# Thread: Irreducible rational function / 3rd degree polynomial

1. ## Irreducible rational function / 3rd degree polynomial

I have 2 question on my algebra homework that have me stumped. Not sure if I am just having a brain fart, or haven't went over it yet (it's extra credit).

1.) Consider an irreducible rational function that is less than zero. Suppose also the numerator and denominator both contain a binomial of degree one. Which of the following statements is true?
a.) In order to find the critical values for the inequality, set the numerator and denominator equal to zero and then solve.
b.) In order to find the critical values for the inequality, set the rational function equal to zero and then solve.
c.) Because the function is irreducible, in order to find the critical vales for the inequality it is sufficient to just set the numerator equal to zero and then solve.
d.) Because the function is irreducible, in order to find the critical values for the inequality it is sufficient to just set the denominator equal to zero and then solve.

2.) Consider a third degree polynomial function consisting of the product of three distinct linear factors. Suppose that the third degree polynomial function is greater than zero. Which of these statements is true?
a.) The inequality has two critical values.
b.) The critical values will generate for intervals.
c.) All of the critical values of the inequality are part of the solution set.
d.) All of the above.

Any suggestions?

2. ## Re: Irreducible rational function / 3rd degree polynomial

Originally Posted by wr00
I have 2 question on my algebra homework that have me stumped. Not sure if I am just having a brain fart, or haven't went over it yet (it's extra credit).

1.) Consider an irreducible rational function that is less than zero. Suppose also the numerator and denominator both contain a binomial of degree one. Which of the following statements is true?
a.) In order to find the critical values for the inequality, set the numerator and denominator equal to zero and then solve.
b.) In order to find the critical values for the inequality, set the rational function equal to zero and then solve.
c.) Because the function is irreducible, in order to find the critical vales for the inequality it is sufficient to just set the numerator equal to zero and then solve.
d.) Because the function is irreducible, in order to find the critical values for the inequality it is sufficient to just set the denominator equal to zero and then solve.

2.) Consider a third degree polynomial function consisting of the product of three distinct linear factors. Suppose that the third degree polynomial function is greater than zero. Which of these statements is true?
a.) The inequality has two critical values.
b.) The critical values will generate for intervals.
c.) All of the critical values of the inequality are part of the solution set.
d.) All of the above.

Any suggestions?
I want to make sure I understand what you are saying in (1)

You have an irreducible polynomial $\frac{f(x)}{g(x)}$ that is strictly less than 0 ?

And then you multiply numerator and denominator by first order binomial terms to get

$r(x) = \dfrac{a_1 x + b_1}{a_2 x + b_2}\frac{f(x)}{g(x)}$

and you want to solve for the roots of $r(x)=0$ ?