Question

How do you find all the zeros of $f(x)={{x}^{3}}+2{{x}^{2}}-x-2$ ?
(a) Factorization
(b) Guessing the roots
(c) Changing the variable
(d) None of the above

Answer 1

Hint: To find the zeros of a given equation we are to try to factorize the equation given ${{x}^{3}}+2{{x}^{2}}-x-2$ . We analyze the factors of the equation and then try to configure them equaling to zero. Thus we are getting the 3 values of x as 1,-1,2, which can be called as the zeros of the given equation. Thus, we find a way of how to get the zeros of the equation.

Complete step by step solution:
Here we are factoring the equation, $f(x)={{x}^{3}}+2{{x}^{2}}-x-2$,
We start with,
${{x}^{3}}+2{{x}^{2}}-x-2$
We know, x+2 can be taken as common. Therefore, we get
$\Rightarrow {{x}^{2}}(x+2)-1(x+2)$
And we know that it can be further written as
$\Rightarrow ({{x}^{2}}-1)(x+2)$
And on further simplifications, as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ , we get
$\Rightarrow (x+1)(x-1)(x+2)$
After further simplification, if we take, $f(x)=0$
So, $(x+2)(x-1)(x+1)=0$ gives us,
$x+2=0$ or $x+1=0$ or $x-1=0$
Now, for $x+2=0$, we have, $x=-2$.
For $x+1=0$ , we get, $x=-1$ .
And for, $x-1=0$ , we get, $x=1$ .
So, $x=-2,1,-1$
Thus, in this way, we get the zeros of the equation as -2, 1 and -1.
And we can also check that the hit and trial method can be used to find the solutions of our given problem.
To describe the hit and trial method,
Let us try with the value -1 to start with,
So, we have our equation as, ${{x}^{3}}+2{{x}^{2}}-x-2$,
putting -1 in the value of x, we get,
$\Rightarrow {{\left( -1 \right)}^{3}}+2.{{\left( -1 \right)}^{2}}-\left( -1 \right)-2$
Now, simplifying the value,
$\Rightarrow -1+2.1+1-2$
After more simplification, we will reach the value we want to evaluate.
$\Rightarrow -1+2+1-2=0$
So, we are getting the value of the function as 0 by putting -8. Then we can conclude that the -8 is the root of the equation.
Similarly if we put the value of x as -2 and 1, we will also get the value of the function as zero. Then we can also conclude that the -1 and 2 are the roots of the equation.

Hence, the correct option is, (a) Factorization and (b) Guessing the roots .

Note:
The zero of a function ${{x}^{3}}+2{{x}^{2}}-x-2$ is any replacement for the variable that will produce an answer of zero. Graphically, the real zero of a function is where the graph of the function crosses the x‐axis; that is, the real zero of a function is the x‐intercept(s) of the graph of the function. Any function which can be drawn in a graph and intersecting the x-axis can give us zeros of the equation.