Question 977187
What are all the rational zeros of the function?

f(x) = 2x^4 – 13x^3 + 12x^2 + 52x – 80<pre>
It's in descending order, so look at the numbers on both ends,
the 80 on the far right and the 2 on the far left.  Ignore signs
for now. 

All the factors of 80 are 1,2,4,5,8,10,16,20,40,80.

All the factors of 2 are 1 and 2.

Make all possible fractions with a numerator as a factor of 80 and
a denominator as a factor of 2.  Some of the fractions will reduce
and become duplicates.

{{{1/1}}},{{{1/2}}},{{{2/1}}},{{{2/2}}},{{{4/1}}},{{{4/2}}},{{{5/1}}},{{{5/2}}},{{{8/1}}},{{{8/2}}},{{{10/1}}},{{{10/2}}},{{{16/1}}},{{{16/2}}},{{{20/1}}},{{{20/2}}},{{{40/1}}},{{{40/2}}},{{{80/1}}},{{{80/2}}}

Then we reduce the ones that will reduce and toss out the duplicates:

{{{1}}},{{{1/2}}},{{{2}}},{{{4}}},{{{5}}},{{{5/2}}},{{{8}}},{{{10}}},{{{16}}},{{{20}}},{{{40}}},{{{80}}},{{{40}}}

Finally we put a  before every one because they could be either positive or
negative.

{{{"" +-  1}}},{{{"" +-  1/2}}},{{{"" +-  2}}},{{{"" +-  4}}},{{{"" +-  5}}},{{{"" +-  5/2}}},{{{"" +-  8}}},{{{"" +-  10}}},{{{"" +-  16}}},{{{"" +-  20}}},{{{"" +-  40}}},{{{"" +-  80}}},{{{"" +-  40}}}

Now we have to use trial and error to find one of those, if there are any, that
will prove to be a rational zero.  We try 1, the easiest one:

1 | 2 -13  12 52 -80
  |<u>     2 -11  1  53</u> 
    2 -11   1 53 -27

So 1 is not a zero, because we get -27 remainder instead of 0


So we try the next easiest one, 2

2 | 2 -13  12  52 -80
  |<u>     4 -18 -12  80</u> 
    2  -9  -6  40   0

So 2 is a zero, and now we have factored f(x) as

f(x) = (x-2)(2x³-9x²-6x+40) 

Now we try the next easiest one, -1.  But we don't try it in the original
polynomial.  Instead we try it in the cubic factor 2x³-9x²-6x+40 that we
got from the synthetic division.


-1 | 2  -9  -6  40
   |<u>    -2  11  -5</u>   
     2 -11   5  35

So -1 is not a zero,


Now we try the next easiest one, -2

-2 | 2  -9  -6  40
   |<u>    -4  26 -40</u>   
     2 -13  20   0

So -2 is a zero, and now we have factored f(x) as

f(x) = (x-2)(x+2)(2x²-13x+20) 

We can now use regular factor methods to factor 2x²-13x+20
as (x-4)(x-5), and now we have factored the polynomial 
completely:

f(x) = (x-2)(x+2)(x-4)(2x-5)

The zeros all rational are found by setting each = 0 and solving.


x-2=0;  x+2=0;  x-4=0;  2x-5=0
  x=2;    x=-2;   x=4     2x=5
                           x={{{5/2}}}

Rational zeros: 2, -2, 4, and 5/2

Edwin</pre>



Edwin</pre>