Question

How many zeros does the cubic polynomial $P\left( a \right)={{a}^{3}}-a$ have?
(a) 0
(b) 1
(c) 2
(d) 3

Answer 1

Hint: We start solving the problem by recalling the definition of the zero of the polynomial $f\left( x \right)$ as the value of ‘x’ for which the value of the function is 0 i.e., $f\left( x \right)=0$. We then equate the given polynomial to zero and factorize it. We then then equate each factor to zero to get the zeros of the polynomial. We then count the total number of zeros we have for the polynomial to report the required answer.

Complete step by step answer:
According to the problem, we need to find the total number of zeros of the cubic polynomial $P\left( a \right)={{a}^{3}}-a$.
Let us recall the definition of zeros of polynomials.
We know that the zero of the polynomial $f\left( x \right)$ is defined as the value of ‘x’ for which the value of the function is 0 i.e., $f\left( x \right)=0$.
So, in order to find zeros of the polynomial $P\left( a \right)={{a}^{3}}-a$. We equate it to zero $P\left( a \right)={{a}^{3}}-a=0$.
Let us factorize the polynomial to get zeros.
$\Rightarrow a\left( {{a}^{2}}-1 \right)=0$.
$\Rightarrow a\left( a+1 \right)\left( a-1 \right)=0$.
$\Rightarrow a=0$, $a+1=0$, $a-1=0$.
$\therefore a=0$, $a=-1$, $a=1$.
So, we have found the zeros of the polynomial as 0, –1, 1.
We have found that there are three zeros for the polynomial $P\left( a \right)={{a}^{3}}-a$.

So, the correct answer is “Option d”.

Note: Whenever we have this type of problems, we first equate the given polynomial to zero and factorize them. We can also solve this problem by using the trial and error method by assuming zero of the polynomial. We can also find the sum and product of the zeros of a polynomial without finding them. Similarly, we can expect problems to find the equation of the polynomial whose zeros are inverse of these.