The number of zeroes of the polynomial x2 + 4x + 2 is

Hint: In this question, we are given a polynomial with its two zeroes and we have to find the values of $a$ .
Now, we know, for a quadratic polynomial given by $a{x^2} + bx + c$ ,
Sum of its zeroes $ = \dfrac{{ - b}}{a}$ and product of its zeroes $ = \dfrac{c}{a}$ .
So, using this, we’ll find the value of $a$ .

Complete step-by-step answer:
We are given a polynomial as ${x^2} + 4x + 2a$ and its zeroes are $\alpha $ and $\dfrac{2}{\alpha }$ .
To find the value of $a$ .
On comparing the polynomial ${x^2} + 4x + 2a$ with $a{x^2} + bx + c$ , we get, $a = 1$ , $b = 4$ and $c = 2a$ .
Now, we know, sum of its zeroes $\alpha + \dfrac{2}{\alpha } = \dfrac{{ - b}}{a}$ , i.e., on putting the values of $a$ and $b$ , we get, $\alpha + \dfrac{2}{\alpha } = \dfrac{{ - 4}}{1}$ .
And the product of its zeroes $\alpha \cdot \dfrac{2}{\alpha } = \dfrac{c}{a}$ i.e., on putting the values of $a$ and $c$ , we get, $\alpha \cdot \dfrac{2}{\alpha } = \dfrac{{2a}}{1}$ .
Now, on solving the equation formed using the product of zeroes, we get, $2 = 2a$ , i.e., $a = 1$ .
Hence, in the polynomial ${x^2} + 4x + 2a$ , whose zeroes are of the form $\alpha $ and $\dfrac{2}{\alpha }$ , the value of $a$ is $1$ .

Note: Remember that the number of zeroes of a polynomial is equal to its degree, i.e., the highest power of the variable of the polynomial.
A polynomial with the number of zeroes greater than two also follows the properties of the sum of zeroes and the product of the zeroes. In that case, we have to take the sum of the zeroes and the product of all its zeroes together.
This question can also be solved by the discriminant method using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , and then equating $\alpha $ and $\dfrac{2}{\alpha }$ , one equal to positive sign and other with the negative sign and solve but that would be very lengthy and calculative.

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