Question

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

Answer 1

`(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`