Question

How do you find all zeros of \[f\left( x \right)={{x}^{4}}+7{{x}^{3}}-{{x}^{2}}-67x-60\]?

Answer 1

Hint: This question belongs to the topic of algebra. We will find the zeros of the given equation in the question using rational root theorem. In this question, we will check that how much possibility of roots or zeros will be there for the equation \[f\left( x \right)={{x}^{4}}+7{{x}^{3}}-{{x}^{2}}-67x-60\]. After that, we will check at those possibilities if they are actual zeros or not. After that, we will get the zeros of the equation.

Complete step by step answer:
Let us solve this question.
This question has asked us to find all the zeros of the equation\[f\left( x \right)={{x}^{4}}+7{{x}^{3}}-{{x}^{2}}-67x-60\].
We are going to use here rational root theorem for solving this question.
Let us first know about rational root theorem.
According to rational root theorem, zeros of \[f\left( x \right)\] is expressed in the form of \[\dfrac{p}{q}\], where p and q are integers. If we take an example of an equation like \[f\left( x \right)={{a}_{1}}{{x}^{n}}+{{a}_{2}}{{x}^{n-1}}+{{a}_{3}}{{x}^{n-2}}+.........+{{a}_{n-1}}x+{{a}_{n}}\], then we can say that the values of p are factors of \[{{a}_{n}}\] and the values of q are factors of \[{{a}_{1}}\] or we can say p is a divisor of constant term and q is a divisor of coefficient of leading term.
So, in the equation \[f\left( x \right)={{x}^{4}}+7{{x}^{3}}-{{x}^{2}}-67x-60\], here zeros will be in the form of \[\dfrac{p}{q}\], where value of p will be factor of -60 and value of q will be factor of 1.
Hence, the possible zeros of the equation \[f\left( x \right)={{x}^{4}}+7{{x}^{3}}-{{x}^{2}}-67x-60\] will be
$-1, +1, -2, +2, -3, +3, -4, +4, -5, +5, -6, +6, -10, +10, -12, +12, -15, +15, -20, +20, -30, +30, -60,$ and $+60$.
Now, we will check if the value of the equation \[f\left( x \right)={{x}^{4}}+7{{x}^{3}}-{{x}^{2}}-67x-60\] is zero by putting the above values as x.
Let us check at x=-1, we get
\[f\left( -1 \right)={{\left( -1 \right)}^{4}}+7{{\left( -1 \right)}^{3}}-{{\left( -1 \right)}^{2}}-67\left( -1 \right)-60=1-7-1+67-60=0\]
As we can see from the above, the value of the equation is getting zero at x=-1. Hence, we can say -1 is a root of the equation.
Now, let us check at x=+1, we get
\[f\left( +1 \right)={{\left( +1 \right)}^{4}}+7{{\left( +1 \right)}^{3}}-{{\left( +1 \right)}^{2}}-67\left( +1 \right)-60=1+7-1-67-60=-120\]
We can see from the above that the value of the equation is not zero at x=+1. So, +1 is not a root of the equation.
Now, let us check for x=-2, we get
\[f\left( -2 \right)={{\left( -2 \right)}^{4}}+7{{\left( -2 \right)}^{3}}-{{\left( -2 \right)}^{2}}-67\left( -2 \right)-60=16-56-4+134-60=30\]
So, -2 is not a root of the equation.
Let us check at x=+2, we get
\[f\left( +2 \right)={{\left( +2 \right)}^{4}}+7{{\left( +2 \right)}^{3}}-{{\left( +2 \right)}^{2}}-67\left( +2 \right)-60=16+56-4-134-60=-126\]
So, +2 is not a root of the equation.
Let us check at x=-3, we get
\[f\left( -3 \right)={{\left( -3 \right)}^{4}}+7{{\left( -3 \right)}^{3}}-{{\left( -3 \right)}^{2}}-67\left( -3 \right)-60=81-189-9+201-60=18\]
So, -3 is not a root of the equation.
Now, let us check for x=+3, we get
\[f\left( +3 \right)={{\left( +3 \right)}^{4}}+7{{\left( +3 \right)}^{3}}-{{\left( +3 \right)}^{2}}-67\left( +3 \right)-60=81+189-9-201-60=0\]
So, we can say that x=+3 is a root of the equation.
Now, let us check for x=-4, we get
\[f\left( -4 \right)={{\left( -4 \right)}^{4}}+7{{\left( -4 \right)}^{3}}-{{\left( -4 \right)}^{2}}-67\left( -4 \right)-60=256-448-16+268-60=0\]
So, we can say that -4 is a root of the equation.
Now, let us check at x=+4, we get
\[f\left( +4 \right)={{\left( +4 \right)}^{4}}+7{{\left( +4 \right)}^{3}}-{{\left( +4 \right)}^{2}}-67\left( +4 \right)-60=256+448-16-268-60=360\]
So, +4 is not a root of the equation.
Let us check for x=-5, we get
\[f\left( -5 \right)={{\left( -5 \right)}^{4}}+7{{\left( -5 \right)}^{3}}-{{\left( -5 \right)}^{2}}-67\left( -5 \right)-60=625-875-25+335-60=0\]
So, -5 is a root of the equation.
Hence, we have seen from the above that -1, 3, -4, and -5 are roots of the equation.
Now, we have found that the zeros of the equation \[f\left( x \right)={{x}^{4}}+7{{x}^{3}}-{{x}^{2}}-67x-60\] are -1, 3, -4, and -5

Note: We should have a better knowledge in the topic algebra. We should know how to find the zeros or roots of any type of equation. We should know about rational root theorems. The rational root theorem says that
If we have given an equation \[f\left( x \right)={{a}_{1}}{{x}^{n}}+{{a}_{2}}{{x}^{n-1}}+{{a}_{3}}{{x}^{n-2}}+.........+{{a}_{n-1}}x+{{a}_{n}}\], then the zeros of \[f\left( x \right)\] will be always in the form of \[\dfrac{p}{q}\], where p and q are integers. The values of p are factors of \[{{a}_{n}}\] and the values of q are factors of \[{{a}_{1}}\] or we can say p is a divisor of constant term and q is a divisor of coefficient of leading term.