Algebra II: Rational Zeros Theorem

[c_r_a_15]

Hi I really need help with this problem! I don't know where to start. Help would be very much appreciated. Thanks!!

Write a polynomial function with rational coefficients that has 6 possible rational zeros according to the rational zero theorem, but no actual rational zeros . You must show the possible rational zeros and the actual answers.

 

[Steven G]

Here is a hint. (x-sqrt(2))(x+sqrt(2))=x^2 - 2. Note that x^2-2 has rational coefficients. What are the zeros? Now play with this idea.

 

[Dr.Peterson]

That's a nice problem! What thoughts do you have, even if they are not places to start?

Here's my thought: First decide on first and last coefficients that would give you 6 possible rational zeros. Then try filling in other coefficients.

To avoid too much trial and error, you might just make it a quadratic that has no real zeros at all!

 

[JeffM]

I suspect that it might help to start at another place.

by the Fundamental Theorem of Algebra, a polynomial with real coefficents and degree n has exactly n complex roots (although some may be duplicated). Furthermore , if n is odd, at least one of those roots is real. So the simplest polynomial with six distinct real roots is of degree 6.

Moreover, if a polynomial of degree 2k has 2k real zeroes, then that polynomial can be factored into k quadratics, each of which has two real zeroes.