Find LCM of Polynomials...4

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feliz_nyc

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Here is my last problem of this type in my self-study of college algebra.

The following sample is my final even number problem from an online textbook I deleted.

Find the LCM of x^2 - 9 and x^2 - x - 12.

I know that x^2 - 9 = (x - 3)(x + 3).
I also know that x^2 - x - 12 factors out to be
(x - 3)(x + 4).

I now have the following list of factors:

(x - 3)(x + 3) and (x - 3)(x + 4)

My LCM has to be a set of factors that x^2 - 9 and x^2 - x - 12 can evenly be divided by.

I say the LCM = (x - 3)(x + 3)(x + 4).

You say?
 
feliz_nyc said:
Here is my last problem of this type in my self-study of college algebra.

The following sample is my final even number problem from an online textbook I deleted.

Find the LCM of x^2 - 9 and x^2 - x - 12.

I know that x^2 - 9 = (x - 3)(x + 3).
I also know that x^2 - x - 12 factors out to be
(x - 3)(x + 4).

I now have the following list of factors:

(x - 3)(x + 3) and (x - 3)(x + 4)

My LCM has to be a set of factors that x^2 - 9 and x^2 - x - 12 can evenly be divided by.

I say the LCM = (x - 3)(x + 3)(x + 4).

You say?
Click to expand...
Now you're getting it!

-Dan
 
Otis said:
There's a sign error in that factorization. Can you see what went wrong? If not, then multiply out (x-3)(x+4) and compare the resulting polynomial to the given polynomial.

[imath]\;[/imath]
Click to expand...
Yes, I see the sign error. The factors should be
(x - 4)(x + 3). So, I can now say that the actual answer for the LCM is (x - 3)(x + 3)(x - 4), which can be expressed as (x - 4)(x^2 - 9).
 
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