SOLUTION: I'm having trouble with "finding least common multiples." I know the first step is to find the prime factors of each expression. Which my expression is:"9(x+2)(2x-1) and 3(x+2)" wh

Algebra ->  Rational-functions -> SOLUTION: I'm having trouble with "finding least common multiples." I know the first step is to find the prime factors of each expression. Which my expression is:"9(x+2)(2x-1) and 3(x+2)" wh      Log On


   

Question 987411: I'm having trouble with "finding least common multiples." I know the first step is to find the prime factors of each expression. Which my expression is:"9(x+2)(2x-1) and 3(x+2)" which I think would end up looking like "(3)(3)(2)(x+2)(2x-1) and (3)(x+2)" the next part is where the first and second expressions are combined... this is what I have trouble with. My paper says, "Step 2: Write each prime factor the greatest amount of times it appears in either expression. Simplify where possible." And then it shows a bunch of work and the least common multiple is stated. I'm soooo confused. This is the work of the example I was given: "4x^2-36=4(x^2-9)=(2)(2)(x-3)(x+3)" and "6x^2 +36x +54=6(x^2 +6x +9)=(2)(3)(x+3)(x+3)" then after step 2: "(2)(2)(3)(x-3)(x+3)=12(x-3)(x+3)^2" What it equals is the least common muliple
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
The instructions you have are plenty clear. Whether as expressions or numbers of a base ten form, the method would be as you already learned by about 6th or 8th grade. 

NUMBER           FACTORS
9(x+2)(2x-1)      9, x+2, 2x-1
3(x+2)            3, x+2
All exponents of each factor is 1.

Each factor in the second product is already present in the first product.
The second product needs another factor 3 and a factor 2x-1.

Common multiple of the two numbers needs to be 9%28x%2B2%29%282x-1%29. You could multiply this if you want.