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Use the rational roots theorem to solve:
The last term is -6, which in absolute value is 6, and which
has these factors 1,2,3,6
The leading term (the term with largest exponent) is , has
coefficient 1, which in absolute value is 1, and which
has only the one factor 1.
Now we form all the fractions with numerator 1,2,3,or 6 and
denominator 1
These are , , , or
, , , .
Their negatives are also possible rational roots, so all the
possible rational roots are:
±, ±, ±, ±
We start out by trying using synthetic division to
see if we get a 0 remainder:
1| 1 -5 5 5 -6
|**__ 1 -4 1 6__
1 -4 1 6 0
Yes we do get 0 remainder, so we know that we have factored
the polynomial
as
So now we can just find the roots of the simpler polynomial:
The first and last numbers happen to be the same as they were
in the original, so we can try the same ones again. We try 1
again:
1| 1 -4 1 6
|__ 1 -3 -2__
1 -3 -2 4
No that leaves a remainder of 4, not 0.
So we try -1
-1| 1 -4 1 6
|__ -1 6 -6__
1 -5 6 0
Yes we do get 0 remainder, so we know that we have factored
the polynomial again.
First we factored
as
Now we have factored the polynomial in the
second parentheses, and we have:
So now we can just find the roots of the simpler polynomial:
But we don't need to do synthetic division again, for
factors as
So now we have factored completely:
Set each factor equal to 0 and so the roots are
, , ,
Edwin

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