ICSE Class 8 Maths Algebra Chapter 06 HCF LCM PDF Download

HCF and LCM

Highest Common Factor

If certain factors are common to two or more expressions then they are called common factors of those expressions. The product of all such common factors is called the highest common factor or HCF of the given expressions.

HCF of Monomials

To find the HCF of monomials, take the following steps.

1. Find the HCF of the numerical coefficients of the monomials.

2. Find the highest power of each of the variables common to the monomials.

3. The product of the number and the powers of the variables obtained in Steps 1 and 2 is the required HCF.

Example

Find the HCF of (i) \(6ab^3\) and \(8ab^2\), and (ii) \(2x^2y^2\), \(5x^3y\) and \(-3y^3\).

Solution

(i) \(6ab^3 = 2 \times 3 \times a \times b \times b \times b\), \(8ab^2 = 2 \times 2 \times 2 \times a \times b \times b\).

The HCF of the numerical coefficients of the monomials = HCF of 6 and 8 = 2.

The highest power of the variable \(a\) common to both = \(a\).

The highest power of the variable \(b\) common to both = \(b \times b = b^2\).

Therefore, the HCF of \(6ab^3\) and \(8ab^2 = 2 \times a \times b^2 = 2ab^2\).

(ii) \(2x^2y^2 = 2 \times x \times x \times y \times y\), \(5x^3y = 5 \times x \times x \times x \times y\), \(-3y^3 = -3 \times y \times y \times y\)

The HCF of 2, 5 and -3 = 1.

The highest power of \(x\) common to the monomials = \(x^0\) (there is no \(x\) in \(-3y^3\))

The highest power of \(y\) common to the monomials = \(y\).

Therefore, the HCF of \(2x^2y^2\), \(5x^3y\) and \(-3y^3 = 1 \times x^0 \times y = y\).

Note If the monomials do not have any factor in common, the HCF is 1.

Teacher's Note

Understanding HCF helps in simplifying fractions and finding common measures in real-world scenarios like dividing items into equal groups.

HCF of Polynomials

Take the following steps to find the HCF of polynomials.

1. Find the HCF of the common numerical coefficients if any.

2. Factorize each of the given expressions and take the factors common to all of them.

3. The product of the number and factors obtained in Steps 1 and 2 is the required HCF.

Example

Find the HCF of \(12(a^2 - ab)\) and \(16(ab - b^2)\).

Solution

The HCF of the numerical coefficients of the given expressions = the HCF of 12 and 16 = 4.

Also, \(a^2 - ab = a(a - b)\) and \(ab - b^2 = b(a - b)\).

We observe that \(a - b\) is a factor common to both expressions.

Therefore, the HCF of the given expressions = \(4(a - b)\).

Solved Examples

Example 1

Find the HCF of \(16a^4b^5c^6\), \(20a^5b^4c^3\) and \(24a^4b^4\).

Solution

\(16a^4b^5c^6 = 2 \times 2 \times 2 \times 2 \times a^4 \times b^5 \times c^6\)

\(20a^5b^4c^3 = 2 \times 2 \times 5 \times a^5 \times b^4 \times c^3\)

\(24a^4b^4 = 2 \times 2 \times 2 \times 3 \times a^4 \times b^4\)

The HCF of 16, 20 and 24 = \(2 \times 2\) = 4.

The highest power of the variable \(a\) common to all = \(a^4\).

The highest power of the variable \(b\) common to all = \(b^4\).

The variable \(c\) is absent in the monomial \(24a^4b^4\).

Therefore, the required HCF = \(4a^4b^4\).

Example 2

Find the HCF of \(a^2 - b^2\), \(a^2 - b^2 + ac - bc\) and \(a^3 - a^2b + ab^2 - b^3\).

Solution

\(a^2 - b^2 = (a + b)(a - b)\).

\(a^2 - b^2 + ac - bc = (a^2 - b^2) + (ac - bc) = (a + b)(a - b) + c(a - b) = (a - b)(a + b + c)\).

\(a^3 - a^2b + ab^2 - b^3 = (a^3 - a^2b) + (ab^2 - b^3) = a^2(a - b) + b^2(a - b) = (a - b)(a^2 + b^2)\).

The factor \((a - b)\) is common to all the given expressions.

Therefore, the required HCF = \(a - b\).

Example 3

Find the HCF of \(x^4 + 2x^2 + 1\), \(x^6 + x^4 - x^2 - 1\) and \(x^4 - 1\).

Solution

\(x^4 + 2x^2 + 1 = (x^2)^2 + 2 \cdot x^2 + 1 + (1)^2 = (x^2 + 1)^2 = (x^2 + 1)(x^2 + 1)\).

\(x^6 + x^4 - x^2 - 1 = (x^6 + x^4) - (x^2 + 1) = x^4(x^2 + 1) - 1(x^2 + 1) = (x^2 + 1)(x^4 - 1) = (x^2 + 1)(x^2 - 1)(x^2 + 1) = (x^2 + 1)((x^2)^2 - 1^2) = (x^2 + 1)(x^2 + 1)(x^2 - 1) = (x^2 + 1)(x + 1)(x - 1)\).

\(x^4 - 1 = (x^2)^2 - (1)^2 = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x(x - 1)(x + 1)(x - 1)\).

Only \((x^2 + 1)\) is common to all the given expressions.

Therefore, the HCF = \(x^2 + 1\).

Teacher's Note

Finding the HCF of polynomials is useful in simplifying complex algebraic fractions and solving equations in advanced mathematics.

Exercise 6A

Find the HCF of the following.

1. (i) \(2x^4y\), \(3xy^4\) (ii) \(2a^2b^3\), \(6a^4b^2\)

2. (i) \(a^4b^2c^5\), \(ab^3c^2\), \(a^2b^4c^4\) (ii) \(abc\), \(bcd\), \(cde\)

(iii) \(12m^2np\), \(16mn^2p\), \(20mnp^2\) (iv) \(24a^5x^3y^4\), \(36a^2x^5y^7\), \(48a^4x^4y^5\)

(v) \(10m^4n^2p^6\), \(15m^3n^3p^5\), \(25m^5n^5p^7\)

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