1. ## Multiplying trinomials

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

My answer: $n^3+4n^2-3n-18$
I went through a tedious solving form because of lack of knowledge.

2. ## Re: Multiplying trinomials

Originally Posted by vaironxxrd
I have to solve for : $(n+3)(n-2)(n+3)$

My answer: $n^3+4n^2-3n-18$
I went through a tedious solving form because of lack of knowledge.
First of all, this is the product of three BINOMIALS. Trinomials have three terms inside each bracket.

3. ## Re: Multiplying trinomials

Originally Posted by Prove It
First of all, this is the product of three BINOMIALS. Trinomials have three terms inside each bracket.

So the long process is needed?

4. ## Re: Multiplying trinomials

Originally Posted by vaironxxrd
So the long process is needed?
What do you mean with the 'long process'?

5. ## Re: Multiplying trinomials

Originally Posted by vaironxxrd
So the long process is needed?
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:

$(n+3)(n-2)(n+3) = (n+3)^2(n-2) = (n^2+6n+9)(n-2) = (n^3 +6n^2+9n) + (-2n^2-12n-18) = n^3+4n^2-3n-18$

6. ## Re: Multiplying trinomials

Originally Posted by Siron
What do you mean with the 'long process'?
Originally Posted by e^(i*pi)
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:

$(n+3)(n-2)(n+3) = (n+3)^2(n-2) = (n^2+6n+9)(n-2) = (n^3 +6n^2+9n) + (-2n^2-12n-18) = n^3+4n^2-3n-18$
well I multiplied the first two terms and then the product of them by the last term.

7. ## Re: Multiplying trinomials

Originally Posted by vaironxxrd
well I multiplied the first two terms and then the product of them by the last term.
That's pretty much the fastest way