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

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Answer 1 · 2018-02-08T03:00:39

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

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.

#color(white)(................)# . . . . . . . . . .

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#