# How do you find the LCM of x^8- 12x^7 +36x^6, 3x^2- 108 and 7x+ 42?

Dec 26, 2016

The LCM is:

$21 {x}^{9} - 126 {x}^{8} - 756 {x}^{7} + 4536 {x}^{6}$

#### Explanation:

For this question, it's probably easiest to factor all of the polynomials first:

${x}^{8} - 12 {x}^{7} + 36 {x}^{6} = {x}^{6} \left({x}^{2} - 12 x + 36\right) = {x}^{6} {\left(x - 6\right)}^{2}$

$3 {x}^{2} - 108 = 3 \left({x}^{2} - 36\right) = 3 \left(x - 6\right) \left(x + 6\right)$

$7 x + 42 = 7 \left(x + 6\right)$

So the LCM of the scalar factors is that of $1$, $3$ and $7$, which is $21$

The simplest product of polynomial factors including all of the linear factors we have found, in their multiplicities is:

${x}^{6} {\left(x - 6\right)}^{2} \left(x + 6\right) = {x}^{6} \left(x - 6\right) \left({x}^{2} - 36\right)$

$\textcolor{w h i t e}{{x}^{6} {\left(x - 6\right)}^{2} \left(x + 6\right)} = {x}^{6} \left({x}^{3} - 6 {x}^{2} - 36 x + 216\right)$

$\textcolor{w h i t e}{{x}^{6} {\left(x - 6\right)}^{2} \left(x + 6\right)} = {x}^{9} - 6 {x}^{8} - 36 {x}^{7} + 216 {x}^{6}$

So to get the LCM of the original polynomials, we just need to multiply this by $21$:

$21 \left({x}^{9} - 6 {x}^{8} - 36 {x}^{7} + 216 {x}^{6}\right) = 21 {x}^{9} - 126 {x}^{8} - 756 {x}^{7} + 4536 {x}^{6}$