Question

How do you use the rational roots theorem to find all the possible zeros of $3{x^3} - {x^2} + 3x - 1$ ?

Answer 1

Hint: As we know that in this question we have to use the rational roots theorem to find all the zeroes of the given polynomial function. We know that the rational roots theorem states that if the polynomial has integer coefficients, then the possible rational roots are of the form $\left( {\dfrac{p}{q}} \right)$ where p is an integral factor of the constant term and q is an integral factor of the leading coefficient. So, we first find out the possible values of p and q by taking.

Complete step by step solution:
According to the question we have the polynomial $f\left( x \right) = 3{x^3} - {x^2} + 3x - 1$.
We can see that the constant term of the polynomial given to us is $ - 1$ and the leading coefficient is $3$.
Now, according to the rational roots theorem, we can say that the possible rational roots are of the form $\left( {\dfrac{p}{q}} \right)$ , where p is a factor of the constant term $ - 1$ and the leading coefficient is $3$.
So, the possible values of p are: $1$ and $ - 1$.
Possible values of q are: $1$, $ - 1$, $3$ and $ - 3$.
Now, substituting the values of p and q to find the value of possible rational roots, we get,
$\dfrac{p}{q} = \dfrac{{ \pm 1}}{{ \pm 1, \pm 3}}$
Hence, the possible rational roots are: $1$ , $ - 1$ , $\dfrac{1}{3}$ and $\dfrac{{ - 1}}{3}$.
So, now we have to check which of these is a root of the polynomial $3{x^3} - {x^2} + 3x - 1$ by substituting these values in place of variables in the polynomial. If we get the value of a polynomial as zero on substituting a value in place of a variable, then that value is a root of the polynomial.
So, substituting the value of x as $1$, we get,
$ \Rightarrow 3{\left( 1 \right)^3} - {\left( 1 \right)^2} + 3\left( 1 \right) - 1$
Computing the powers of $1$ and opening brackets, we get,
$ \Rightarrow 3\left( 1 \right) - \left( 1 \right) + 3\left( 1 \right) - 1$
$ \Rightarrow 3 - 1 + 3 - 1 = 4$
Since the value of the polynomial for $x = 1$ is not zero. So, $1$ is not a root of the polynomial $3{x^3} - {x^2} + 3x - 1$ .
So, substituting the value of x as $ - 1$, we get,
$ \Rightarrow 3{\left( { - 1} \right)^3} - {\left( { - 1} \right)^2} + 3\left( { - 1} \right) - 1$
Simplifying the expression, we get,
$ \Rightarrow - 3 - 1 - 3 - 1$
$ \Rightarrow - 8$
Since the value of the polynomial for $x = - 1$ is not zero. So, $ - 1$ is not a root of the polynomial $3{x^3} - {x^2} + 3x - 1$ .
So, substituting the value of x as $\dfrac{1}{3}$, we get,
$ \Rightarrow 3{\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{1}{3}} \right)^2} + 3\left( {\dfrac{1}{3}} \right) - 1$
Computing the powers of $\left( {\dfrac{1}{3}} \right)$ and simplifying further, we get,
$ \Rightarrow 3\left( {\dfrac{1}{{27}}} \right) - \left( {\dfrac{1}{9}} \right) + 1 - 1$
$ \Rightarrow \dfrac{1}{9} - \dfrac{1}{9} + 1 - 1 = 0$
Since the value of the polynomial for $x = \dfrac{1}{3}$ is zero. So, $\dfrac{1}{3}$ is a root of the polynomial $3{x^3} - {x^2} + 3x - 1$ .
So, substituting the value of x as $\left( { - \dfrac{1}{3}} \right)$, we get,
$ \Rightarrow 3{\left( { - \dfrac{1}{3}} \right)^3} - {\left( { - \dfrac{1}{3}} \right)^2} + 3\left( { - \dfrac{1}{3}} \right) - 1$
$ \Rightarrow 3\left( { - \dfrac{1}{{27}}} \right) - \left( {\dfrac{1}{9}} \right) - 1 - 1$
$ \Rightarrow - \dfrac{1}{9} - \dfrac{1}{9} - 1 - 1$
$ \Rightarrow - 2 - \dfrac{2}{9} = \dfrac{{ - 20}}{9}$
Since the value of the polynomial for $x = \dfrac{{ - 1}}{3}$ is not zero. So, $ - \dfrac{1}{3}$ is not a root of the polynomial $3{x^3} - {x^2} + 3x - 1$ .
Hence, the rational root of the polynomial $3{x^3} - {x^2} + 3x - 1$ is $\dfrac{1}{3}$.

Note: Before solving this kind of question we should have the full knowledge of the rational roots theorem and how to apply those. We should note that it is a very important theorem, it tells us that with the given polynomial function with integer or whole number coefficients, a list of possible solutions can be found by listing the factors of the constant or the last term, over the factors of the coefficient of the leading term. Then, we have to check which of these possible rational roots are actually the roots of the polynomial provided to us by substituting the values in place of the variable.