document.write( "Question 1153628: I can do each of the requirements on any given polynomial, but to determine the
\n" ); document.write( "probabilities is giving me issues. I could do all 1,458 problems, but I'd like
\n" ); document.write( "to know a better way, please.
\n" ); document.write( "The question is:
\n" ); document.write( "Randomly construct a polynomial
\n" ); document.write( "f(x) = a4x^4 + a3x^3 + a2x^2 + a1x + a0
\n" ); document.write( "A) Each of the five coefficients is equally likely to be positive one, negative one, or zero.
\n" ); document.write( "B) What is the probability that f(x) is an odd function (i.e. f(−x) = −f(x))?
\n" ); document.write( "C) What is the probability that lim as x→∞ of f(x) = −∞?
\n" ); document.write( "D) What is the probability that x = 0 is a root of f(x)?
\n" ); document.write( "E) What is the probability that x = 1 is a root of f(x)?
\n" ); document.write( "F) What is the probability that x = −1 is a root of f(x)? \n" ); document.write( "
Algebra.Com's Answer #776436 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
I can do each of the requirements on any given polynomial, but to determine the
\n" ); document.write( "probabilities is giving me issues. I could do all 1,458 problems, but I'd like
\n" ); document.write( "to know a better way, please.
\n" ); document.write( "The question is:
\n" ); document.write( "Randomly construct a polynomial
\n" ); document.write( "
\n" ); document.write( "A) Each of the five coefficients is equally likely to be positive one, negative one, or zero.
\n" ); document.write( "B) What is the probability that f(x) is an odd function (i.e. f(−x) = −f(x))?
\n" ); document.write( "
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document.write( "That is when and only when all the terms with even powers of x are 0.\r\n" );
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document.write( "\"a%5B4%5D=a%5B2%5D=a%5B0%5D=0\", it doesn't matter what the other a's are.\r\n" );
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document.write( "The probability of any one coefficient being 0 is 1/3.\r\n" );
document.write( "So the probability of all three of those being 0 is\r\n" );
document.write( "(1/3)³ = 1/3³ = 1/27\r\n" );
document.write( "  \r\n" );
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C) What is the probability that lim as x→∞ of f(x) = −∞?
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document.write( "That means the graph falls on the extreme right after the last turning point.\r\n" );
document.write( "That happens when and only when the leading coefficient a4 is negative,\r\n" );
document.write( "or in this case, a4 = -1.\r\n" );
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document.write( "That's 1 way out of 3, or a probability of 1/3.

\n" ); document.write( "D) What is the probability that x = 0 is a root of f(x)?
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document.write( "That happens when and only when the polynomial is divisible by x,\r\n" );
document.write( "i.e., when x can be factored out and x can be set = 0 using the zero-factor\r\n" );
document.write( "property.  That happens when and only when the constant term a0 = 0.\r\n" );
document.write( "The probability that a0 can be zero is 1 time out of 3, or 1/3.

\n" ); document.write( "E) What is the probability that x = 1 is a root of f(x)?
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document.write( "That is when, if we substitute 1 for x we get 0.\r\n" );
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document.write( "or\r\n" );
document.write( "\"a%5B4%5D%2Ba%5B3%5D%2Ba%5B2%5D%2Ba%5B1%5D%2Ba%5B0%5D=0\"\r\n" );
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document.write( "There are three cases when that happens.  It's when the five coefficients \r\n" );
document.write( "consist of\r\n" );
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document.write( "Case 1: all zeros.\r\n" );
document.write( "Case 2: one 1, one -1, and three 0's.\r\n" );
document.write( "Case 3: two 1's, two -1's and one 0.\r\n" );
document.write( " \r\n" );
document.write( "Case 1. We choose all the a's all 0's\r\n" );
document.write( "        That's 1 way for Case 1.\r\n" );
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document.write( "Case 2. We can choose the coefficient that are to be 1 in 5 ways\r\n" );
document.write( "        We can then choose the coefficient that are to be -1 in 4 ways.\r\n" );
document.write( "        The remaining three coefficients will be 0's.\r\n" );
document.write( "        That' 5×4 = 20 ways\r\n" );
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document.write( "Case 3. We can select the two coefficients that are to be 1's in 5C2=10 ways.\r\n" );
document.write( "        We can then select the two remaining coefficients to be -1's in 3C2=3 ways.\r\n" );
document.write( "        The remaining coefficient will be 0.\r\n" );
document.write( "        That's 10×3 = 30 ways\r\n" );
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document.write( "The total number of 'successful' ways is 1+20+30 = 51 ways.\r\n" );
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document.write( "The total number of 'possible' ways is 35 = 243 ways.\r\n" );
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document.write( "So the probability is 51 out of 243 ways or 51/243 which reduces to 17/41.

\n" ); document.write( "F) What is the probability that x = −1 is a root of f(x)?
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document.write( "If we substitute -x for x in f(x), we have\r\n" );
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document.write( "then f(-x) has the same roots with the opposite signs.  So it will have\r\n" );
document.write( "-1 as a root.  What that amounts to is changing the signs of the coefficients\r\n" );
document.write( "of only the terms with odd powers of x.\r\n" );
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document.write( "So for each case in part E above for which f(x) has 1 as a solution, if we\r\n" );
document.write( "change the signs of the coefficients of the odd powers of x in each one, it will\r\n" );
document.write( "will produce a polynomial with -1 as a root.  They are in 1 to 1 correspondence.\r\n" );
document.write( "So the answer to E is the same as the answer to D, which is 17/41.\r\n" );
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document.write( "Edwin

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