x + ab

ANSWERS

        1.    II

        2.    III

        3.    I

        4.    III

        5.    I

        6.    III

        7.    I

        8.    III

        9.     III

        10.  I

Maximum time- 30 minutes

Maximum marks- 20

1.  Add: a + b + ab; b – c + bc and c + a + ac.      (2)

2.  Verify the identity (x + a)(x + b) = x² + (a + b)x + ab for a = 2, b = 3 and x = 4.      (3)

3.  Find the volume of cuboid whose dimensions are (x² – 2); (2x + 4) and (x - 3).      (4)

4.  Write the terms and coefficients of 3 – xy + yz – xz.      (2)

5.  Simplify: (a + b +c)(a + b – c).      (2)

6.  Simplify the expression x(2x-1) + 5 and its value at x = -2.      (3)

7.  Using suitable identities find (xy + 3p)².      (2)

8.  Subtract 5x² – 6y² + 8y – 5 from 7x² – 5xy + 10y² + 5x – 4y.      (2)

Algebraic Expressions and Identities Important Questions For Class 8 (Chapter 9)

Algebraic Expressions and Identities Important Questions For Class 8 (Chapter 9)

. Using suitable algebraic identity, solve 10922

Solution:

Use the algebraic identity: (a + b)² = a² + 2ab + b²

Now, 1092 = 1000 + 92

So, 10922 = (1000 + 92)2

(1000 + 92)2 = ( 1000 )² + 2 × 1000 × 92 + ( 92 )²

= 1000000 + 184000 + 8464

Thus, 10922 = 1192464.

2. Identify the type of expressions:

(i) x2y + xy2

(ii) 564xy

(iii) -8x + 4y

(iv) x2 + x + 7

(iv) xy + yz + zp + px + 9xy

Solution:

(i) x2y + xy2 = Binomial

(ii) 564xy = Monomial

(iii) -8x + 4y = Binomial

(iv) x2 + x + 7 = Trinomial

(iv) xy + yz + zp + px + 9xy = Polynomial

3. Identify terms and their coefficients from the following expressions:

(i) 6x2y2 – 9x2y2z2+ 4z2

(ii) 3xyz – 8y

(iii) 6.1x – 5.9xy + 2.3y

Solution:

(i) 6x2y2 – 9x2y2z2+ 4z2

Terms = 6x2y2, -9x2y2z2, and 4z2

Coefficients = 6, -9, and 4

(ii) 3xyz – 8y

Terms = 3xyz, and -8y

Coefficients = 3, and -8

(iii) 6.1x – 5.9xy + 2.3y

Terms = 6.1x, – 5.9xy, and 2.3y

Coefficients = 6.1, – 5.9 and 2.3

4. Find the area of a square with side 5x2y

Solution:

Given that the side of square = 5x2y

Area of square = side2 = (5x2y)2 = 25x4y2

5. Calculate the area of a rectangle whose length and breadths are given as 3x2y m and 5xy2 m respectively.

Solution:

Given,

Length = 3x2y m

Breadth = 5xy2 m

Area of rectangle = Length × Breadth

= (3x2y × 5xy2) = (3 × 5) × x2y × xy2 = 15x3y3 m2

Long Answer Type Questions:

6. Simplify the following expressions:

(i) (x + y + z)(x + y – z)

(ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)

(iii) 2x2(x + 2) – 3x (x2 – 3) – 5x(x + 5)

Solution:

Notes: “+” × “+” = “+”, “-” × “-” = “+”, and “+” × “-” = “-”.

(i) (x + y + z)(x + y – z)

= x2 + xy – xz + yx + y2 – yz + zx + zy – z2

Add similar terms like xy and yx, xz and zx, and yz and zy. Then simplify and rearrange.

= x2 + y2 – z2 + 2xy

(ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)

= x3 – 3x2y2 – xy3 + 2x2y2 – xy3 + 5x3

Now, add the similar terms and rearrange.

= x3 + 5x3 – 3x2y2 + 2x2y2 – xy3 – xy3

= 6x3 – x2y2 – 2xy3

(iii) 2x2(x + 2) – 3x (x2 – 3) – 5x(x + 5)

= 2x3 + 4x2 – 3x3 + 9x – 5x2 – 25x

= 2x3 – 3x3 – 5x2 + 4x2 + 9x – 25x

= -x3 – x2 – 16x

7. Add the following polynomials.

(i) x + y + xy, x – z + yx, and z + x + xz

(ii) 2x2y2– 3xy + 4, 5 + 7xy – 3x2y2, and 4x2y2 + 10xy

(iii) -3a2b2, (–5/2) a2b2, 4a2b2, and (⅔) a2b2

Solution:

(i) x + y + xy, x – z + yx, and z + x + xz

= (x + y + xy) + (x – z + yx) + (z + x + xz)

= x + y + xy + x – z + yx + z + x + xz

Add similar elements and rearrange.

= 2xy + xz + 3x + y

(ii) 2x2y2– 3xyz + 4, 5 + 7xy – 3x2y2, and 4x2y2 + 10xy

= (2x2y2– 3xy + 4) + (5 + 7xy – 3x2y2) + (4x2y2 + 10xy)

= 2x2y2 – 3xy + 4 + 5 + 7xy – 3x2y2 + 4x2y2 + 10xy

Add similar elements and rearrange.

= 3x2y2 + 14xy + 9

(iii) -3a2b2, (–5/2) a2b2, 4a2b2, and (⅔) a2b2

Answer= (⅚)a2b2 (Self Assessment)

8. Subtract the following polynomials.

(i) (7x + 2) from (-6x + 8)

(ii) 3xy + 5yz – 7xz + 1 from -4xy + 2yz – 2xz + 5xyz + 1

(iii) 2x2y2– 3xy + 4 from 4x2y2 + 10xy

Solution:

(i) (7x + 2) from (-6x + 8)

= (-6x + 8) – (7x + 2)

= -6x + 8 – (7x + 2)

= -6x + 8 – 7x – 2

= -13x + 6

(ii) 3xy + 5yz – 7xz + 1 from -4xy + 2yz – 2xz + 5xyz + 1

= 5xz + 5xyz – 7xy – 3yz (Self Assessment)

(iii) 2x2y2– 3xy + 4 from 4x2y2 + 10xy

= 21xy – 4 (Self Assessment)

9. Calculate the volume of a cuboidal box whose dimensions are 5x × 3x2 × 7x4

Solution:

Given,

Length = 5x

Breadth = 3x2

Height = 7x4

Area of cuboid = Length × Breadth × Height

= 5x × 3x2 × 7x4

Multiply 5, 3, and 7

= 105xx2x4

Use exponents rule: xa × xb = x(a+b)

So, 105xx2x4 = 105x1+2+4 = 105x7

10. Simplify 7x2(3x – 9) + 3 and find its values for x = 4 and x = 6

Solution:

7x2(3x – 9) + 3

Solve for 7x2(3x – 9)

= 7x2 × 3x × -7x2 × 9 (using distributive law: a(b – c) = ab – ac)

= 7 × 3x2x – 7 × 9x2

= 21x3 – 63x2

So, 7x2(3x – 9) + 3

= 21x3 – 63x2 + 3

Now, for x = 4,

21x3 – 63x2

= 21 × 43 – 63 × 42

= 1344 – 1008

= 336

Now, for x = 6,

21x3 – 63x2

= 21 × 63 – 63 × 62

= 2268