Solution:
Given:
[tex]\frac{x^2+3x-4}{x^2+7x+12}[/tex][tex]\begin{gathered} \frac{x^2+3x-4}{x^2+7x+12}=\frac{x^2-x+4x-4}{x^2+3x+4x+12} \\ Factorizing \\ \frac{x(x-1)+4(x-1)}{x(x+3)+4(x+3)}=\frac{(x+4)(x-1)}{(x+4)(x+3)} \\ =\frac{x-1}{x+3} \end{gathered}[/tex]
To make the condition equal,
[tex]\begin{gathered} \frac{x-1}{x+3}\div\frac{x-1}{x+1}=\frac{x+1}{x+3} \\ \\ Hence,\text{ the middle term for the expression to be true is:} \\ \frac{x-1}{x+1} \end{gathered}[/tex]
Hence, the polynomial that can be made for the middle expression is any value multiplied to the numerator and denominator that makes it always true.
Thus,
[tex]\begin{gathered} \frac{(x-1)}{(x+1)}\times\frac{(x+4)}{(x+4)} \\ =\frac{(x-1)(x+4)}{(x+1)(x+4)} \\ =\frac{x^2+4x-x-4}{x^2+4x+x+4} \\ =\frac{x^2+3x-4}{x^2+5x+4} \end{gathered}[/tex]
Therefore,
