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consecutive integers in the form of quadratic equations

find the product of three consecutive integers whose product is 336
i did x (x+1) (x+2) =336
which is : x(x^+2x+x+2) =336
combine like terms..x(x^+3x+2) =336
then xcubed +2x^+x^+2x...combine like terms...xcubed +3x^ +2x=336
then bring the 336 to the left making the equation xcubed +3x^+2x-336=0
I have tried a million ways to factor this..grouping,trial and error..???

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Alvin S. | Information Technology Professional and Math & Statistics TutorInformation Technology Professional and ...
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x, x+1, x+2 are the consecutive integers product of which is equal to 336
x * x +1 * x + 2 = x (x^2 + 3x + 2) = x^3 + 3 x^2 + 2x = 336
x^3 + 3 x^2 + 2x - 336 = 0
From here you have to test potential roots of f(x) =  x^3 + 3 x^2 + 2x - 336.  You can use prime factorization to start.  Normal suspects of x = 2 or x = 3 did not work, so I expanded the options.
If f(root) = 0 then root is a solution.  In this situation f(6) = 0 which implies (x - 6) is a factor of x^3 + 3 x^2 + 2x - 336.  Using polynominal division (or synthetic division) we find the factors of x^3 + 3 x^2 + 2x - 336 are (x-6) and (x^2 +9x +56).   since the only integer solutions to (x-6) (x^2 +9x +56) = 0 is x = 6, we surmise the three consecutive integers are 6, 7 and 8 (which double checking is equal to 336)
Hope this helps.


I have no idea what you mean by the f root..what we are being taught is once we get to the polynominal which you and I agree is x^3+3x^2+2x-336=0 we have to factor something that when multiplied gives us the product 336 but when added gives us 2. how did you get the x-6...ect...?
Combined a couple of concepts to factor this unusual polynominal.   Normal factoring techniques will not get us to the promised land.
Tough to do in an email, but I'll give it a shot.
Assume f(x) = x^3+3x^2+2x-336
(1) The zeros of f(x) are the roots or solutions to the equation f(x) = 0.  Essentially the x-intercept.  In this case x = 6 is a zero of f(x).   Why? because f(6) = 0.  How did I get there?  I conducted a prime factorization and tested some factors.
(2) From the remainder theorem, if the polynominal f(x) is divided by x - c (where c is a zero of  f(x)), then the Remainder is f(c) which is turn equal to zero.  (We have found polynominal factors).   In our case, if I know x = 6 (or (x - 6)) is a factor of our f(x) then I know f(x) = (x-c) q(x) with no remainder.   In our problem f(x) = (x-6)(x^2+9x+56) where q(x) the other factor is equal to (x^2+9x+56).  How did I find q(x)?  I used polynomial division.
(3) so solving the rest of the problem is easier. Solving (x-6)(x^2+9x+56) = 0, gives us x = 6 and x = some non-integer solution.   If we have x = 6, that represents the first integer, thus x = 7 is the second integer, and 8 is the 3rd.
Makes sense soon as you said "unusual" I felt better..this was not a simple,typical one.