Question

If $x-2$ and $x-\frac{1}{2}$ both are the factors of the polynomial $n{x}^2-5x+m$, then show that $m=n=2$.

Answer 1

Let $f\left(x\right)=$$n{x}^2-5x+m$.

Given,

$x-2$ and $x-\frac{1}{2}$ are factors of $f\left(x\right)$.

Then using remainder theorem we have,

$f\left(2\right)=0$ and $f\left(\frac{1}{2}\right)=0$.

Now,

$f\left(2\right)=0$ gives,

$4n-10+m=0$

or, $m+4n=10$......(1).

Again,

$f\left(\frac{1}{2}\right)=0$ gives,

or, $\frac{n}{4}-\frac{5}{2}+m=0$

or, $4m-10+n=0$

or, $4m+n=10$.......(2).

Now solving (1) and (2) we get,

$m=2=n$.