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Q.
If the polynomials $f(x)=x^2+15 x+m$ and $g(x)=x^2+16 x+n$ have a common factor, then find the value of $(m-n)^2$___
Remainder and Factor Theorems
Solution:
Let the common factor be $x-a$ for both $f(x)$ and $g(x)$.
Now $x-a$ is a factor
So, $f(a)=a^2+15 a+m=0$
$ a^2+15 a+m=0 $
$a^2=-15 a-m$....(i)
Also, $g(a)=a^2+16 a+n=0$
$ a^2+16 a+n=0 $
$ a^2=-16 a-n$...(ii)
From (i) and (ii),
$ -15 a-m=-16 a-n $
$\Rightarrow m-n=a $
$\Rightarrow(m-n)^2=a^2 $
$ \Rightarrow(m-n)^2=-15 a-m$
$ \text { or }(m-n)^2=-16 a-n$
$\Rightarrow(m-n)^2=-15(m-n)-m $
$\text { or } $ $ (m-n)^2=-16(m-n)-n $
$\Rightarrow(m-n)^2=-15 m+15 n-m $
$\text { or }(m-n)^2=-16 m+16 n-n $
$\Rightarrow(m-n)^2=-16 m+15 n $
$\text { or } (m-n)^2=-16 m+15 n$