# Finding a least common multiple (LCM)

My Algebra 2 book explains how to find a least common multiple:

Find the least common multiple of $4x^2 - 16$ and $6x^2 - 24x + 24$.

Solution

Step 1 Factor each polynomial. Write numerical factors as products of primes.

$4x^2 - 16 = 4(x^2 - 4) = (2^2)(x + 2)(x - 2)$ $6x^2 - 24x + 24 = 6(x^2 - 4x + 4) = (2)(3)(x-2)^2$

Step 2 Form the LCM by writing each factor to the highest power it occurs in either polynomial.

LCM = $(2^2)(3)(x + 2)(x - 2)^2 = 12(x + 2)(x - 2)^2$

I don't understand their wording, and I don't want to go onto the rest of my assignment that includes finding the least common denominators until I know how to do it correctly, instead of going back and doing it over when I find out I'm doing it wrong.

• Which step do you not understand? 1 or 2? Or both?
– J.R.
Jan 22 '14 at 23:01
• I see that they factored them. So you put the regular numbers as products of primes (the $2^2$), and then... you have the $2^2$ and the normal 2, so you just put the highest exponent one? Jan 22 '14 at 23:04
• (Along with the rest of the factors from everything) Jan 22 '14 at 23:05
• And also, if I had one with $24x^2$ and $8x^2 - 16x$, are the "prodcuts of primes" for 24 "$2^3$ and 3"? Jan 22 '14 at 23:07
• you had used the intuitions that one receives from the use of integers Jan 22 '14 at 23:15

It's kind of a weird way of saying it. $$\mathrm{lcm}(P(x),Q(x)) = \frac{P(x)\cdot Q(x)}{\gcd(P(x),Q(x))}$$
Where $$\gcd(P(x),Q(x))$$ is just the product of the terms that appear in both factorizations.
For instance $$\mathrm{lcm}(((x+1)(x+1)(x-1)),((x+1)(x-1)(x-1)))=\frac{(x+1)^3(x-1)^3}{(x+1)(x-1)}=(x+1)^2(x-1)^2$$
It works just like with integers. If you want to find the LCM of $84=2^2\cdot 3 \cdot 7$ and $90=2\cdot 3^2 \cdot 5$ you collect the highest power of each prime, getting $2^2 \cdot 3^2 \cdot 5 \cdot 7 = 1260$ The LCM has to incorporate all the polynomial factors to the highest power in any of the things you are taking the LCM of.