Expert: Sherry Wallin - 12/5/2009

Question
I have been presented with the problem x^3 + 6x^2y + 12xy^2 + 8y^3. I feel like the answer is right in front of my face, but I just can't figure it out! Help would be much appreciated!

Answer
This is not a problem, it is an expression. You need to tell me what it is they want you to do. Is it to factor the expression? This is what I suspect, but I am not a mind reader :). In the event I have guessed correctly then this factors as (x+2y)^3. And you should be asking me how do I know this?
Have you ever heard of Pascal's triangle? If not, try googling it, it is easy to make and it is relevant to where all the coefficients come from. This is also part of the binomial theorem. You know that (a+b)^1 = a + b and (a+b)^2 = a^2+2ab+b^2 and (a+b)^3 = a^3 +a^2b+ab^2+b^3. You can check this by multiplying (a+b)(a+b)(a+b) which is exactly what (a+b)^3 is. So what you have above is a slight variation of (x+y)^3. I'm not sure what level of mathematics you are in and if you have studied combinations or not or factorials but the coefficients come from combinations. Since it is a 3rd degree polynomial the first coefficient is 3C0 = 3!/3!0!, the 2nd coefficient is 3C1 = 3!/2!1!, the 3rd coefficient is 3C2 = 3!/1!2! and the last coefficient is 3C3 = 3!/3!0! which means the coefficients are 1,3,3,1 respectively. The coefficient on x^3 clearly the 1 matches, but how do we get a 3 on the 6x^2y term? It must be multiplied by 2. How do we get the 12xy^2 term when all we have is a 3? It must be multiplied by a 4 or 2^2. And finally how do we get 8y^3 when we have a 1, we multiply by 8 or 2^3. Every one of these last 3 coefficients have a factor of 2 in them so try (x+2y)^3 and see what you get:
(x+2y)(x+2y) = x^2 +4xy + 4y^2 & (x+2y)(x^2 + 4xy + 4y^2) = x(x^2 + 4xy + 4y^2) +2y(x^2 + 4xy + 4y^2)
= x^3 +4x^2y + 4xy^2 + 2x^2y + 8xy^2 + 8y^3 = x^3 + 6x^2y + 12xy^2 + 8y^3 what you wanted to factor.

Math Prof