**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 right**. **Try 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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