Question

Find the LCM of each set of polynomials. If needed, show work and explain.
19. x^2 - y^2, x^3 + x^2 y <--- What is it not x^2 (x + y) (x + y) (x - y) for the answer???
20. 2t^2 + t - 3, 2t^2 + 5t + 3

Answer 1

So we have two separate equations.\[\large x^2-y^2\]This is the `Difference of Squares` which can be factored into conjugates.\[\large (x-y)\color{orangered}{(x+y)}\]

If we look at the other equation,\[\large x^3+x^2y\]Factoring an x^2 out of each term gives us,\[\large x^2\color{orangered}{(x+y)}\]

What factor do these have in common? Hint, I colored them orange :) lol

Answer 2

(x + y)

Answer 3

yayyyy

Answer 4

So

Answer 5

Oh sorry, you're not looking for a common factor, you're looking for the least common multiple. Ok ok my mistake.

So since they both already have the (x+y) factor, they only need the others. We don't need 2 copies of (x+y).

LCM: \(\large x^2(x-y)(x+y)\)

Answer 6

I understand, because they both have an x + y in common, so there's no need to write it twice

Answer 7

yes good c:

Answer 8

so that's why we just write it once

Answer 9

now onto to 20.

Answer 10

Do you understand how to factor the first polynomial in 20?

Answer 11

it's (t - 1) (2t + 3) and the 2nd polynomial is (t + t) (2t+ 3). Is the answer
(2t + 3) (t - 1) (t + 1)

Answer 12

Yes! Good job.

Answer 13

Are you sure it's the answer

Answer 14

@Mertsj I have the answer for 21., and I want to know if I am right?
Problem: x^2 - 7n + 12, n^2 - 2n - 8

Answer 15

My answer is: (x - 4) (n - 3) (n + 2)

Answer 16

Are you supposed to have two different variables?

Answer 17

That x is suppose to an n

Answer 18

Yes it is correct.

Answer 19

I am on a roll today man! :D

Answer 20

I'll say!!

Answer 21

At first I was shaking on this, but now I get it

Answer 22

Excellent

Answer 23

I know right! :D

Answer 24

I am glad I have you and some others for Math help! :D

Answer 25

Me too.

Answer 26

Really! :(