First of all you don't have all the candidates for rational roots,
for  and  are also candidates for rational roots.

You can use DesCartes' rule of signs.  There is only one sign change
from left to right, so you know there is one positive root.  You might
try to find it, since it might be rational (though it might not be!).

Try the easiest, 1:

1 | 4   8  -1  -14 -24
  |     4  12   11  -3 
    4  12  11   -3 -27 

Nope!

Try the next easiest, 2:

2 | 4   8  -1  -14 -24
  |     8  32   62  96 
    4  16  31   48  72

Now we know there is no need to try any
larger number than 2, because all the numbers
on the bottom are positive.  That means
2 is an UPPER BOUND to the positive roots.

Also since the remainder was negative (-27) when
we tried 1 and positive (72) when we tried 2, that
the positive root is between those two.  The
only candidate for a rational root between those
is 3/2, which you left out!!!

3/2 | 4   8  -1  -14 -24
    |     6  21   30  24 
      4  14  20   16   0

Hurray! We have found a root , and x- is a factor

So now you have factored the original polynomial as

(x - )(4x³ + 14x² + 20x +16)

and you can factor a 2 out of that cubic polynomial

(x - )2(2x³ + 7x² + 10x + 8)

or

2(x - )(2x³ + 7x² + 10x + 8)

So now we are working with only the cubic polynomial:

2x³ + 7x² + 10x + 8

It has no sign changes so it has no positive roots.

So its candidates for rational roots are only negatives:

-1/2, -1, -2, -3, -4, -8

We try the easiest, -1

-1 | 2  7  10  8
   |   -2  -5 -5 
     2  5   5  3

Nope!

[Note: the rule for LOWER bounds on negative roots is 
not the same as the rule for UPPER bounds on positive
roots, for the signs on the bottom have to ALTERNATE
in order to know that we have a LOWER bound.]

We try the next easiest, -2

-2 | 2  7  10  8
   |   -4  -6 -8 
     2  3   4  0

Hurray! We have found another root -2, and x+2 is a factor

So we have now factored the original polynomial as

2(x - )(x + 2)(2x² + 3x + 4)

The quadratic will not factor so we have to use the
quadratic formula to find the other two roots, by
setting the quadratic equal to zero:

2x² + 3x + 4 = 0











So the roots are

, -2, , 

Are you allowed to use a TI-83 or TI-84 calculator on tests?
If so you can shorten your work by not having to try so many
numbers as candidates for rational roots.

Edwin