Question

I'm confused how to find the least common multiple of x^2-4x, x^2+4x. I know they factor out to x(x-4) and x(x+4) but dont know how to go any further

Answer 1

do you know how to find the gcd of the two functions?

Answer 2

the gcd is the terms that are common between the two functions

Answer 3

in this case, x

Answer 4

Yes, i got that far.

Answer 5

to find the lcm, multiple the functions together and divide by the gcd

Answer 6

*multiply

Answer 7

so the lcm would be x(x-4)x(x+4)/x, which is equal to x(x-4)(x+4)

Answer 8

ok that makes no sense lol

Answer 9

Maybe you can clarify it a little better for me, looking at a problem x^2-6x, x^2 i know the lcm is x^2(x-15) how is this? i know this is correct bc of my text.

Answer 10

i think your text has a typo. the answer should be x(x^2-16)

Answer 11

hmmmmm

Answer 12

thanks!

Answer 13

:)

Answer 14

lcm of x^2-6x and x^2 should be x^2(x-6)

x^2-6x = x(x-6)
x^2 = x.x

gcd = x

carry out the method suggensted by eashy :

LCM = x(x-6)*x^2/x = x^2(x-6)

Answer 15

one idea is to spose you are try to add fractions, with those as denominators:

\[\frac{1}{x^2-4x}+\frac{1}{x^2+4x}=n\]
factor them
\[\frac{1}{x(x-4)}+\frac{1}{x(x+4)}=n\]
start clearing out the fractions by multiplying thru by something useful, like an x, and simplifying the results
\[\frac{\cancel x}{\cancel x~(x-4)}+\frac{\cancel x}{\cancel x~(x+4)}=x~n\]
\[\frac{\cancel{(x-4)}}{\cancel{(x-4)}}+\frac{(x-4)}{(x+4)}=x(x-4)~n\]
\[\frac{(x-4)\cancel{(x+4)}}{\cancel{(x+4)}}=x(x-4)(x+4)~n\]
the LCM gets created on the right hand side next to that "n" once you have cleared all the fractions