Question

What are all the possible rational zeros for $f\left( x \right)=5{{x}^{3}}-11{{x}^{2}}+7x-1$ and how do you find all zeros ?

Answer 1

Hint: In order to solve this problem, we will use rational root theorem in order to find possible rational zeros. Since this is a polynomial function, we can also solve by factorization method to find possible rational zeros. With this we can get all values of $x$ that is all rational zeros.

Complete step by step solution:
We have our polynomial function as:
$f\left( x \right)=5{{x}^{3}}-11{{x}^{2}}+7x-1$
By the rational root theorem, any rational zeros of \[f\left( x \right)~\]are expressible in the form $\dfrac{p}{q}$ for integers \[p,q\] with $p$ a divisor of the constant term $-1$ and $q$ a divisor of the coefficient $5$ of the leading term.
That means that the only possible rational zeros are:
$\pm \dfrac{1}{5},\pm 1$
Now in order to solve for the rational zeros of our polynomial function, we have to factorize our function.
Therefore, we get:
$f\left( x \right)=5{{x}^{3}}-5{{x}^{2}}-6{{x}^{2}}+6x+x-1$
On taking out common terms of $x$, we have:
\[f\left( x \right)=5{{x}^{2}}\left( x-1 \right)-6x\left( x-1 \right)+1\left( x-1 \right)\]
On simplifying, we get:
\[f\left( x \right)=\left( x-1 \right)\left( 5{{x}^{2}}-6x+1 \right)\]
On more factorization, we get:
\[f\left( x \right)=\left( x-1 \right)\left( 5x\left( x-1 \right)-1\left( x-1 \right) \right)\]
On simplifying, we get:
\[f\left( x \right)=\left( x-1 \right)\left( x-1 \right)\left( 5x-1 \right)\]
On Further simplification, we get:
\[f\left( x \right)={{\left( x-1 \right)}^{2}}\left( 5x-1 \right)\]
Now, on making the polynomial function on the left hand side to be zero, we get the value of $x$.
Therefore, we get:
$\Rightarrow x-1=0$ and
$\Rightarrow 5x-1=0$
Therefore, the possible rational zeros are:
$x=1$ and $x=0.2$
Therefore, we can conclude that the actual zeros are:
$1$ with multiplicity $2$
$\dfrac{1}{2}$ with multiplicity $1$


Note:
A rational number is a number that can be expressed in the form $\dfrac{p}{q}$ for some integers $p$ and $q$.
A zero of an expression \[f\left( x \right)~\] is a value of $x$ such that \[f\left( x \right)=0\].
So a rational zero of an expression \[f\left( x \right)~\] is basically a fraction $\dfrac{p}{q}$ such that \[f\left( \dfrac{p}{q} \right)=0\].
For example, \[2{{x}^{2}}-3x-5~\]has rational zeros \[x=-1\] and \[x=5\], since substituting either of these values for $x$ in the expression results in the value $0$.
If the quadratic cannot be factored using the two numbers that add to this and multiple to be this method, then use the quadratic formula, $-b\pm \sqrt{{{b}^{2}}-4ac}$ to factor an equation in the form of \[a{{x}^{2}}+bx+c\].
Also we have to keep in mind that the sum of the coefficients of \[f\left( x \right)~\] is zero.