Question

The number of polynomials having zeros \[-2\] and \[5\]is
(A) $ 1 $
(B) $ 2 $
(C) $ 3 $
(D) More than 3

Answer 1

Hint: Let $ p\left( x \right)=a{{x}^{2}}+bx+c $ be the polynomial whose zeroes are $ -2 $ and $ 5 $ . Now we have to find out the sum of zeroes and the product of zeroes. From the sum of zeroes and the product of zeroes, we will get two equations. From those two equations, we will get the quadratic equation. Then multiply the polynomial by any arbitrary constant. Then we get the required number of polynomials having zeros $ -2 $ and $ 5 $ .

Complete step by step answer:
Let us assume that the required polynomial whose zeroes are $ -2 $ and $ 5 $ is
 $ p\left( x \right)=a{{x}^{2}}+bx+c $
Now let us find out the sum of zeroes,
From the basic concept, we know that sum of zeroes of a quadratic polynomial $ p\left( x \right)=a{{x}^{2}}+bx+c $ is
 $ \text{sum of zeroes}=\dfrac{-b}{a} $
 $ \dfrac{-b}{a}=-2+5 $
 $ \dfrac{-b}{a}=\dfrac{3}{1} $
 $ \dfrac{-b}{a}=\dfrac{-\left( -3 \right)}{1} $ ………………..(1)
From the basic concept we know that the product of zeroes of a quadratic polynomial $ p\left( x \right)=a{{x}^{2}}+bx+c $ is
 $ \text{Product of zeroes}=\dfrac{c}{a} $
 $ \dfrac{c}{a}=-2\left( 5 \right) $
 $ \dfrac{c}{a}=\dfrac{-10}{1} $ ………………(2)
From the equations (1) and (2), we will have
 $ a=1,b=-3,c=-10 $
Now we can conclude that equation of the polynomial is
 $ \begin{align}
  & p\left( x \right)=a{{x}^{2}}+bx+c \\
 & ~\Rightarrow p(x)=1.{{x}^{2}}-3x-10 \\
 & \Rightarrow p(x)={{x}^{2}}-3x-10 \\
\end{align} $
But we know that, if we multiply or divide any polynomial by any arbitrary constant. Then, the zeroes of the polynomial never change.
Therefore,
 $ p\left( x \right)=k{{x}^{2}}-3kx-10 $
 [Where, k is a real number]
 $ p\left( x \right)=\dfrac{{{x}^{2}}}{k}-\dfrac{3}{k}x-\dfrac{10}{k} $ [Where, k is a non-zero real number]
Hence, the required number of polynomials having zeros $ -2 $ and $ 5 $ are infinite i.e. more than $ 3 $ .
Therefore option D is the correct option for the given question.

Note:
 We have to assume the general polynomial for the given zeros of the polynomial. The calculation must be done carefully while forming the polynomial with the values of the sum of zeroes and the product of zeroes. Multiplying with the arbitrary constant must be done carefully.