"Yes, this is a simplification technique that may be described as"
"factoring out the" \ lowest \ "powers of common exponentiated"
"factors. It is very powerful, and makes some simplifications a"
"lot easier and faster, and can get you to the end result very"
"directly."
"[It works for any exponents -- positive, negative, fractional, any.]"
"In the case you have, at the line prior to line 3, you had:"
\qquad f'(x) \ = \ ( 2 x - 5 )^4 \cdot 5 ( x^2 + x + 1 )^4 ( 2 x + 1 )
\qquad \qquad \qquad \qquad \qquad \qquad + ( x^2 + x + 1 )^5 \cdot 4 ( 2 x - 5 )^3 cdot 2.
"Maybe to illustrate the technique more easily, let me name"
"the common exponentiated factors here:"
\qquad \qquad \qquad \qquad \qquad \qquad A = 2 x - 5, \qquad B = x^2 + x + 1.
"Now we can rewrite" \ \ f'(x) \ \ "as:"
\qquad f'(x) \ = \ ( 2 x - 5 )^4 \cdot 5 ( x^2 + x + 1 )^4 ( 2 x + 1 )
\qquad \qquad \qquad \qquad \qquad \qquad + ( x^2 + x + 1 )^5 \cdot 4 ( 2 x - 5 )^3 cdot 2
\qquad \qquad \qquad \quad = \ A^4 \cdot 5 B^4 ( 2 x + 1 ) + B^5 \cdot 4 A^3 cdot 2
\qquad \qquad \qquad \quad = \ A^4 B^4 \cdot 5 ( 2 x + 1 ) + A^3 B^5 \cdot 4 cdot 2 .
"Now factor out the" \ lowest \ "powers of the quantites" \ A, B ":"
\qquad f'(x) \ = \ A^4 B^4 \cdot 5 ( 2 x + 1 ) + A^3 B^5 \cdot 4 cdot 2 .
\qquad \qquad \qquad \quad = \ A^3 B^4 [ A \cdot 5 ( 2 x + 1 ) + B \cdot 4 cdot 2 ] .
"Now let's return" \ \ A, B \ \ "to their original values, as we assigned"
"them above:"
\qquad f'(x) \ = \ A^4 B^4 \cdot 5 ( 2 x + 1 ) + A^3 B^5 \cdot 4 cdot 2
:. \qquad f'(x) \ = \ ( 2 x - 5 )^3 ( x^2 + x + 1 )^4 [ ( 2 x - 5 ) \cdot 5 ( 2 x + 1 )
\qquad \qquad \qquad \qquad \qquad \qquad \qquad + ( x^2 + x + 1 ) \cdot 4 cdot 2 ] .
"(Sorry for the line breaks, system can go no farther !!)"
"This is how you go from line (2) to line (3), in the example you"
"provided. I think this is what you wanted explained. But just"
"to be sure, let me continue, and finish the problem as you gave"
"it."
"So, continuing:"
:. \qquad f'(x) \ = \ ( 2 x - 5 )^3 ( x^2 + x + 1 )^4 [ ( 2 x - 5 ) \cdot 5 ( 2 x + 1 )
\qquad \qquad \qquad \qquad \qquad \qquad \qquad + ( x^2 + x + 1 ) \cdot 4 cdot 2 ] .
\qquad \qquad \qquad \qquad \qquad = \ ( 2 x - 5 )^3 ( x^2 + x + 1 )^4 [ 5 ( 2 x - 5 ) ( 2 x + 1 )
\qquad \qquad \qquad \qquad \qquad \qquad \qquad + 8 ( x^2 + x + 1 ) ]
\ = \ ( 2 x - 5 )^3 ( x^2 + x + 1 )^4 [ 5 ( 4 x^2 - 8 x - 5 ) + 8 x^2 +8 x + 8 ]
\ = \ ( 2 x - 5 )^3 ( x^2 + x + 1 )^4 [ 20 x^2 - 40 x - 25 + 8 x^2 +8 x + 8 ]
\ = \ ( 2 x - 5 )^3 ( x^2 + x + 1 )^4 ( 28 x^2 - 32 x - 17 );
"which is as your example, as given, ended."
"I hope this helps !!"
