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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Originally Posted by vaironxxrd 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. First of all, this is the product of three BINOMIALS. Trinomials have three terms inside each bracket. Your answer is correct.
Originally Posted by Prove It First of all, this is the product of three BINOMIALS. Trinomials have three terms inside each bracket. Your answer is correct. So the long process is needed?
Originally Posted by vaironxxrd So the long process is needed? What do you mean with the 'long process'?
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: $\displaystyle (n+3)(n-2)(n+3) = (n+3)^2(n-2) = (n^2+6n+9)(n-2) = (n^3 +6n^2+9n) + (-2n^2-12n-18) = n^3+4n^2-3n-18$
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: $\displaystyle (n+3)(n-2)(n+3) = (n+3)^2(n-2) = (n^2+6n+9)(n-2) = (n^3 +6n^2+9n) + (-2n^2-12n-18) = n^3+4n^2-3n-18$ well I multiplied the first two terms and then the product of them by the last term.
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
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