Question

How do you find all possible rational zeros of $f(x)=2{{x}^{3}}-5{{x}^{2}}+3x-1$ ?

Answer 1

Hint: The given polynomial is an expression in one variable so to find all the possible rational zeros of the given polynomial, $f(x)=2{{x}^{3}}-5{{x}^{2}}+3x-1$ we use the rational zeros theorem to check if it has any rational zeros or not.

Complete step-by-step solution:
The given polynomial is $f(x)=2{{x}^{3}}-5{{x}^{2}}+3x-1$
To find if there are any possible rational zeros, we use the rational zero theorem.
The rational zero theorem states that if there is a polynomial with integer coefficients,
${{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+.....+{{a}_{0}}$
Given that the coefficients of the highest term and the last term should not be equal to zero, ${{a}_{n}}\ne 0;{{a}_{0}}\ne 0$ Then any rational zeros of the given polynomial can be represented by or expressed in the form of $\dfrac{p}{q}$ Such that $p,q\;$ are integers and $p$ is the divisor(numerator) of ${{a}_{0}}$( denominator), and $q$ is the divisor(numerator) of ${{a}_{n}}$( denominator);${{a}_{n}}$ is the coefficient of the highest degree term.
Here in our given polynomial, $f(x)=2{{x}^{3}}-5{{x}^{2}}+3x-1$
$\Rightarrow {{a}_{0}}=-1;{{a}_{n}}=2$
The constant term is $-1$ and the factors of $-1$ are $p=\pm 1$
The constant term is $2$ and the factors of $2$ are $q=\pm 1,\pm 2$
Then $\left\{ \dfrac{p}{q}=\pm 1,\pm \dfrac{1}{2} \right\}$
Now we can determine which of the following possible zeros are actual zeros by substituting these above values for $x$ in the polynomial $f(x)=2{{x}^{3}}-5{{x}^{2}}+3x-1$
For $\dfrac{p}{q}=+1$ ;
$\Rightarrow f(+1)=2{{(1)}^{3}}-5{{(1)}^{2}}+3(1)-1$
$\Rightarrow f(+1)=2-5+3-1=-1$
For $\dfrac{p}{q}=-1$ ;
$\Rightarrow f(-1)=2{{(-1)}^{3}}-5{{(-1)}^{2}}+3(-1)-1$
$\Rightarrow f(-1)=-2-5-3-1=-11$
For $\dfrac{p}{q}=+\dfrac{1}{2}$ ;
$\Rightarrow f\left( +\dfrac{1}{2} \right)=2{{\left( \dfrac{1}{2} \right)}^{3}}-5{{\left( \dfrac{1}{2} \right)}^{2}}+3\left( \dfrac{1}{2} \right)-1$
$\Rightarrow f\left( +\dfrac{1}{2} \right)=\dfrac{1}{4}-\dfrac{5}{4}+\dfrac{3}{2}-1=-\dfrac{1}{2}$
For $\dfrac{p}{q}=-\dfrac{1}{2}$ ;
$\Rightarrow f\left( -\dfrac{1}{2} \right)=2{{\left( -\dfrac{1}{2} \right)}^{3}}-5{{\left( -\dfrac{1}{2} \right)}^{2}}+3\left( -\dfrac{1}{2} \right)-1$
$\Rightarrow f\left( -\dfrac{1}{2} \right)=-\dfrac{1}{4}+\dfrac{5}{4}-\dfrac{3}{2}-1=-\dfrac{3}{2}$
Since none of the values of $\dfrac{p}{q}$ are the rational zeros of the polynomial,
Hence, this cubic polynomial $f(x)=2{{x}^{3}}-5{{x}^{2}}+3x-1$ has no rational zeros.

Note: Whenever we are asked to find the rational zeros for a certain polynomial, use the rational zeros theorem and after finding the $\dfrac{p}{q}$ form, never forget to substitute all the values of the $\dfrac{p}{q}$ form to see which value of $x$ gives a zero. Students tend to assume the values of the $\dfrac{p}{q}$ form as the wrong solution.