Question

5. Find the LCM of ² - 81 and ² - 12x + 27. (x + 9) (x − 9) (x+3) (x - 9) (x − 3) (x +9) (x 9) (x − 3) (x - 9)​

Answer 1

The LCM of
`(x^2 -81)` and
`(x^2 -12x+27)` is (x+9)(x−9)(x−3), which matches option 4: (x+9)(x−9)(x−3).

To find the LCM (Least Common Multiple) of the polynomials
`x^2 -81` and
`x^2 -12x+27`, let's factorize these polynomials first:

`x^2 -81` can be factorized using the difference of squares formula:

`x^2 -81=(x+9)(x-9).`

`x^2 -12x+27` factors as a perfect square trinomial:

`x^2 -12x+27=(x-3)(x-9).`

Now, to find the LCM, we need to consider each factor and take the maximum power for each unique factor.

The factors in
`x^2 -81` are (x+9) and (x−9).

The factors in
`x^2 -12x+27` are (x−3) and (x−9).

The LCM will contain each unique factor raised to its highest power:

x+9 appears in both factorizations, so its highest power is x+9.

x−9 appears in both factorizations, so its highest power is x−9.

x−3 appears only in the factorization of
`x^2 -12x+27`, so it remains as x−3.

Therefore, the LCM of
`(x^2 -81)` and
`(x^2 -12x+27)` is (x+9)(x−9)(x−3), which matches option 4: (x+9)(x−9)(x−3).

Question

Find the LCM of (x^2-81) and(x^2-12x+27)

1. (x - 9)​

2. (x − 3) (x -9)

3. (x + 9) (x − 9) (x+3)

4. (x + 9) (x − 9) (x-3)