# How do you find the LCM for x^2-x-6 and x^2+6x+8?

Jun 1, 2015

First factor the starting polynomials:

${x}^{2} - x - 6 = \left(x - 3\right) \left(x + 2\right)$

${x}^{2} + 6 x + 8 = \left(x + 4\right) \left(x + 2\right)$

The LCM is a polynomial which is a multiple of both of the starting polynomials. So for each linear factor we have found, it must occur as many times as in each of the starting polynomials. But you do not need to include it more times than it occurs in either.

So since $\left(x + 2\right)$ occurs once in each of our starting polynomials, it should occur once in the LCM.

$L C M \left({x}^{2} - x - 6 , {x}^{2} + 6 x + 8\right)$

$= \left(x - 3\right) \left(x + 2\right) \left(x + 4\right)$

$= \left({x}^{2} - x - 6\right) \left(x + 4\right)$

$= {x}^{3} + 3 {x}^{2} - 10 x - 24$