How do you multiply $(2m^2 + n)(3n^2 + 6mn - m^2)$ ?

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Answer 1 · 2015-07-14T01:29:43

$(2m^2 + n)(3n^2 + 6mn -m^2) = – 2 m^4 + 12 m^3n + 6m^2n^2 – m^2n + 6mn^2 + 3n^3$

We use the distributive property of multiplication.

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

$= (6m^2n^2 + 12 m^3n – 2 m^4) + (3n^3 + 6mn^2 – m^2n)$

Now we remove parentheses and arrange the terms in descending powers of $m$ and ascending powers of $n$ .

$(2m^2 + n)(3n^2 + 6mn -m^2) = – 2 m^4 + 12 m^3n + 6m^2n^2 –m^2n + 6mn^2 + 3n^3$

How do you multiply (2m^2 + n)(3n^2 + 6mn - m^2) (2m^2 + n)(3n^2 + 6mn - m^2)?