# How do you find the product 4n(2n^3p^2-3np^2+5n)+4p(6n^2p-2np^2+3p)?

Feb 8, 2018

8n^4 p^2−12n^2p^2+20n^2+24n^2p^2−8np^3+12p^2

#### Explanation:

Simplify
4n (2n^3p^2−3np^2+5n)+4p (6n^2p−2np^2+3p)

Clear the parentheses by distributing the coefficients

8n^4 p^2−12n^2p^2+20n^2+24n^2p^2−8np^3+12p^2

There are no like terms, so this is as far as you can go.

You can factor by grouping, but if you do that, there was no point in distributing in the first place.

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Maybe you consider this step as a simplification:

Factor out an additional $n$ from the first group, and an additional $p$ from the second group

Simplify
4n (2n^3p^2−3np^2+5n)+4p (6n^2p−2np^2+3p)

Factor out one more $n$ and one more $p$

Simplify
4n^2 (2n^2p^2−3p^2+5)+4p^2 (6n^2−2np+3)

Distributing these coefficients brings back the first answer, which was:
8n^4 p^2−12n^2p^2+20n^2+24n^2p^2−8np^3+12p^2