"We are given the function:" We are given the function:
\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { ( x - 5 ) ( x^2 - 2 x + 1 ) } / { ( x - 7 ) ( x^2 + 2 x + 3 ) } \quad.
"Before proceeding, let's rewrite the function a little, primarily"
"by some factoring:"
\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { ( x - 5 ) ( x - 1 )^2 } / { ( x - 7 ) ( x^2 + 2 x + 1 + 2 ) }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ { ( x - 5 ) ( x - 1 )^2 } / { ( x - 7 ) ( ( x + 1 )^2 + 2 ) } \quad.
"Thus:"
\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { ( x - 5 ) ( x - 1 )^2 } / { ( x - 7 ) ( ( x + 1 )^2 + 2 ) } \quad.
"So, we may write, in terms that should be clear:"
\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { ( x - 5 ) cdot "positive" } / { ( x - 7 ) ( "positive" + 2 ) }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { ( x - 5 ) cdot "positive" } / { ( x - 7 ) cdot "positive" } \quad.
"Thus:"
\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { ( x - 5 ) cdot "positive" } / { ( x - 7 ) cdot "positive" } \quad.
"Now suppose:" \quad 5 < a < 7.
"So clearly now:"
\qquad \qquad \qquad \qquad \qquad \qquad f(a) \ = \ { ( a - 5 ) cdot "positive" } / { ( a - 7 ) cdot "positive" }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { "negative" \ cdot \ "positive" } / { "positive" \ cdot \ "positive" }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { "negative" } / { "positive" }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ "negative" \quad.
"So we have shown:"
\qquad \qquad \qquad \qquad \qquad \quad 5 < a < 7 \quad => \quad f(a) \ = \ "negative"
\qquad "i.e.": \qquad f(a) \ = \ "positive" \quad => \quad 5 < a < 7 \quad \ "is impossible".
"And so, finally:"
\qquad \qquad \qquad { ( x - 5 ) ( x^2 - 2 x + 1 ) } / { ( x - 7 ) ( x^2 + 2 x + 3 ) } \quad "is positive" \quad =>
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 5 < x < 7 \quad \ "is impossible". \qquad \qquad \ \ square