SOLUTION: The sum of all integral values of y for which 18(y^2-7y+10)/(y^3-6y^2+3y+10) has an integral value is... A)-17 B)-19 C)-12 D)-14 E)-20

Algebra ->  Exponents -> SOLUTION: The sum of all integral values of y for which 18(y^2-7y+10)/(y^3-6y^2+3y+10) has an integral value is... A)-17 B)-19 C)-12 D)-14 E)-20      Log On


   

Question 1132671: The sum of all integral values of y for which 18(y^2-7y+10)/(y^3-6y^2+3y+10) has an integral value is...
A)-17 B)-19 C)-12 D)-14 E)-20
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

%2818%2A%28y%5E2-7%2Ay%2B10%29%29%2F%28y%5E3-6%2Ay%5E2%2B3%2Ay%2B10%29

First we factor the numerator as 18(y-2)(y-5)

Factor the denominator by the ±p/q method

Try 2 as as a zero of the denominator, using
synthetic division:

2 | 1 -6  3  10
  |    2 -8 -10
    1 -4 -5   0

So the denominator factors as (y-2)(y²-4y-5) which
factors completely as (y-2)(y-5)(y+1)

So the expression becomes:

%2818%28y-2%29%28y-5%29%29%2F%28%28y-2%29%28y-5%29%28y%2B1%29%29

We cancel, but observe that y cannot be 2,5,-1,
for they would make the denominator 0.



18%2F%28y%2B1%29

For that to be an integer, the denominator must be a factor
of the numerator 18.

The integer factors of 18 are ±1,±2,±3,±6,±9,±18

So the denominator y+1 must be one of those in order
for the expression to be an integer:

We take positive and negative values separately:

The positive values: 
y+1 = 1, 2, 3, 6, 9, 18
  y = 0, 1, 2, 5, 8, 17  

But as we said above, we must eliminate 2 and 5 from the 
list of permissible values of y, so when y is positive, 
it must be one of these values:

y = 0, 1, 8, 17 

The negative values: 

y+1 = -1,-2,-3,-6, -9,-18
  y = -2,-3,-4,-7,-10,-19  

All those are permissible values of y.

So all values of y which cause the original
expression to be an integer are

0, 1, 8, 17,-2,-3,-4,-7,-10,-19   

Sum = 0+1+8+17-2-3-4-7-10-19 = -19

Edwin