finding all possible rtnl zeroes of f(x) = x^7+37x^5-6x^2+12
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Re: finding the zeros
It is really a simple thing. You can see it by solving this equation: ax + b = 0 The solution is x = -b/a. This is a beautiful thing. It has something to do with the constant term divided by something to do with the coefficient on the 'x' term. The extenstion to quadratics and higher-order polynomials is not necessarily obvious, but let's just go with it for now. We can explore it later.
ax^3 + bx^2 + cx + d = 0 IF it has rational roots, they should be related to this: x = d/a. Again, something to do with the constant term divided by something to do with the coefficient on the highest-order 'x' term.
The Rational Root Theorem tells us what the "somethign to do with" part is. It is FACTORS.
In your problem, the constant term is 12. This has factors 1, 2, 3, 4, 6, and 12.
The coefficient on the x^7 (the highest order) is one (1). That has only one factor, 1.
Now divide every combination: 1/1, 2/1, 3/1, 4/1, 6/1, 12/1 -- Is that it?
Not quite. It doesn't give us any information about the SIGN of the rational root, so we have to include all the negatives, as well. -1, -2, -3, -4, -6, and -12. Now we're done.
Remember, this does NOT tell us that there IS a Rational Root. It tells us only that IF there is a Rational Root, it is one of these.
It is really a simple thing. You can see it by solving this equation: ax + b = 0 The solution is x = -b/a. This is a beautiful thing. It has something to do with the constant term divided by something to do with the coefficient on the 'x' term. The extenstion to quadratics and higher-order polynomials is not necessarily obvious, but let's just go with it for now. We can explore it later.
ax^3 + bx^2 + cx + d = 0 IF it has rational roots, they should be related to this: x = d/a. Again, something to do with the constant term divided by something to do with the coefficient on the highest-order 'x' term.
The Rational Root Theorem tells us what the "somethign to do with" part is. It is FACTORS.
In your problem, the constant term is 12. This has factors 1, 2, 3, 4, 6, and 12.
The coefficient on the x^7 (the highest order) is one (1). That has only one factor, 1.
Now divide every combination: 1/1, 2/1, 3/1, 4/1, 6/1, 12/1 -- Is that it?
Not quite. It doesn't give us any information about the SIGN of the rational root, so we have to include all the negatives, as well. -1, -2, -3, -4, -6, and -12. Now we're done.
Remember, this does NOT tell us that there IS a Rational Root. It tells us only that IF there is a Rational Root, it is one of these.