How do you find the product of #(3n^2+2n-1)(2n^2+n+9)#?

1 Answer
Jul 3, 2016

Answer:

#(3n^2+2n-1)(2n^2+n+9)=6n^4+7n^3+27n^2+17n-9#

Explanation:

We use the distributive property for multiplication as follows:

#(3n^2+2n-1)(2n^2+n+9)#

= #3n^2(2n^2+n+9)+2n(2n^2+n+9)-1(2n^2+n+9)#

= #(6n^4+3n^3+27n^2)+(4n^3+2n^2+18n)-2n^2-n-9#

= #6n^4+(3+4)n^3+(27+2-2)n^2+(18-1)n-9#

= #6n^4+7n^3+27n^2+17n-9#