Question

What are all possible rational zeros for $f\left( x \right) = 3{x^3} - {x^2} - 3x + 1$ and how do you find all zeros?

Answer 1

Hint:
Given a polynomial and the zeros of the polynomial are determined by finding the values of $x$ for which the polynomial is equal to zero. Then we will apply the rational root theorem which states that the rational root is of the form $r = \pm \dfrac{p}{q}$ where the denominator must divide by the leading coefficient and the numerator is divided by the constant.

Complete step by step solution:
Here in the polynomial, $q = 3$ and $p = 1$. First, we will find the factors of the trailing constant.
$p = 1$
Therefore, the possible rational roots of the polynomial are $ \pm \dfrac{1}{3}, \pm 1$
Now we will substitute the value of roots one by one into the polynomial to check for which value the polynomial is zero.
By substituting $x = 1$ into the polynomial.
$ \Rightarrow 3{\left( 1 \right)^3} - {\left( 1 \right)^2} - 3\left( 1 \right) + 1$
$ \Rightarrow 3 - 1 - 3 + 1 = 0$
Here, $f\left( 1 \right) = 0$. Thus, $x = 1$ is a root of the polynomial which means $x - 1$ is a factor of the polynomial.
By substituting $x = - 1$ into the polynomial.
$ \Rightarrow 3{\left( { - 1} \right)^3} - {\left( { - 1} \right)^2} - 3\left( { - 1} \right) + 1$
$ \Rightarrow - 3 - 1 + 3 + 1 = 0$
Thus, $x = - 1$ is a root of the polynomial which means $x + 1$ is a factor of the polynomial.
Now substituting $x = \dfrac{1}{3}$ into the polynomial.
$ \Rightarrow 3{\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{1}{3}} \right)^2} - 3\left( {\dfrac{1}{3}} \right) + 1$
$ \Rightarrow \dfrac{1}{9} - \dfrac{1}{9} - 1 + 1 = 0$
Thus, $x = \dfrac{1}{3}$ is a root of the polynomial which means $x - \dfrac{1}{3}$ is a factor of the polynomial.
Now substituting $x = - \dfrac{1}{3}$ into the polynomial.
$ \Rightarrow 3{\left( { - \dfrac{1}{3}} \right)^3} - {\left( { - \dfrac{1}{3}} \right)^2} - 3\left( { - \dfrac{1}{3}} \right) + 1$
$ \Rightarrow - \dfrac{1}{9} - \dfrac{1}{9} + 1 + 1 = \dfrac{{16}}{9}$
Here, $f\left( { - \dfrac{1}{3}} \right) \ne 0$. Thus, $x = - \dfrac{1}{3}$ is not a root of the polynomial.

Final answer: Hence all zeros of the polynomial are $x = - 1,1,\dfrac{1}{3}$. Thus, the function has 1 negative zero and 1 positive real zeros and 1 rational zero.

Note:
In such types of questions students mainly don't get an approach on how to solve it. In such types of questions students mainly forget to apply the rational root theorem by which factorizes the whole expression. Sometimes we have to factor out a number from an expression then the expression forms in a way so that it can be factorized and identity can be applied. In such types of questions, to factor, the polynomial synthetic division can also be used instead of the long division process. The main approach to solve this question is to form a question in such a way and then apply identity.