Question

What is the Rational zeros test?

Answer 1

Hint: After looking at the term “rational” means it has something to do with rational numbers. And the word ‘zeros’ means there is some involvement of polynomials and the roots of it are to be considered. Since, rational numbers are involved, we can figure out that this test helps to find the rational roots of a polynomial $P(x)$.

Complete step-by-step solution:
Consider a polynomial $P(x)$, then it looks something like this:
$P(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0$
Where $a_n\neq 0$
The ‘root’ of a polynomial means any value ‘k’ such that $P(k)=0$.
Now, consider a polynomial with non zero constant term, means in the above general equation of the polynomial we have:
$a_0\neq 0$
Then the Rational zeros test says that the rational roots of such a polynomial have the following form always:
$\dfrac{p}{q}=\dfrac{\pm \text{Factors of }a_0}{\pm\text{Factors of }a_n}$
This would become more clear with the help of an example.
Take a polynomial:
$P(x)=x^2-1$
We have $a_0=-1$ and $a_1=1$
Factors of 1 and -1 are 1 and -1 only.
So, the possible roots of the equation will be:
$\dfrac{p}{q}=\dfrac{\pm (-1)}{\pm 1 }$
So, the possible rational roots will be 1,-1. And now we check whether these are roots or not.
$P(1)=1^2-1=0$
$P(-1)=(-1)^2-1=0$
Both 1 and -1 are roots of this polynomial. It is always not necessary that all the rational numbers of this form will be the roots of this equation, but there is only a possibility that these might be the roots.

Note: After calculating the possibilities cancel out the duplicate factors i.e. if two roots are there $\dfrac{6}{3}$ and 2 then both are equal. Do not interchange the terms in the numerator and denominator by mistake. List out the factors one by one and then divide them to find the rational roots.