I have to solve for : $\displaystyle (n+3)(n-2)(n+3)$

My answer: $\displaystyle n^3+4n^2-3n-18$

I went through a tedious solving form because of lack of knowledge.

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- Oct 1st 2011, 10:48 PMvaironxxrdMultiplying trinomials
I have to solve for : $\displaystyle (n+3)(n-2)(n+3)$

My answer: $\displaystyle n^3+4n^2-3n-18$

I went through a tedious solving form because of lack of knowledge. - Oct 1st 2011, 10:51 PMProve ItRe: Multiplying trinomials
- Oct 1st 2011, 11:06 PMvaironxxrdRe: Multiplying trinomials
- Oct 1st 2011, 11:45 PMSironRe: Multiplying trinomials
- Oct 2nd 2011, 02:47 AMe^(i*pi)Re: Multiplying trinomials
I'm not sure what you mean by the long process but the easiest way is to multiply two of them and then their product by the third. In your example:

$\displaystyle (n+3)(n-2)(n+3) = (n+3)^2(n-2) = (n^2+6n+9)(n-2) = (n^3 +6n^2+9n) + (-2n^2-12n-18) = n^3+4n^2-3n-18$ - Oct 2nd 2011, 06:47 AMvaironxxrdRe: Multiplying trinomials
- Oct 2nd 2011, 07:05 AMe^(i*pi)Re: Multiplying trinomials