Expert: Steve Holleran - 4/11/2007

Question
I'm lacking the common logical aspect to answer these questions:

#1 Using syntheic division, Solve the equation, given that a least on of the
solutions to each equation is an integer b/w -5 and 5.

6x^3+13x^2-4=0

#2 Let f(x) =x^4-1, g(x)=x^3-3x^2+5 and h(x)=4x^4-3x^2+3x-1. Find the following
function values by using two different methods: g(-1) and h(-1/2)


3. Find all possible rational zeros for each polynomial function.

P(x) = x^3-2x^2-5x+6

P(x) = 3x^3 +16x^2 -8


I hope this is not asking much...but Your help is greatly appreciated...  

Answer
Hi WMR,

1.  Okay, we need the Rational Root Theorem here.  It says that IF a polynomial has rational roots, they have to be of the form:  

(factors of constant) / (factors of lead coeff)

Here, the constant is 4 and the lead is 6, so the possibilities are:

+/- [1, 2, 4, 1/6, 1/3, 1/2, 2/3, 4/3].

But you are given that at least one of the zeros is an integer, so let's check the whole numbers first:

I found that -2 works:

  -2   |  6     13     0     -4
       |
       |       -12      -2    4
       --------------------------
          6      1      -2     0

Now you can either factor this to find the others, or try some of the other possibilities in synthetic division.

I factored it and got x = -2/3 and x = 1/2 as the others.


2.  To find function values, you can either:

A) Substitute the value into the function

B) Run the value through synthetic division and your result will be the remainder.

For g(-1), I would substitute:

g(-1) = (-1)^3 - 3(-1)^2 + 5

     = -1 - 3 + 5 = 1.

For h(-1/2) we can do synthetic:


  -1/2  |  4     0     -3     3     -1
        |
        |       -2      1     1     -2
        ------------------------------
           4    -2      -2    4      -3

So h(-1/2)= -3


3.   Here, you do the same Rational Root Theorem list that we did on #1:

P(x) = x^3 - 2x^2 - 5x + 6 has possibilities

+/- [1, 2, 3, 6]  (factors of 6 over factors of 1)



P(x) = 3x^3 + 16x^2 - 8 has possibilities

+/- [1, 2, 4, 8, 1/3, 2/3, 4/3, 8/3]
(factors of 8 over factors of 3)

I certainly hope this is clear, and that it helps you out.
Steve Holleran