Synthetic Division in FactoringDate: 04/18/2002 at 01:25:13 From: Brandon Subject: Factoring completely Factor completely: 2x^3 + 17x^2 + 58x + 25. I don't know what to do or where to begin. Date: 04/18/2002 at 09:01:23 From: Doctor Rick Subject: Re: Factoring completely Hi, Brandon. This is not an easy type of problem to solve! The approach I would take is to use synthetic division to search for one root. If you are not familiar with this method, see the following answer in our Dr. Math Archives: Synthetic Division http://mathforum.org/library/drmath/view/53056.html Synthetic division can be viewed in two ways: as a way to evaluate a polynomial in x at a particular value x = a, or as a way to divide the polynomial by (x-a). Use this method to test possible roots. If the cubic has any rational roots (and if it doesn't, you won't be able to find an exact factorization by this method), they must be of the form of plus or minus a factor of 25 (the constant term) divided by a factor of 2 (the coefficient of the highest power). Thus the numerator can be 1, 5, or 25, and the denominator can be 1 or 2, and the sign can be positive or negative. Synthetic division has a property that is very useful in searching for roots: If all the numbers below the line (in the notation used in the archived explanation) are positive, then you know that there are no roots greater than the number you tested. On the other hand, if the numbers below the line alternate positive and negative, then there are no roots less than the number you tested. Thus, I suggest that you start by testing x = 0. This will tell you right away whether the polynomial may have negative roots, positive roots, or both. Continue in the same way (something like a "binary search"): each time you choose a possible root to test, choose one that is in the middle of the range of possible roots that you haven't already eliminated. It may turn out that this one test eliminates all possible roots less than the test number, or all possibilities greater than the test number, thus giving you only half as many possibilities to test. Once you find a value of x for which the polynomial evaluates to zero (in other words, a factor (x-a) of the polynomial), you have also found the other factor: just use the numbers below the line of your synthetic division as the coefficients of a polynomial. In your case, the resulting polynomial will be a quadratic. Then your task is to factor this quadratic, if it can be done with real coefficients. (Or, if you have learned about polynomials with complex coefficients, you can always factor the quadratic.) You can use familiar methods for factoring a quadratic, or you can use the sure-fire method of setting the quadratic equal to zero and finding roots with the quadratic formula. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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