Download ICSE Class 8 Mathematics Textbooks
Review the ICSE Class 8 Maths Chapter 15 Simplification of Algebraic Fractions PDF Download designed for Class 8 Mathematics students. Published under the latest ICSE guidelines for 2026-27, this chapter-wise resource supports daily study and targeted revision.
Access Chapter 15 Simplification of Algebraic Fractions for Class 8 Mathematics
View or download the dedicated Chapter 15 Simplification of Algebraic Fractions resource below. This chapter-by-chapter structuring ensures easy navigation for daily study routines. For comprehensive exam preparation, pair this reading with our verified ICSE Solutions.
Chapter 15: Simplification Of Algebraic Fractions
You have worked on numerical fractions like \(\frac{3}{12}, \frac{5}{7}, -\frac{4}{9}, 1\frac{3}{4}\) and algebraic fractions like \(\frac{x}{3}, \frac{2y+3}{7}, \frac{x+2y-1}{5}, \frac{5}{2x-1}\) etc. Working with algebraic fractions often involves the concept of H.C.F. and L.C.M. of polynomials (with integral coefficients). In this chapter, we shall introduce the idea of H.C.F. and L.C.M. of polynomials and strengthen the concept of simplification of algebraic expressions.
H.C.F. And L.C.M. Of Polynomials
H.C.F. of two or more polynomials (with integral coefficients) is the highest common factor of the given polynomials.
In particular,
H.C.F. of two or more monomials = (H.C.F. of their numerical coefficients) \(\times\) (H.C.F. of their literal coefficients)
H.C.F. of literal coefficients = product of each common literal raised to the lowest power.
Remark: Here it is understood that the numerical coefficients of the monomials (under consideration) are integers and the powers of the literals involved in the monomials are positive integers.
L.C.M. of two or more polynomials (with integral coefficients) is the smallest common multiple of the given polynomials.
In particular,
L.C.M. of two or more monomials = (L.C.M. of their numerical coefficients) \(\times\) (L.C.M. of their literal coefficients)
L.C.M. of literal coefficients = product of each literal raised to the highest power.
Example 1.
Find H.C.F. and L.C.M. of the monomials \(24x^4y^3\) and \(32x^2y^4z\).
Solution.
H.C.F. of numerical coefficients = H.C.F. of 24 and 32 = 8
H.C.F. of literal coefficients = product of each common literal raised to the lowest power
\[= x^2y^3\]
\(\therefore\) H.C.F. of the given monomials = \(8 \times x^2y^3 = 8x^2y^3\)
L.C.M. of numerical coefficients = L.C.M. of 24 and 32 = 96
L.C.M. of literal coefficients = product of each literal raised to the highest power
\[= x^4y^4z\]
\(\therefore\) L.C.M. of the given monomials = \(96 \times x^4y^4z = 96x^4y^4z\).
Example 2.
Find H.C.F. and L.C.M. of the monomials \(12a^3b^2, 16ab^3c^2\) and \(20a^2b^2c\).
Solution.
H.C.F. of numerical coefficients = H.C.F. of 12, 16 and 20 = 4
H.C.F. of literal coefficients = product of each common literal raised to the lowest power
\[= ab^2\]
\(\therefore\) H.C.F. of the given monomials = \(4 \times ab^2 = 4ab^2\)
L.C.M. of numerical coefficients = L.C.M. of 12, 16 and 20 = 240
L.C.M. of literal coefficients = product of each literal raised to the highest power
\[= a^3b^3c^2\]
\(\therefore\) L.C.M. of the given monomials = \(240 \times a^3b^3c^2 = 240a^3b^3c^2\).
Method To Find H.C.F. And L.C.M. Of Polynomials
To find H.C.F. and L.C.M. of given polynomials - factorise each polynomial, then
(i) H.C.F. = (H.C.F. of numerical coefficients) \(\times\) (each common factor raised to the lowest power)
(ii) L.C.M. = (L.C.M. of numerical coefficients) \(\times\) (each factor raised to the highest power)
Example 3.
Find the H.C.F. and L.C.M. of the polynomials \(4x^2 + 4xy, 6x^2 - 6y^2\).
Solution.
\(4x^2 + 4xy = 4x(x + y)\)
and \(6x^2 - 6y^2 = 6(x^2 - y^2) = 6(x + y)(x - y)\)
Factorise given polynomials
H.C.F. of numerical coefficients = H.C.F. of 4 and 6 = 2
\(\therefore\) H.C.F. of the given polynomials = \(2(x + y)\)
L.C.M. of numerical coefficients = L.C.M. of 4 and 6 = 12
\(\therefore\) L.C.M. of the given polynomials = \(12x(x + y)(x - y)\).
Example 4.
Find the H.C.F. and L.C.M. of \(5a^2 + 15a, a^2 + 6a + 9\) and \(2a^2 - 18\).
Solution.
Factorising the given polynomials, we get
\(5a^2 + 15a = 5a(a + 3),\)
\(a^2 + 6a + 9 = (a)^2 + 2 \times a \times 3 + (3)^2 = (a + 3)^2\)
and \(2a^2 - 18 = 2(a^2 - 9) = 2(a + 3)(a - 3)\)
H.C.F. of numerical coefficients = H.C.F. of 5, 1 and 2 = 1
\(\therefore\) H.C.F. of the given polynomials = \(1 \times (a + 3) = a + 3\)
L.C.M. of numerical coefficients = L.C.M. of 5, 1 and 2 = 10
\(\therefore\) L.C.M. of the given polynomials = \(10a(a + 3)^2(a - 3)\).
Example 5.
Find the H.C.F. and L.C.M. of \(12x^2 - 75, 4x^2 - 20x + 25\) and \(6x^2 - 13x - 5\).
Solution.
Factorising the given polynomials, we get
\(12x^2 - 75 = 3(4x^2 - 25) = 3(2x + 5)(2x - 5),\)
\(4x^2 - 20x + 25 = (2x)^2 - 2 \times 2x \times 5 + (5)^2 = (2x - 5)^2\)
and \(6x^2 - 13x - 5 = 6x^2 + 2x - 15x - 5\)
We need two integers whose sum is -13 and product is \(6 \times (-5)\) i.e. -30.
By trial, \(2 + (-15) = -13\) and \(2 \times (-15) = -30\)
\(= 2x(3x + 1) - 5(3x + 1)\)
\(= (3x + 1)(2x - 5).\)
H.C.F. of numerical coefficients = H.C.F. of 3, 1 and 1 = 1
\(\therefore\) H.C.F. of the given polynomials = \(1 \times (2x - 5)\)
\(= 2x - 5\)
L.C.M. of numerical coefficients = L.C.M. of 3, 1 and 1 = 3
\(\therefore\) L.C.M. of the given polynomials = \(3(2x + 5)(2x - 5)^2(3x + 1)\).
Teacher's Note
Understanding H.C.F. and L.C.M. of polynomials helps students grasp how common factors work, similar to finding the greatest common divisor when sharing items equally among groups in real life.
Exercise 15.1
Find the H.C.F and L.C.M. of the following (1 to 4) monomials:
1. (i) \(6x^2y\) and \(4xy^2\) (ii) \(12a^3b^2\) and \(18a^2b^5\)
2. (i) \(12x^3y^3\) and \(28x^2y^2z\) (ii) \(48a^2bc\) and \(56ab^2c^3d\)
3. (i) \(2m^3n^2, 3mn^3\) and \(5m^2n\) (ii) \(6a^2b^2, 15ab^4\) and \(12a^4b^2\)
4. (i) \(9p^3q^2r, 36pq^3r^3\) and \(12p^2q^3r^3\) (ii) \(10x^3y^2z^4, 15x^2y^3\) and \(20yz^2\)
Find the H.C.F and L.C.M. of the following (5 to 8) polynomials:
5. (i) \(x^2 + 3xy\) and \(x^2 - 9y^2\) (ii) \(9x^2 - 16y^2\) and \(15x^2 - 20xy\)
6. (i) \(4a^2 - 25\) and \(4a^2 - 20a + 25\) (ii) \(6x^2 + 12xy\) and \(4x^2 - 16y^2\)
7. (i) \(x^2 + 6x + 9\) and \(2x^2 + 7x + 3\) (ii) \(4a^2 + 4a + 1\) and \(6a^2 + 7a + 2\)
8. (i) \(3x^2 + 6x, x^2 + 5x + 6\) and \(2x^2 + 8x + 8\)
(ii) \(4x^4 - 36, 2x^3 - 12x + 18\) and \(2x^2 + x - 21\)
Algebraic Fractions
Fractions involving polynomials either in numerator or denominator (or both) are called algebraic fractions.
For example:
(i) \(\frac{2x}{7}, \frac{3x-2}{5}, \frac{x+2y-7}{11}, \frac{2x^4+5}{3}\) etc. are algebraic fractions with integral denominators.
(ii) \(\frac{a}{b}, -\frac{3}{x}, \frac{1}{x+2}, \frac{2x+3}{x-5}, \frac{x^2-1}{2x+1}, \frac{3x^2+5}{x^2+4}\) etc. are algebraic fractions involving variables in the denominators.
This is a preview of the first 3 pages. To get the complete book, click below.
Official ICSE Textbook PDF: Class 8 Mathematics Chapter 15 Simplification of Algebraic Fractions
Class 8 Mathematics Chapter 15 Simplification of Algebraic Fractions Official E-Book
Secure your copy of the ICSE Textbook for Class 8 Mathematics Chapter 15 Simplification of Algebraic Fractions. Widely adopted across educational institutions, final question papers map directly to the framework outlined in this chapter.
English Medium ICSE Textbooks for Class 8
Explore our exhaustive library of ICSE books in English Medium spanning all subjects in Class 8. Every chapter features comprehensive explanations followed by extensive end-of-chapter exercises.
Additional Study Resources for Class 8 Mathematics
Elevate your study routine by reviewing our comprehensive NCERT Solutions and revision notes available on our platform free of charge.
FAQs
You can download the latest, teacher-verified PDF for ICSE Class 8 Maths Chapter 15 Simplification of Algebraic Fractions PDF Download for free on StudiesToday.com. These digital editions are updated as per 2026-27 session and are optimized for mobile reading.
Yes, our collection of Class 8 Mathematics ICSE books follow the 2026 rationalization guidelines. All deleted chapters have been removed and has latest content for you to study.
Downloading chapter-wise PDFs for Class 8 Mathematics allows for faster access, saves storage space, and makes it easier to focus in 2026 on specific topics during revision.
ICSE books are the main source for ICSE exams. By reading ICSE Class 8 Maths Chapter 15 Simplification of Algebraic Fractions PDF Download line-by-line and practicing its questions, students build strong understanding to get full marks in Mathematics.