Question

The number of polynomials having zeros as 2 and 5 is
(a) 1
(b) 2
(c) 3
(d) More than 3

Answer 1

Hint: We solve this problem by using the construction of a polynomial from its zeros.
If \[{{x}_{1}},{{x}_{2}},{{x}_{3}},......,{{x}_{n}}\] are the zeros of a polynomial then the polynomial is given as
\[f\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)......\left( x-{{x}_{n}} \right)\]
Where, \[a\ne 0\]
By using the above formula we find the number of polynomials having 2 and 5 as zeros.

Complete answer:
We are asked to find the number of polynomials having zeros as 2 and 5
We know that if \[{{x}_{1}},{{x}_{2}},{{x}_{3}},......,{{x}_{n}}\] are the zeros of a polynomial then the polynomial is given as
\[f\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)......\left( x-{{x}_{n}} \right)\]
Where, \[a\ne 0\]
By using the above formula we get the polynomial of zeros 2 and 5 as
\[\begin{align}
  & \Rightarrow f\left( x \right)=a\left( x-2 \right)\left( x-5 \right) \\
 & \Rightarrow f\left( x \right)=a\left( {{x}^{2}}-7x+10 \right) \\
\end{align}\]
Here, we can see that for \[a=1\] we get the polynomial as
\[\Rightarrow f\left( x \right)={{x}^{2}}-7x+10\]
Similarly, let us assume that the value of \[a=2\] then we get the polynomial as
\[\begin{align}
  & \Rightarrow f\left( x \right)=2\left( {{x}^{2}}-7x+10 \right) \\
 & \Rightarrow f\left( x \right)=2{{x}^{2}}-14x+20 \\
\end{align}\]
Now, let us check whether the above polynomial has zeros 2 and 5
We know that the zeros of polynomial \[a{{x}^{2}}+bx+c\] are given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
By using this formula to above polynomial we get
\[\begin{align}
  & \Rightarrow x=\dfrac{14\pm \sqrt{{{14}^{2}}-4\left( 20 \right)\left( 2 \right)}}{2\times 2} \\
 & \Rightarrow x=\dfrac{14\pm \sqrt{196-160}}{4} \\
 & \Rightarrow x=\dfrac{14\pm 6}{4} \\
\end{align}\]
Here, we get two zeros for positive and negative signs
By taking the positive sign we get
\[\begin{align}
  & \Rightarrow x=\dfrac{14+6}{4} \\
 & \Rightarrow x=5 \\
\end{align}\]
Now, by taking the negative value we get
\[\begin{align}
  & \Rightarrow x=\dfrac{14-6}{4} \\
 & \Rightarrow x=2 \\
\end{align}\]
Here, we can see that the roots are 2 and 5
Therefore, we can say that the polynomial having zeros 2 and 5 is
\[\Rightarrow f\left( x \right)=a\left( {{x}^{2}}-7x+10 \right)\forall a\in \mathbb{R}\]
Here, we can see that there will be infinite number of polynomials depending on the value of \[a\]
Therefore, we can conclude that there are more than 3 polynomials of zeros 2 and 5.

So, option (d) is the correct answer.

Note:
Here, we can give the other explanation for this problem.
Here, we are asked to find the number of polynomials but not mentioned anything about degree of polynomial
We know that we can form infinite polynomials of degree 3 with given two zeros as follows
\[\Rightarrow p\left( x \right)=a\left( x-2 \right)\left( x-5 \right)\left( x-k \right)\]
Here, we can see that for all possible combinations of \[a,k\] we get infinite polynomials.
Therefore, we can conclude that there are more than 3 polynomials of zeros 2 and 5