Help Understanding Chain Rule Example Problem in Calculus

[RslM]

Homework Statement

May I please have help understanding this example problem in my textbook.

I am working on the chain rule for finding derivatives in Calculus. What I am currently having trouble with is factoring. My textbook gives the follwing example:

f'(x) =(3x - 5)⁴(7 - x)¹⁰

f'(x) = (3x - 5)⁴ x 10(7 - x)⁹(-1) + (7 - x)¹⁰4(3x - 5)³(3)

= -10(3x - 5)⁴(7 - x)⁹ + (7 - x)¹⁰12(3x - 5)³

Here it says 2(3x - 5)³(7 - x)⁹ is factored out.

= 2(3x - 5)³(7 - x)⁹[-5(3x - 5) + 6(7 - x)]

=2(3x - 5)³(7 - x)⁹(-15x + 25 + 42 - 6x)

And it continues onto simplification.

The area I have trouble with is when 2(3x - 5)³(7 - x)⁹ is factored out.

I am not fully understanding why

= 2(3x - 5)³(7 - x)⁹[-5(3x - 5) + 6(7 - x)]

is the result. How was it derived?

Thank you

Homework Equations

The Attempt at a Solution

 

[DrClaude]

Maybe it will be simpler for you to understand if we rewrite

= -10(3x - 5)⁴(7 - x)⁹ + (7 - x)¹⁰12(3x - 5)³

as
$$
(2) (-5) a^4 b^9 + (2) (6) b^{10} a^3
$$
A factor of 2 is common to both terms, and you can factor out lowest power of $a$ and of $b$, that is $2 a^3 b^9$. Divide the first term by this factor,
$$
\frac{(2)(-5) a^4 b^9 }{2 a^3 b^9} = -5 a,
$$
and do the same with the second term. You get
$$
(2) (-5) a^4 b^9 + (2)(6) b^{10} a^3 = 2 a^3 b^9 \left( -5 a + 6 b \right)
$$