SOLUTION: Use the rational roots theorem to solve: p(x)=x^4-5x^3+5x^2+5x-6

Algebra ->  Graphs -> SOLUTION: Use the rational roots theorem to solve: p(x)=x^4-5x^3+5x^2+5x-6      Log On


   



Question 148851: Use the rational roots theorem to solve:
p(x)=x^4-5x^3+5x^2+5x-6

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

Use the rational roots theorem to solve:
p%28x%29=x%5E4-5x%5E3%2B5x%5E2%2B5x-6

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 x%5E4, 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 1%2F1, 2%2F1, 3%2F1, 6%2F1 or

          1, 2, 3, 6.

Their negatives are also possible rational roots, so all the
possible rational roots are:

          ±1, ±2, ±3, ±6 

We start out by trying 1 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

  p%28x%29=x%5E4-5x%5E3%2B5x%5E2%2B5x-6

   as

p%28x%29=%28x-1%29%28x%5E3-4x%5E2%2B1x%2B6%29

So now we can just find the roots of the simpler polynomial:

     x%5E3-4x%5E2%2B1x%2B6

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

  p%28x%29=x%5E4-5x%5E3%2B5x%5E2%2B5x-6

   as

  p%28x%29=%28x-1%29%28x%5E3-4x%5E2%2B1x%2B6%29

Now we have factored the polynomial in the 
second parentheses, and we have:

p%28x%29=%28x-1%29%28x%2B1%29%28x%5E2-5x%2B6%29

So now we can just find the roots of the simpler polynomial:

     x%5E2-5x%2B6

But we don't need to do synthetic division again, for
x%5E2-5x%2B6 factors as %28x-2%29%28x-3%29

So now we have factored p%28x%29 completely:

p%28x%29=%28x-1%29%28x%2B1%29%28x-2%29%28x-3%29

Set each factor equal to 0 and so the roots are 

1, -1, 2, 3

Edwin