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
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))?
That is when and only when all the terms with even powers of x are 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
C) What is the probability that lim as x→∞ of f(x) = −∞?
That means the graph falls on the extreme right after the last turning point.
That happens when and only when the leading coefficient a4 is negative,
or in this case, a4 = -1.
That's 1 way out of 3, or a probability of 1/3.
D) What is the probability that x = 0 is a root of f(x)?
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 a0 = 0.
The probability that a0 can be zero is 1 time out of 3, or 1/3.
E) What is the probability that x = 1 is a root of f(x)?
That is when, if we substitute 1 for x we get 0.
or
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 35 = 243 ways.
So the probability is 51 out of 243 ways or 51/243 which reduces to 17/41.
F) What is the probability that x = −1 is a root of f(x)?
If we substitute -x for x in f(x), we have
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
