# How do you find the LCM for d^2+6d+9 and 2(d^2-9)?

Jun 2, 2015

${d}^{2} + 6 d + 9 = {\left(d + 3\right)}^{2}$

$2 \left({d}^{2} - 9\right) = 2 \left({d}^{2} - {3}^{2}\right) = 2 \left(d - 3\right) \left(d + 3\right)$

$2 \left(d - 3\right) {\left(d + 3\right)}^{2}$ will be divisible by both of these as it includes all the factors of each in the multiplicity that they occur in each. It is the simplest polynomial that does and therefore is the LCM.

$2 \left(d - 3\right) {\left(d + 3\right)}^{2}$

$= 2 \left({d}^{2} - 9\right) \left(d + 3\right)$

$= 2 \left({d}^{3} + 3 {d}^{2} - 9 d - 27\right)$

$= 2 {d}^{3} + 6 {d}^{2} - 18 d - 54$