Question

# The number of polynomials having zeros -2 and 5 isA. 1B. 2C. 3D. more than 3

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Hint: To find out the number of polynomials, obtain sum and product of roots individually from the data given and then substitute the variables in generalized form to obtain the final polynomial. Hence, by this we can get the total number of polynomials.

Quadratic function is a polynomial function with one or more variables in which the highest degree term is of the second degree.
As we need to find the number of polynomials having zeros, let us apply the quadratic equation to solve this.
Let p(x) be the required polynomial, in which zeroes of the polynomial are -2 and 5.
The polynomial is of the form,
$p\left( x \right) = a{x^2} + bx + c$
To find the variables a and b:
Sum of the zeroes = $\dfrac{{ - b}}{a}$ , where
b is coefficient of $x$
a is coefficient of ${x^2}$
$- 2 + 5 = \dfrac{{ - b}}{a}$
$3 = \dfrac{{ - b}}{a}$
Hence,
$\dfrac{3}{1} = \dfrac{{ - b}}{a}$
Therefore, the values of a and b is
$a = 1$ and $b = - 3$ .
Product of the zeroes = constant term ÷ coefficient of x2 i.e.
Product of zeroes = $\dfrac{c}{a}$ , where
c is the constant term.
a is coefficient of ${x^2}$ .
$\left( { - 2} \right)5 = \dfrac{c}{a}$
$- 10 = \dfrac{c}{a}$
Hence,
$\dfrac{{ - 10}}{1} = \dfrac{c}{a}$
Therefore, the value of c is
$- 10 = c$
Hence, we got
$a = 1$ , $b = - 3$ and $c = - 10$ .
Substituting the values of a, b and c in the polynomial
$p\left( x \right) = a{x^2} + bx + c$
$p\left( x \right) = 1 \cdot {x^2} - 3x - 10$
Therefore, p(x) is
$p\left( x \right) = {x^2} - 3x - 10$
Therefore, we can conclude that x can take any value.
Hence, option D is the correct answer.
So, the correct answer is “Option D”.

Note: The key point to find the number of polynomials having zeros is to apply the Quadratic function for the polynomials by finding the sum and product of zeroes. If we multiply and divide any polynomial by arbitrary, then the zeros of the polynomial never change.