Question

How do you find all the zeros of $f(x)=2{{x}^{3}}+{{x}^{2}}-8x-4$ ?
(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 $2{{x}^{3}}+{{x}^{2}}-8x-4$. 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 -2, 2 and $-\dfrac{1}{2}$, which can be called as the zeros of the given equation. Thus, we find how to get the zeros of the equation.

Complete step by step solution:
We are trying to factorize the equation, $f(x)=2{{x}^{3}}+{{x}^{2}}-8x-4$,
We start with,
$2{{x}^{3}}+{{x}^{2}}-8x-4$
We know that 2x+1 can be taken out as common. Therefore, we get
$\Rightarrow {{x}^{2}}(2x+1)-4(2x+1)$
And we know that it can be further written as
$\Rightarrow ({{x}^{2}}-4)(2x+1)$
And on further simplifications, as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ , we get
$\Rightarrow (x+2)(x-2)(2x+1)$
After more simplification, if we take, $f(x)=0$
So, $(x+2)(x-2)(2x+1)=0$ gives us,
$x+2=0$ or $x-2=0$ or $2x+1=0$
Now, for $x+2=0$, we have, $x=-2$.
For $x-2=0$ , we get, $x=2$ .
And for, $2x+1=0$ , we get, $x=-\dfrac{1}{2}$ .
So, $x=-2,2,-\dfrac{1}{2}$
Thus, in this way, we get the zeros of the equation as -2, 2 and $-\dfrac{1}{2}$ .
And we can also check that the hit and trial method can be used to find the solutions of our given problem.
Let us try with the value -2 to start with,
So, we have our equation as, $2{{x}^{3}}+{{x}^{2}}-8x-4$,
putting -2 in the value of x, we get,
\[\Rightarrow 2.{{\left( -2 \right)}^{3}}+{{\left( -2 \right)}^{2}}-8.\left( -2 \right)-4\]
 Now, simplifying the value,
$\Rightarrow 2.\left( -8 \right)+4+16-4$
After more simplification, we will reach the value we want to evaluate.
$\Rightarrow -16+8+16-8=0$
So, we are getting the value of the function as 0 by putting -2. Then we can conclude that the -2 is the root of the equation.
Similarly if we put the values of x as 2 and $-\dfrac{1}{2}$, we will get the value of the function as zero. Then we can also conclude that the 2 and $-\dfrac{1}{2}$ are the roots of the equation.

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

Note:
The zero of a function $2{{x}^{3}}+{{x}^{2}}-8x-4$ 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.