Simplify Derivative of a Product of Two Functions

[Euler]

My book has an example that loses me at the jump from step1 to step2. Right now I am on a homework problem that I cannot simplify, but I just understood the example step I could do it.

It reads: If \(\displaystyle y=(3x-8)^7(4x+9)^5\), find y' and simplify

STEP 1: \(\displaystyle y'=7(3x-8)^6(3)(4x+9)^5+(3x-8)^7(5)(4x+9)^4(4)\)
STEP 2: \(\displaystyle y' = (3x-8)^6(4x+9)^4[21(4x+9)+20(3x-8)]\)

My question is, how do they arrive at step two from step 1?

 

[stapel]

They did it by factoring.

. . .y' = 7 (3x - 8)⁶ (3) (4x + 9)⁵ + (3x - 8)⁷ (5) (4x + 9)⁴ (4)

. . . . .= 21 (3x - 8)⁶ (4x + 9)⁵ + 20 (3x - 8)⁷ (4x + 9)⁴

. . . . .= 21 (3x - 8)⁶ (4x + 9) (4x + 9)⁴ + 20 (3x - 8)⁶ (3x - 8) (4x + 9)⁴

. . . . .= (3x - 8)⁶ (4x + 9)⁴ [21(4x + 9)] + (3x - 8)⁶ (4x + 9)⁴ [20(3x - 8)]

Hope that helps a bit.

Eliz.