Question

How do you find all the rational zeros of function?

Answer 1

Hint: To find all the rational zeros of a function. First of all, we will write all the terms in the function in the descending order of their power. After that, we will write the constant term and the coefficient of the highest power of x. Then we will find the positive and negative factors of the constant term and the coefficient of the highest power of x. Now, we will say the factors of the last term as “p” and the factors of the coefficient of the highest power of x as “q” then the possible rational zeros are the ratio of p by q. Now, to check how many of these rational zeros are possible, we will put these rational zeros in the given function and see how many of these zeros on putting in the function are giving value as 0.

Complete step-by-step solution:
To understand what we have hinted in hint, we are going to take a function as follows:
$2{{x}^{3}}+{{x}^{2}}-4x+1$
Now, the constant term (or the last term) of the above function is 1. And the coefficient of the highest power of x (or the first term) is 2. So, we are writing the factors of 1 as follows:
$\Rightarrow 1=\pm 1$
Let us call these factors “p”.
Now, let us write the factors of 2 we get,
$\Rightarrow 2=\pm 1,\pm 2$
Let us call the above factors as “q”.
Now, the possible rational zeros of the above function is equal to:
$\Rightarrow \dfrac{p}{q}$
The possible factors $\left( \dfrac{p}{q} \right)$ by taking one of the factors of -4 in the numerator and then fix that factor in the numerator and vary different factors of 2 in the denominator are as follows:
\[\pm \dfrac{1}{1},\pm \dfrac{1}{2}\]
The actual zeros are found by substituting the above possible zeros in the given function and if after substitution we are getting the value 0 then that rational zero is the actual zero.
Substituting $x=1$ in the given function we get,
$\begin{align}
  & \Rightarrow f\left( 1 \right)=2{{\left( 1 \right)}^{3}}+{{\left( 1 \right)}^{2}}-4\left( 1 \right)+1 \\
 & \Rightarrow f\left( 1 \right)=2+1-4+1 \\
 & \Rightarrow f\left( 1 \right)=4-4=0 \\
\end{align}$
Substituting $x=-1$ in the given function we get,
$\begin{align}
  & \Rightarrow f\left( -1 \right)=2{{\left( -1 \right)}^{3}}+{{\left( -1 \right)}^{2}}-4\left( -1 \right)+1 \\
 & \Rightarrow f\left( -1 \right)=-2+1+4+1 \\
 & \Rightarrow f\left( -1 \right)=6-2=4 \\
\end{align}$
Substituting $x=\dfrac{1}{2}$ in the given function we get,
$\begin{align}
  & \Rightarrow f\left( \dfrac{1}{2} \right)=2{{\left( \dfrac{1}{2} \right)}^{3}}+{{\left( \dfrac{1}{2} \right)}^{2}}-4\left( \dfrac{1}{2} \right)+1 \\
 & \Rightarrow f\left( \dfrac{1}{2} \right)=2\left( \dfrac{1}{8} \right)+\dfrac{1}{4}-2+1 \\
 & \Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{1}{4}+\dfrac{1}{4}-1 \\
 & \Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{1}{2}-1=\dfrac{1-2}{2}=-\dfrac{1}{2} \\
\end{align}$
Substituting $x=-\dfrac{1}{2}$ in the given function we get,
$\begin{align}
  & \Rightarrow f\left( -\dfrac{1}{2} \right)=2{{\left( -\dfrac{1}{2} \right)}^{3}}+{{\left( -\dfrac{1}{2} \right)}^{2}}-4\left( -\dfrac{1}{2} \right)+1 \\
 & \Rightarrow f\left( -\dfrac{1}{2} \right)=2\left( -\dfrac{1}{8} \right)+\dfrac{1}{4}+2+1 \\
 & \Rightarrow f\left( -\dfrac{1}{2} \right)=-\dfrac{1}{4}+\dfrac{1}{4}+3 \\
 & \Rightarrow f\left( -\dfrac{1}{2} \right)=3 \\
\end{align}$
As you can see that only $x=1$ gives 0 on substitution so $x=1$ is the only actual zero of the given function.

Note: The mistake that could be possible in the above solution is not to check the rational zeros which we have calculated from the $\dfrac{p}{q}$ form by substituting those zeros in the original function and see which rational zero is giving 0 on substitution in the given function. And the other mistake is the calculation mistake in the above problem because there is some rigorous calculation involved while substituting the possible zeros in the given function so make sure you won’t make such mistakes.