Use the Rational Zeros Theorem to write a list of all potential rational zeros

f(x) = 2x3 - 5x2 + 7x - 3

Question

Use the Rational Zeros Theorem to write a list of all potential rational zeros

f(x) = 2x3 - 5x2 + 7x - 3

whpalmer4 · Answer

The rational zeros theorem says that if you have a polynomial of this form, you can make a list of all potential rational zeros from the coefficients of the constant term and the highest exponent term.

whpalmer4 · Answer

To do so, you need to factor both of those coefficients.

Write down all the factors of the constant term.  These will be possible values of $p$.

Write down all the factors of the leading term.  These will be possible values of $q$.

Your possible zeros will be all combinations of $\pm p/q$.

whpalmer4 · Answer

The reason this works is that the constant term is built up by the coefficients of the constant term and the leading term when multiplying.  We are just factoring to find the possible numbers that could have gotten us the final constant term.  Unfortunately, if the constant term has many factors, there are many combinations to try, and not all of them will be actual zeros.

whpalmer4 · Answer

$$(2a+1)(a-6) = 2a^2-12a+a-6 = 2a^2-11a- 6$$Our constant term came just from the leading and constant terms
$$(2a^2-11a-6)(3a-4) = 6a^3-8a^2-33a^2+44a-18a+24 $$$$= 6a^3-41a^2+26a+24$$Again, the constant term came just from the product of leading and constant terms.

If we apply the RZT, our values of $p$ are 1,2,3,4,6,8,12,24 and values of $q$ are 1,2,3,6.  Easy to see how we end up with many more candidates than actual solutions!

anonymous · Answer

There is only one real root, as shown below:$$\frac{1}{6} \left(5-\frac{17}{\left(-28+3 \sqrt{633}\right)^{1/3}}+\left(-28+3 \sqrt{633}\right)^{1/3}\right)  $$Don't want to ruffle any feathers here, but, this is not a reasonable class room problem to present to a student or, there is a mistake in the problem expression itself.

whpalmer4 · Answer

Oh, not sure I'm in complete agreement with that — the student isn't asked to actually find any zeros, just make a list of possible rational zeros, and with the equation given, that's pretty easy.  Admittedly, it would be better if it was a similarly simple list where you could actually try them all out and find at least one that was a zero :-)

anonymous · Answer

@whpalmer4

Thank you for your response.