Question

What are all the possible rational zeros for $f\left( x \right)=5{{x}^{3}}+{{x}^{2}}-5x-1$ and how do you find all zeros? \[\]

Answer 1

Hint: We recall the rational root or rational zero theorem. We first find the integral factors of leading coefficient that is5 and the integral factors of constant term that is 1. If $p$ is a factor of 5 and $q$is a factor 1 the possible rational zeros are in the form $\dfrac{p}{q}$. We take ${{x}^{2}}$ common from the first two terms and $-1$ from the last two terms and factorize to get all the zeros. \[\]

Complete step by step answer:
We know from rational zero theorem if all of the coefficients ${{a}_{n}},{{a}_{n-1}},...,{{a}_{1}},{{a}_{0}}$of the ${{n}^{\text{th}}}$ degree polynomial ${{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+...+{{a}_{1}}x+{{a}_{0}}$ are integers where the leading ${{a}_{n}}$ and the constant term ${{a}_{0}}$ are noon-zero then the possible rational zeros are in the form of $\dfrac{p}{q}$ where $p$ is any integral factor of ${{a}_{0}}$ and $q$is any integral factor of ${{a}_{n}}$. We are given the following polynomial of degree 3 in the question
\[f\left( x \right)=5{{x}^{3}}+{{x}^{2}}-5x-1\]
We see that the leading coefficient is the constant that is multiplied to ${{x}^{3}}$ that is 5 and the constant term is $-1$. The factors of 5 are $q=\pm 1,\pm 5$ since $5=1\times 5=\left( -1 \right)\times \left( -5 \right)$. The factors of $-1$ are $\pm 1$ since $-1=-1\times 1$. So the possible rational zeros are given by
\[\dfrac{p}{q}=\dfrac{\pm 1}{\pm 1,\pm 5}\]
We take combinations from numerator and denominators and find the possible rational zeros as
\[\begin{align}
  & \dfrac{p}{q}=\dfrac{1}{1},\dfrac{1}{-1},\dfrac{-1}{1},\dfrac{-1}{1},\dfrac{1}{5},\dfrac{1}{-5},\dfrac{-1}{5},\dfrac{-1}{-5} \\
 & \Rightarrow \dfrac{p}{q}=1,-1,\dfrac{1}{5},\dfrac{-1}{5} \\
\end{align}\]
Let us factorize the given polynomial by taking ${{x}^{2}}$ common from the first two terms and $-1$ from the last two terms. We have
\[\begin{align}
  & f\left( x \right)=5{{x}^{3}}+{{x}^{2}}-5x-1 \\
 & \Rightarrow f\left( x \right)={{x}^{2}}\left( 5x+1 \right)-1\times \left( 5x+1 \right) \\
\end{align}\]
We take $5x+1$ common from the above step to have;
\[\Rightarrow f\left( x \right)=\left( 5x+1 \right)\left( {{x}^{2}}-1 \right)\]
We use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the above step for $a=x,b=1$ to have
\[\Rightarrow f\left( x \right)=\left( 5x+1 \right)\left( x+1 \right)\left( x-1 \right)\]
We obtain the zeros from the equation $f\left( x \right)=0$. So we have
\[\begin{align}
  & f\left( x \right)=0 \\
 & \Rightarrow \left( 5x+1 \right)\left( x+1 \right)\left( x-1 \right)=0 \\
 & \Rightarrow 5x+1=0\text{ or }x+1=0\text{ or }x-1=0 \\
 & \Rightarrow x=\dfrac{-1}{5}\text{ or }x=-1\text{ or }x=1 \\
\end{align}\]

So the zeros of the given polynomial $f\left( x \right)$ are
\[x=-1,1,\dfrac{-1}{5}\].

Note: We note that the zeros of polynomials $P\left( x \right)$ are all the values of $x$ such that $P\left( x \right)=0$. We also note that rational zero theorem gives us the possible zeros of the polynomial, not the actual zeros. Here in this problem $x=\dfrac{1}{5}$ was a possible zero but not an actual zero. We also note that if the degree is $n$ there are exactly $n$ zeros may or may not be real.