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

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Answer 1 · 2016-07-03T16:07:15

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

We use the distributive property for multiplication as follows:

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

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

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

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

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

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