Factoring with 2 letters!

Q:  Factor completely:  9m³n – 19m²n² + 2mn³

A:  So, to factor… I think to myself: do they all have something in common?  Are they all divisible by 2 or 3? No.  None of the “numbers” have common factors, so that is out.  Well, they all have an “m” and they all have an “n”, so I will bring out one m and one n out of each term like so:

mn (9m² – 19mn + 2n²)

Now, the question is: can I factor the remaining (9m² – 19mn + 2n²)?  I think I can:

(9m² – 19mn + 2n²)

(……….)(……….)

(3m – 2n)(3m -n)?? If I FOIL (or distribute this back out), I get: 9m² – 9mn + 2n².. Not quite rightTry again.  Remember, as I said before.. Factoring involves a bit of guessing and good luck:

(m – 2n)(9m -n)…?  FOIL this out to get: 9m² – 19mn + 2n²…. So, it worked.  Second time’s a charm.

Therefore:

9m³n – 19m²n² + 2mn³ = mn (9m² – 19mn + 2n²) = nm(m – 2n)(9m -n).

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