Question

Find the LCM of the following polynomials: a8 - 18 and (09 - 64)(a + b)​

Answer 1

Answer:Second expression = a3 - a

                         = a(a2 - 1), by taking common ‘a’

                         = a(a2 – 12), by using the formula of a2 – b2

                         = a(a + 1) (a - 1), we know a2 – b2 = (a + b) (a – b)

The common factors of the two expressions are ‘a’ and (a + 1); (a - 1) is the extra factor in the second expression.

Therefore, the required L.C.M. of a2 + a and a3 – a is a(a + 1) (a - 1)

2. Find out the L.C.M of x2 - 4 and x2+ 2x by factorization.

Solution:

First expression = x2 - 4

                     = x2 - 22, by using the formula of a2 – b2

                     = (x + 2) (x - 2), we know a2 – b2 = (a + b) (a – b)

Second expression = x2 + 2x

                         = x(x + 2), by taking common ‘x’    

The common factor of the two expressions is ‘(x + 2)’.

The extra common factor in the first expression is (x - 2) and in the second expression is x.

Therefore, the required L.C.M = (x + 2) × (x - 2) × x

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

3. Find out the L.C.M of x3 + 2x2 and x3 + 3x2 + 2x by factorization.

Solution:

First expression = x3 + 2x2

                     = x2(x + 2), by taking common ‘x2’

                     = x × x × (x + 2)

Second expression = x3 + 3x2 + 2x

                         = x(x2 + 3x + 2), by taking common ‘x’

                         = x(x2 + 2x + x + 2), by splitting the middle term 3x = 2x + x

                         = x[x(x + 2) + 1(x + 2)]

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

                         = x × (x + 2) × (x + 1)

In both the expressions, the common factors are ‘x’ and ‘(x + 2)’; the extra common factors are ‘x’ in the first expression and ‘(x + 1)’ in the second expression.

Therefore, the required L.C.M. = x × (x + 2) × x × (x + 1)

                                        = x2(x + 1) (x + 2)

Step-by-step explanation: