Question 1153628
I can do each of the requirements on any given polynomial, but to determine the
probabilities is giving me issues. I could do all 1,458 problems, but I'd like
to know a better way, please.
The question is: 
Randomly construct a polynomial
{{{"f(x)" = a[4]x^4 + a[3]x^3 + a[2]x^2 + a[1]x + a[0]}}}
A) Each of the five coefficients is equally likely to be positive one, negative one, or zero.
B) What is the probability that f(x) is an odd function (i.e. f(−x) = −f(x))?
<pre>
That is when and only when all the terms with even powers of x are 0.

{{{a[4]=a[2]=a[0]=0}}}, it doesn't matter what the other a's are.

The probability of any one coefficient being 0 is 1/3.
So the probability of all three of those being 0 is
(1/3)³ = 1/3³ = 1/27
  
</pre>C) What is the probability that lim as x→∞ of f(x) = −∞?<pre>
That means the graph falls on the extreme right after the last turning point.
That happens when and only when the leading coefficient a<sub>4</sub> is negative,
or in this case, a<sub>4</sub> = -1.

That's 1 way out of 3, or a probability of 1/3.</pre>
D) What is the probability that x = 0 is a root of f(x)?<pre>
That happens when and only when the polynomial is divisible by x,
i.e., when x can be factored out and x can be set = 0 using the zero-factor
property.  That happens when and only when the constant term a<sub>0</sub> = 0.
The probability that a<sub>0</sub> can be zero is 1 time out of 3, or 1/3.</pre>
E) What is the probability that x = 1 is a root of f(x)?<pre>
That is when, if we substitute 1 for x we get 0.
{{{a[4](1)^4+a[3](1)^3+a[2](1)^2+a[1](1)+a[0] = 0}}}
or
{{{a[4]+a[3]+a[2]+a[1]+a[0]=0}}}

There are three cases when that happens.  It's when the five coefficients 
consist of

Case 1: all zeros.
Case 2: one 1, one -1, and three 0's.
Case 3: two 1's, two -1's and one 0.
 
Case 1. We choose all the a's all 0's
        That's 1 way for Case 1.

Case 2. We can choose the coefficient that are to be 1 in 5 ways
        We can then choose the coefficient that are to be -1 in 4 ways.
        The remaining three coefficients will be 0's.
        That' 5×4 = 20 ways

Case 3. We can select the two coefficients that are to be 1's in 5C2=10 ways.
        We can then select the two remaining coefficients to be -1's in 3C2=3 ways.
        The remaining coefficient will be 0.
        That's 10×3 = 30 ways

The total number of 'successful' ways is 1+20+30 = 51 ways.

The total number of 'possible' ways is 3<sup>5</sup> = 243 ways.

So the probability is 51 out of 243 ways or 51/243 which reduces to 17/41.</pre>
F) What is the probability that x = −1 is a root of f(x)?<pre>
If we substitute -x for x in f(x), we have

{{{"f(-x)"=a[4](-x)^4 + a[3](-x)^3 + a[2](-x)^2 + a[1](-x) + a[0]=
a[4]x^4 - a[3]x^3 + a[2]x^2 - a[1]x + a[0]}}}

then f(-x) has the same roots with the opposite signs.  So it will have
-1 as a root.  What that amounts to is changing the signs of the coefficients
of only the terms with odd powers of x.

So for each case in part E above for which f(x) has 1 as a solution, if we
change the signs of the coefficients of the odd powers of x in each one, it will
will produce a polynomial with -1 as a root.  They are in 1 to 1 correspondence.
So the answer to E is the same as the answer to D, which is 17/41.

Edwin</pre>