 
 
 LCD = (x-5)(x+1)2
LCD = (x-5)(x+1)2
 
 
 
 
 Now we analyze the polynomial
Now we analyze the polynomial  candidates for zeros are ±1, ±2, ±13, ±26
We begin by trying +1 using synthetic division with
the factor theorem:
1|1  5  20 -26
 |   1   6  26
  1  6  26   0
That was fortunate.  The very first candidate for a
zero succeeded!
So
candidates for zeros are ±1, ±2, ±13, ±26
We begin by trying +1 using synthetic division with
the factor theorem:
1|1  5  20 -26
 |   1   6  26
  1  6  26   0
That was fortunate.  The very first candidate for a
zero succeeded!
So  factors as
 factors as  The quadratic does not factor, so the final result is:
The quadratic does not factor, so the final result is:
 Edwin
Edwin