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Questions on Algebraic Expressions | Algebraic Identities | Algebraic Formulas

Algebraic Expressions | Algebraic Identities | Algebraic Formulas

Hi students, welcome to Amans Maths Blogs (AMB). On this page, you will get the short notes and important question of CBSE Class 8 Chapter 9 Algebraic Expressions and Algebraic Identities or Algebraic Formulas. These short notes and important questions are taken from the NCERT text books and other books. If you want to get the NCERT Solution for class 8 Maths Algebraic Expressions and Algebraic Identities or Algebraic Formulas, then click here.  If you need sample questions based on algebraic expression or identities or algebraic formulas for practice, then you are at right place. First, we start with short notes about the Algebraic Expressions and Algebraic Identities or Algebraic Formulas.

Algebraic Expressions Definition

When constants and variables are combined with mathematical operator to express a mathematical terms known as algebraic expressions.

Algebraic Expressions Example

4x2y, 7xy + 2xy2, 3x2y2 + 2xy – 5 etc. All are algebraic expressions.

Algebraic Expressions Terms

Algebraic expression is a combination of terms and each term is the product of a constant and variables. In the example of algebraic expression 3x2y2 + 2xy – 5, 3x2y2, 2xy and -5 are algebraic expressions terms.

Types of Algebraic Expressions Terms

In algebraic expression terms, there are two types. Like and unlike terms

Like Terms of Algebraic Expressions

In an algebraic expression, the terms which have same literals coefficients are called like terms. For example, in the algebraic expression 3xy + 5x + 7xy – 2y + 12, 3xy and 7xy are like terms.

Unlike Terms of Algebraic Expressions

In an algebraic expression, the terms which have not same literals coefficients are called unlike terms. For example, in the algebraic expression 3xy + 5x + 7xy – 2y + 12, 5x, -2y are unlike terms.

Mathematical Operations of Algebraic Expressions

Like we do mathematical operations on numbers, we can also do mathematical operations on the algebraic expressions. 

Addition of Algebraic Expressions

In the addition of the algebraic expressions, we add the like terms only and other unlike terms are kept same. For example, in the algebraic expressions (3xy + 5x) + (7xy – 2y + 12) = 10xy + 5x – 2y + 12.

Subtraction of Algebraic Expressions

In the subtraction of the algebraic expressions, we subtract the like terms only and other unlike terms are kept same. For example, in the algebraic expressions (3xy + 5x) – (7xy – 2y + 12) = -4xy + 5x + 2y – 12.

Multiplication of Algebraic Expressions

In the multiplication of the algebraic expressions, we multiply the numerical coefficients and multiply the literal coefficients using laws of exponents and then we add or subtract the algebraic expressions. For example, in the algebraic expressions (2x + 3)(3x + 2y – 9) = 6x2 + 4xy – 18x + 9x + 6y – 27 = 6x2 + 4xy – 9x + 6y – 27.

Division of Algebraic Expressions

In the division of the algebraic expressions, we do the long division of the algebraic expressions same as the long division of numbers. For example, in the algebraic expressions (x2 + 12x + 35)/(x + 5) = (x + 7).

Value of an Algebraic Expressions

To find the value of an algebraic expression, substitute the given value to the variables in the algebraic expression. The value of algebraic expression depends on the different value of the variables. For example, we need to find the value of 3x2 + 4 at x = 1. We put x = 1 in 3x2 + 4. Thus, we get 3(1)2 + 4 = 7.

Polynomial

An algebraic expression of the form of anxn + an-1xn-1 + an-2xn-2 + … + a3x3 + a2x2 + a1x + a0 is called as polynomial, where n is non-negative integers and an, an-1, an-2, …,a2, a1, a0 all are real numbers known as coefficient of algebraic expression terms. The algebraic expression 3x3 + 4x2 – 5x + 8 is a polynomial.

Degrees of Polynomial

In a polynomial anxn + an-1xn-1 + an-2xn-2 + … + a3x3 + a2x2 + a1x + a0, the highest exponent of a term is known as the degree of the polynomial. In the polynomial 3x3 + 4x2 – 5x + 8, the highest exponent is 3. Thus, the degree of the polynomial is 3.

Types of Polynomial

A polynomial is categories on the basis of number of terms and degrees of the polynomial. 

Types of Polynomial On The Basis of Number of Terms

The algebraic expression containing one term is a MONOMIAL. For example, 3x3 is a monomial.

The algebraic expression containing two unlike terms is a BINOMIAL. For example, 3x3  + 3x2 is a binomial.

The algebraic expression containing three unlike terms is a TRINOMIAL. For example, 3x3  + 3x2 + 6 is a trinomial.

The algebraic expression containing more than three unlike terms is a POLYNOMIAL. For example, 3x3  + 3x2 + 6x + 9 is a polynomial.

Types of Polynomial On The Basis of Degrees

The polynomial with degree 1 is a known as linear polynomial. Its general form is (ax + b). For example, 3x + 5 is a linear polynomial.

The polynomial with degree 2 is a known as quadratic polynomial. Its general form is (ax2 + bx + c). For example, 3x2 + 6x + 5 is a quadratic polynomial.

The polynomial with degree 3 is a known as cubic polynomial. Its general form is (ax3 + bx2 + cx + d). For example, 5x3 + 3x2 + 6x + 5 is a cubic polynomial.

The polynomial with degree 4 is a known as bi-quadratic polynomial. Its general form is (ax4 + bx3 + cx2 + dx + e). For example, 7x4 – 5x3 + 3x2 + 6x + 5 is a bi-quadratic polynomial.

The polynomial with degree 5 or more than 5 is a known as polynomial with degree 5 or more. For example, 6x5 – 7x4 – 5x3 + 3x2 + 6x + 5 is a polynomial with degree 5. The general form of n degree polynomial is (anxn + an-1xn-1 + an-2xn-2 + … + a3x3 + a2x2 + a1x + a0).

Constant Polynomial

The polynomial with the constant term is known as constant polynomial. The degree of a constant polynomial is zero. For example: p(x) = 6 is a constant polynomial.

Zero Polynomial

If a= an-1 = an-2= … = a= a= a= a0 = 0 in the polynomial anxn + an-1xn-1 + an-2xn-2 + … + a3x3 + a2x2 + a1x + a0, then it is called as zero polynomial. The degree of the zero polynomial is not defined. The polynomial p(x) = 0 is a zero polynomial.

Zeroes/Roots of Polynomial

The number or numerical value for which the value of a polynomial is zero is called zero/roots of the polynomial. For example: If we put x = 1 in the polynomial x2 – 4x + 3, we get 1 – 4 + 3 = 0. Thus, x = 1 is a zero or roots of the polynomial x2 – 4x + 3.

Algebraic Identities or Algebraic Formulas

An algebraic identities or algebraic formula is special product of two or more algebraic expressions. There are following algebraic identities.

Algebraic Identities 1

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

Algebraic Formulas 2

(x – a)(x – b) = x2 – (a + b)x + ab

Algebraic Formulas 3

(a + b)2 = a2 + 2ab + b2 

Algebraic Identities 4

(a – b)2 = a2 – 2ab + b2 

Algebraic Formulas 5

(a + b)(a – b)= a2 – b2 

Algebraic Identities 6

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Algebraic Formulas 7

(a + b)2 = (a – b)2 + 4ab

Algebraic Identities 8

(a – b)2 = (a + b)2 – 4ab

Algebraic Formulas 9

(a + b)3 = a3 + b3 + 3ab(a + b) = a3 + b3 + 3a2b + 3ab2

Algebraic Identities 10

(a – b)3 = a3 – b3 – 3ab(a – b) = a3 – b3 – 3a2b + 3ab2

Algebraic Formulas 11

a3 + b3 = (a + b)3 – 3ab(a + b) = (a + b)(a2 – ab + b2)

Algebraic Identities 12

a3 – b3 = (a – b)3 + 3ab(a – b) = (a – b)(a2 + ab + b2)

Algebraic Formulas 13

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Algebraic Identities 14

a3 + b3 + c3 – 3abc = (1/2)(a + b + c)[(a – b)2 + (b – c)2 + (c – a)2

Algebraic Formulas 15

If a + b + c = 0, then a3 + b3 + c3 = 3abc

Algebraic Identities 16

(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

Questions on Algebraic Expressions

Question 1: 

Multiply (x – y) by (x2 + y2).

Answer 1: (x3 – xy2 – x2y – y3)

Question 2: 

Divide (p2 + 6p + 8) by (p + 4)

Answer 2: (p + 2)

Question 3: 

Divide (2x3 + x2 – 3x – 3) by (2x – 1)

Answer 3: 

Question 4: 

Find a if (2x – 3) is a factor of 2x4 – x3 – 3x2 – 2x + a.

Answer 4: 

Question 5: 

Multiply (p – 8)(p – 7).

Answer 5: p2 – 15p + 56

Question 6: 

Multiply (105 x 103) using algebraic identities.

Answer 6: 10815

Question 7: 

Expand (4a + 3b + 2c)2

Answer 7: 16a2 + 9b2 + 4c2 + 24ab + 12bc + 16ca

Question 8: 

Expand and simplify (p + q + r)– (p – q – r)2

Answer 8: 

Question 9: 

Simplify (a + 3b)3 + (a – 3b)3

Answer 9: 

Question 10: 

Evaluate (p2 + 4)(p2 – 1/4)

Answer 10: p4 + (15/4)p2 – 1

Question 11: 

Simplify (2a – b + c)– (2a + b – c)2

Answer 11: 8a(c – b)

Question 12: 

Find the value of (x3 – 8y3) if x – 2y = 2 and xy = 8

Answer 12: 104

Question 13: 

Find the value of (97 1/2)2

Answer 13: 9506 1/4

Question 14: 

Find the product (n – 1/n)(n + 1/n)(n2 + 1/n2)

Answer 14: (n4 – 1/n4)

Question 15: 

If x – 1/x = 5, then find the value of x3 – 1/x3.

Answer 15: 140

Question 16: 

Find the value of (16)3 + (-20)3 + (4)3

Answer 16: -3840

Question 17: 

If x – 1/x = 3, then find the value of (x2 + 1/x2)

Answer 17: 11

Question 18: 

If x2 + 1/x2 = 27, then find the value of (x 1/x)

Answer 18: 5 or -5

Question 19: 

Find the product of (x2 + x + 1) and (x2 – x + 1)

Answer 19: (x4 + x2 + 1)

Question 20: 

If x + 1/x = 9, then find the value of (x4 + 1/x4)

Answer 20: 6239

Question 21: 

If (a + b + c) = 6 and a2 + b2 + c2 = 26, then find the value of ab + bc + ca.

Answer 21: 5

Question 22: 

If p + 1/q = 1, q + 1/r = 1, then find the value of r + 1/p.

Answer 22: 1

Question 23: 

If (a+ b + c) = 0, then find the value of a(b – c)3 + b(c – a)3 + c(a – b)3.

Answer 23: 0

Question 24: 

If zero of (3x + 4) and (px + 9) are equal, then find the value of p.

Answer 24: 27/4

Question 25: 

If x2 – x – 42 = (x + k)(x + 6), then find the value of k.

Answer 25: -7

Question 26: 

Find the product of (x – 1)(x + 1)(x2 + 1)(x4 + 1)

Answer 26: x8 – 1

Question 27: 

Simplify (a + b + c)2 – (a – b – c)2

Answer 27: 4a(b + c)

Question 28: 

If x + 1/x = \sqrt{5}, then find the value of (x2 + 1/x2) and (x4 + 1/x4).

Answer 28: 3 and 7

Question 29: 

If 2x – 3y = 8 and xy = 2, then find the value of (4x2 + 9y2)

Answer 29: 88

Question 30: 

If p = 2 – a, then find the value of a3 + 6ap + p3 – 8

Answer 30: 0

Question 31: 

If a + b + c = 9 and ab + bc + ca = 26, then find the value of a3 + b3 + c3 – 3abc.

Answer 31: 27

Question 32: 

Find the value of (p2 + q2) if (p + q) = 3 and pq = 2

Answer 32: 5

Question 33: 

Check whether (x2 + 3) is a factor of 4x4 + 7x2 – 15.

Answer 33: YES

Question 34: 

What should be added to (4p2 – 12p + 6) to make it a perfect square?

Answer 34: 3

Question 35: 

What should be subtracted from (4x2 – 20x + 30) to make it a perfect square?

Answer 35: 5

Question 36: 

Find the value of (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a).

Answer 36: 0

Question 37: 

For what value of a for which (x + 2) is a factor of (4x4 + 2x3 – 3x2 + 8x + 5a).

Answer 37: -4

Question 38: 

If 2x – 3y = 2 and 2x + 3y = 14, then find the value of xy.

Answer 38: 8

Question 39: 

For what values of a, is (2x3 + ax2 + 11x + a + 3) is perfectly divisible by (2x – 1). 

Answer 39: -7

Question 40: 

What must be subtracted from (4x4 – 2x3 – 6x2 + x – 5) so that the result is exactly divisible by (2x2 + x – 1)

Answer 40: -6

Question 41: 

If both (x – 2) and (x – 1/2) are factor of px2 + 5x + r, then find the value of (p – r)

Answer 41: 0

Question 42: 

Find the remainder when (5x3 – x2 + 6x – 2) is divided by (1 – 5x).

Answer 42: -4/5

Question 43: 

What must be subtracted from (8x4 + 14x3 – 2x2 + 7x – 10) so that the difference is exactly divisible by (4x2 + 3x – 2)?

Answer 43: 14x – 12

Question 44: 

Find the value of b in (x + 6)(x + b) = x2 + 2x – 24

Answer 44: -2

Question 45: 

Find the value of (102)3.

Answer 45: -2

Question 46: 

When kx3 + 9x2 + 4x – 8 is divided by (x + 3) leaves the reminder -20, then find the value of k.

Answer 46: 3

Question 47: 

Check whether 32x10 – 33x2 + 1 is divisible by (x – 1) and (x + 1).

Answer 47: YES and NO

Question 48: 

If f(x) = x4 – 2x3 + 3x2 – ax + b leaves remainder 5 and 19 on division by (x – 1) and (x + 1) respectively, then find the remainder when f(x) is divided by (x – 2).

Answer 48: 10

Question 49: 

Find the reminder when x4a + x2ay2b + y4b is divided by x2a + xayb + y2b.

Answer 49: 0

Question 50: 

Find the expansion (x2 + 4)(x2 – 4)(x4 + 16).

Answer 50: x8 – 256

Question 51: 

Factorize a3 + (a – b)3 – b3

Answer 51: -3ab(b – a)

Question 52: 

If (a/b) + (b/a) = 2, then find the value of (a/b)10 – (b/a)10

Answer 52: 0

Question 53: 

If x2 + y2 – xy = 3 and y – x = 1, then find the value of xy/(x2 + y2).

Answer 53: 2/5

Question 54: 

If abc = 6 and a + b + c = 6, then 1/ac + 1/bc + 1/ab = x, then find x.

Answer 54: 1

Question 55: 

Factorize x2 + y2 – z2 – 2xy

Answer 55: (x – y – z)(x – y + z)

Question 56: 

If x – 1/(2x) = 2, then find the value of (4x4 + 1)/4x2.

Answer 56: 5

Question 57: 

If x2 + 8x + k is a perfect square, then find the value of k.

Answer 57: 16

Question 58: 

If x + 1/x = 2, then find the value of x2010 + x2009.

Answer 58: 2

Question 59: 

Find the square root of  3^{6n^{2}}(36)^{2a}(16)^{b}.

Answer 59:  3^{3n^{2}}(6)^{2a}(2)^{2b}.

Question 60: 

If 2x + 1/(3x) = 6, then find the value of 9x2 + 1/(4x2).

Answer 60: 78

Question 61: 

If x + 1/(2x) = 4, then find the value of 4x2 + 1/(x2).

Answer 61: 60

Question 62: 

Find the value of algebraic expression 4xy/7 + 7z, if x = Additive identity of whole number, y = multiplicative identity of natural number, z = additive inverse of 7.

Answer 62: -49

Question 63: 

Simplify (2x + 3)(3x – 2) – (x – 4)(2x + 1)

Answer 63: 4x2 + 12x – 2

Question 64: 

If 4x2 + 9y2 = 20 and xy = 2, then find the value of (2x + 3y).

Answer 64: 8 or -8

Question 65: 

Find the product (p – q/5 + 1)(p + q/5 + 1)

Answer 65: p2 – q2/25 – 2q/5 – 1

Question 66: 

Subtract 2(a4 – 3a2) from 2a2(a3 – a) – 3a(a4 + 2a)

Answer 66: -a5 – 2a4 – 2a3

Question 67: 

Subtract the sum of (-6b + c  + a) and (-3a + 2c + 3b) from the sum of (5a – 8b + 2c) and (3c – 4b – 2a)

Answer 67: 5a – 9b + 2c

Question 68: 

Find the value of x5/125 + x3/25 – x2/5 at x = 5

Answer 68: 25

Question 69: 

If A = x3 – x2 + x – 1, B = 2(x2 + 1), C = 1 + x + x2 + x3, then find the value of (A + B + C) at x = 2.

Answer 69: 30

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