Question

If the degree of polynomial \[p(y)\] is $a$, then the maximum number of zeros of \[p(y)\] would be:
$A)a + 1$
$B)a - 1$
$C)a$
$D)2a$

Answer 1

Hint: First, we will know what is polynomial; polynomial is the expression that will consist of the coefficients and variables.
In the polynomial, the first term poly means many and the second term nominal means terms, hence which is the many terms with variables.
An example of the polynomial is ${x^2} + x + 1$. If the degree of the polynomial is one $x + 1$ then it is called linear degree or straight line.

Complete step-by-step solution:
Since from the given question that the degree of the polynomial \[p(y)\] is $a$ which means it has at the most degree $a$.
Since from the known result in the polynomial, the number of the zeros of the given polynomial is less than or equals the degree of the given polynomial. This means it cannot produce more than the values from the degree of the polynomial.
From the given that degree of the polynomial function \[p(y)\] is $a$, now we will proceed with the number of the zeros of the given polynomial \[p(y)\], it can be represented as $a, a - 1,a - 2,...,2,1,0$ (at most a, so no more than the value a and it will be decreased function is shown).
Hence these values $a,a - 1,a - 2,...,2,1,0$ are the zeros of the polynomial to the given \[p(y)\].
Now we will need to find the maximum number in the zeros of the polynomial, which is definitely $a$.
Suppose take $a - 1$ is the maximum but $a - 1$ can have a greater value which is $a$ because $a - 1 < a$.
Hence $a$is the maximum number of zeros \[p(y)\].
Therefore option $C)a$ is correct.

Note: Since $a + 1,2a$ is greater than the value $a$, but in the zeros of the polynomial, the at most value is $a$. Since a constant polynomial is the degree zero function $4$. The linear polynomial is the degree one function $x + 1$. Quadratic means degree two, cubic means degree three, quartic means degree four. These are the polynomial types with the degree increasing.