consecutive integers in the form of quadratic equations

Question

hi..

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..???

help!

Answer 1

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.

 

Regards,

 

Alvin