Factor completely. m^3 – 9m^2n + 18n^2m?

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2007-12-03 04:45:12 · 3 answers · asked by Boogie 2 in Science & Mathematics ➔ Mathematics

m^3 – 9m^2n + 18n^2m

You have to factorice m

= m(m^2 - 9mn + 18n^2)

= m(m - 3n)(m - 6n)

Because (-3n)x(-6n)=18n^2 and -3n-6n=-9n

2007-12-06 07:02:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋

m(m^2 - 9mn + 18n^2)
m(m - 6n)(m - 3n)

2007-12-03 12:51:10 · answer #2 · answered by Anonymous · 0⤊ 0⤋

= m(m^2 -9mn+18n^2)
= m(m -6n)(m-3n)

2007-12-03 12:49:32 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋