Question

Question 4
The number of polynomials having zeroes as –2 and 5 is
(A) 1
(B) 2
(C) 3
(D) more than 3

Answer 1

let $p\left(x\right)=a{x}^2+bx+c,$ be the required polynomial whose zeroes are -2 and 5
$\therefore$ sum of zeroes = $\frac{-b}{a}$
$\Rightarrow \frac{-b}{a}=-2+5=3=\frac{3}{1}=\frac{-(-3)}{1}$

and product of zeroes =$\frac{c}{a}$
$\Rightarrow \frac{c}{a}=-2\times 5=-10=\frac{-10}{1}$
From above we can conclude that
a = 1, b = -3 and c= - 10
$p\left(x\right)=a{x}^2+bx+c=1x^2–3x-10$
$=x^2–3x-10$
But we know that, if we multiply or divide a polynomial by any constant then the zeroes of polynomial do not change
$\therefore p\left(x\right)=k{x}^2–3kx–10k$ [where, k is a real number]
$\Rightarrow p\left(x\right)=\frac{x^2}{k}-\frac{3}{k}x-\frac{10}{k}$ [where, k is a non – zero real number]
Hence, the required number of polynomials are infinite.
So option $D\left(morethan3\right)$, is correct.