# Textbook Example

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

## The Attempt at a Solution

Mentor
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
$$(2) (-5) a^4 b^9 + (2)(6) b^{10} a^3 = 2 a^3 b^9 \left( -5 a + 6 b \right)$$