WBBSE Solutions For Class 8 Maths Algebra Chapter 8 Lowest Common Multiple

Algebra Chapter 8 Lowest Common Multiple

Lowest Common Multiple Introduction

In arithmetic, you have learned how to find the H.C.F. i.e., the Highest Common Factor, and the L.C.M. i.e., the Lowest Common Multiple of numbers. Similarly, we can find the H.C.F. and L.C.M. of two or more algebraic expressions. In this chapter, we aim to find the L.C.M. of algebraic expressions by the method of factorization.

Multiple

When an expression is divisible by another expression, then the first expression is called a multiple of the second expression. For example, x2y2 is a multiple of the expressions, x, y, xy, x2y, xy2, etc.

WBBSE Class 8 Lowest Common Multiple Notes

Common Multiple

When an expression is divisible by each of two or more expressions, then the first expression is called the common multiple of those expressions. For example, xy, x2y2, xy2, etc. are divisible by each of x and y and hence they are common multiples of x and y.

The Lowest Common Multiple

Among the common multiples of some quantities, the lowest one (which is of the lowest power) is called the Lowest Common Multiple or L.C.M. of those quantities.

For Example :

L.C.M. of ab, a2b, and ab2 is a2b2. Here a2b2 is divisible by each of the quantities ab, a2b, and ab2. Moreover, a2b2 is of minimum power among other quantities which are divisible by ab, a2b, and ab2 (for example, a3b2, a2b3, a464, etc.).

Determination of L.C.M. by factorization

1. First of all, the given expressions are resolved into factors.

2. The L.C.M. is the product of all kinds of factors in their highest powers.

3. If numerical coefficients appear before the alphabetic expressions then after finding the L.C.M. of the alphabetic expressions the arithmetical L.C.M. of the numerical coefficients is written before it.

Read And Learn More WBBSE Solutions For Class 8 Maths

Algebra Chapter 8 Lowest Common Multiple Some Examples of L.C.M.

Example 1

Find the L.C.M. of x2yz, xy2z, and xyz2.

Solution :

Given: x2yz, xy2z, And xyz2

Here, the factor with the highest power of x is the x2

The factor with the highest power of y is y2

The factor with the highest power of z is z2

Hence, the required

L.C.M. =x2y2z2

Understanding Lowest Common Multiple (LCM)

Example 2

Find the L.C.M. of 4a26c2, 8ab2c3 and 16a4b3c.

Solution :

Given: 4a26c2, 8ab2c3 And 16a4b3c.

The L.C.M. of 4, 8, and 16 is 16.

The factor with the highest power of a is a4

The factor with the highest power of b is b3

The factor with the highest power of c is c3

Hence, the required L.C.M. = 16a4b3c3

WBBSE Solutions For Class 8 Maths Algebra Chapter 8 Lowest Common Multiple

Example 3

Find the L.C.M. of a3b – ab3 and a3b2 + a2b3.

Solution:

Given  a3b – ab3 And a3b2 + a2b3

First expression = a3b – ab3

= ab(a2 – b2) = ab(a +b) (a – b)

Second expression

= a362 + a2b3

= a2b2(a + b)

Hence, the required L.C.M.

= a2b2(a + b) (a – b)

= a2b2(a2 – b2)

The required L.C.M = a2b2(a2 – b2)

Example 4

Find the L.C.M. of x2  4x + 3 and x2  5x + 6.

Solution :

Given x2  4x + 3 And x2  5x + 6

First expression = x2 – 4x + 3

= x2-3x-x + 3

= x(x – 3) – l(x – 3)

= (x – 3) (x – 1)

Second expression

= x2 – 5x + 6

= x2 – 3x – 2x + 6

= x(x – 3) – 2(x – 3)

= (x-3)(x-2)

Hence, the required L.C.M. = (x – 1) (x – 2) (x – 3)

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Step-by-Step Guide to Finding LCM

Example 5

Find the L.C.M. of 8(a2– 4), 12(a3 + 8) and 36(a2 – 3a – 10).

Solution :

Given 8(a2– 4), 12(a3 + 8) And 36(a2 – 3a – 10).

First expression

= 8(a2 – 4)

= 23 x {(a)2 – (2)2}

= 23x (a+2) (a- 2)

Second expression

= 12(a3 + 8)

= 22 x 3 x {(a)3 + (2)3}

= 22 x 3 x (a + 2) (a2 – 2a + 4)

Third expression

= 36 (a2 – 3a – 10)

= 22 x 32 x {a2 – 5a + 2a – 10}

= 22 x 32 x {a(a – 5) + 2(a – 5)}

= 22 x 32 x (a – 5) (a + 2)

Hence, the required L.C.M.

= 23 x 32 x (a + 2) (a – 2) (a2-2a + 4) (a – 5)

= 72 (a + 2) (a – 2) (a – 5) (a2 – 2a + 4)

The required L.C.M = 72 (a + 2) (a – 2) (a – 5) (a2 – 2a + 4)

Example 6

Find the L.C.M. of x3– 1, x4– 1,x4 + x2+ 1.

Solution:

Given x3– 1, x4– 1,x4 + x2+ 1

First expression

= x3 – 1

= (x)3 – (1)3

= (x – 1) (x2 + x + 1)

Second expression

= x4 – 1

= (x2)2 – (l)2

= (x2 + 1) (x2 – 1)

= (x2+1)(x+1)(x-1)

Third expression

= x4 + x2 + 1

= (x2)2 + 2.x2.1+ (1)2 -x?

= (x2 + 1)2 – (x)2

= (x2+1+x)(x2+ 1-x)

= (x2 + x + 1) (x2 – x + 1)

Hence, the required L.C.M.

= (x – 1) (x2 + x +1) (x +1) (x2 + 1) (x2 – X + 1)

= (x – 1) (x + 1) (x2 + 1) (x2 + x + 1) (x2 – x + 1)

The required L.C.M = (x – 1) (x + 1) (x2 + 1) (x2 + x + 1) (x2 – x + 1)

Practice Problems on LCM for Class 8

Example 7

Find the L.C.M. of x2 – y2, x3 – y3, 3X2 – 5x.y+ 2y2.

Solution :

Given x2 – y2, x3 – y3, 3X2 – 5x.y+ 2y2

First expression

= x2 – y2

= (x + y) (x – y)

Second expression

= x3 – y3

= (x – y) (x2 + xy +y2)

Third expression

= 3X2 – 5x2 + 2y2

= 3X2 – 3xy – 2x2 + 2y2

= 3x(x -y)~ 2y(x – y)

= (x-y) (3x – 2y)

Hence, the required L.C.M.

= (x+y) (x-y) (x2 + xy +y2) (3x- 2y)

= (x+y) (x-y)(3x- 2y)(x2 + xy +y2)

The required L.C.M = (x+y) (x-y)(3x- 2y)(x2 + xy +y2)

Example 8

Find the L.C.M. of x2 – y2 – z2 + 2yz, (x + y – z)2 and x2 + z2 – y2 + 2xz.

Solution :

Given x2 – y2 – z2 + 2yz, (x + y – z)2 And x2 + z2 – y2 + 2xz.

First expression = x2 – y2 – z2 + 2yz = x2 – (y2 – 2yz + z2)

= (x)2 -xy-z)2 = (x + y-z) (x-y +z)

Second expression

= (x + y – z)

Third expression

= x2 + z2 – y2 + 2xz

= x2 + 2xz + z2 -y2

= (x + z)2 – (y)2

= (x + z + y) (x + z – y)

= (x + y + z) (x – y + z)

Hence, the required L.C.M.

= (x + y- z)2 (x-y+z) (x+y+z)

= (x-y+z) (x+y+z) (x + y- z)2

The required L.C.M = (x-y+z) (x+y+z) (x + y- z)2

Example 9

Find the L.C.M. of x3 – 16x, 2x3 + 9x2 + 4x and x + 4.

Solution :

Given x3 – 16x, 2x3 + 9x2 + 4x And x + 4.

First expression

= x3 – 16x

= xfx2 – 16)

= x{(x)2 – (4)2}

= x(x + 4)(x – 4)

Second expression

= 2x3 + 9x2 + 4x

= x(2x2 + 9x + 4)

= x(2x2 + 8x + x + 4)

= x{2x(x + 4) + l(x + 4)}

= x(x + 4)(2x + 1)

Third expression

= x + 4

Hence, the required L.C.M.

= x(x + 4)(x – 4)(2x + 1)

The required L.C.M = x(x + 4)(x – 4)(2x + 1)

Example 10

Find the L.C.M. of a2 – 62 + c2 + 2ac, a2 – 62 – c2 + 26c and ab + ac + b2 – c2.

Solution :

Given a2 – 62 + c2 + 2ac, a2 – 62 – c2 + 26c And ab + ac + b2 – c2.

First expression

= a2 – b2 + c2 + 2ac

= a2 + 2ac + c2 – b2

= (a + c)2 – (b)2

= (a + c + b)(a + c – b)

= (a + b + c)(a – b + c)

Second expression

= a2 – b2 – c2 + 2bc

= a2 – (b2 – 2bc + c2)

= (a)2 – (b – c)2

= (a + b – c)(a – b + c)

Third expression

= ab + ac + b2 – c2

= a(b + c) + (b + c)(b – c)

= (b + c)(a + b – c)

Hence, the required L.C.M.

= (a + b + c)(a – b + c)(a + b – c)(b + c)

= (a + b + c)(a – b + c)(a + b – c)(b + c)

The required L.C.M = (a + b + c)(a – b + c)(a + b – c)(b + c)

Conceptual Questions on Applications of LCM

Example 11

Find the L.C.M. of x2 – xy – 2y2, 2x2 – 5xy + 2y2 and 2x2 + xy – y2.

Solution :

Given x2 – xy – 2y2, 2x2 – 5xy + 2y2 And 2x2 + xy – y2.

First expression

= x2 – xy – 2y2

= x2 – 2xy + xy – 2y2

= x(x – 2y) + y(x – 2y)

= (x- 2y)(x +y)

Second expression

= 2X2 – 5xy + 2y2

= 2x2 – 4xy – xy + 2y2

= 2x(x – 2y) – y(x – 2y)

= (x-2y)(2x-y)

Third expression

= 2x2 + xy – y2

= 2x2 + 2xy – xy -y2

= 2x(x + y) – y(x + y)

= (x + y)(2x – y)

Hence, the required L.C.M.

= (x – 2y)(x + y)(2x – y)

The required L.C.M = (x – 2y)(x + y)(2x – y)

Examples of Finding LCM Using Prime Factorization

Example 12

Find the L.C.M. of 3(x2 – 9), 9(x3 + 27) and 27(x2 – 3x + 9).

Solution :

Given 3(x2 – 9), 9(x3 + 27) and 27(x2 – 3x + 9)

First expression

= 3(x2 – 9)

= 3{(x)2 – (3)2}

= 3(x + 3)(x – 3)

Second expression

= 9(x3 + 27)

= 9{(x)3 + (3)3}

= 32 (x + 3)(x2 – 3x + 9)

Third expression

= 27(x2 – 3x + 9)

= 33 (x2 – 3x + 9)

Hence, the required L.C.M.

= 33 (x + 3)(x – 3)(x2 – 3x + 9)

= 27(x + 3)(x – 3)(x2 – 3x + 9)

The required L.C.M = 27(x + 3)(x – 3)(x2 – 3x + 9)

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