# Thread: Help simplifying with binomials

1. ## Help simplifying with binomials

I am actually differentiating functions, but I get stuck on the part involving simplifying. Below is what the book says, but I cannot follow how they got from the first to the second part...

= 28(3x+1)^3 (7x-4)^3 + 9(7x-4)^4 (3x+1)^2

= (3x+1)^2 (7x-4)^3 [28(3x+1) +9(7x-4)]

Any help would be much appreciated!
Thanks.

2. ## Still not quite getting it.

Thanks for the full process math_helper, but my books also shows the full process. I just don't quite get the change in the two lines in my original post. I can kind of see what changes are made, but I cannot see where the exponent (^3) goes that was originally after 28(3x+1), or the exponent (^4) that was originally after 9(7x-4).

3. Originally Posted by instinctive1
I am actually differentiating functions, but I get stuck on the part involving simplifying. Below is what the book says, but I cannot follow how they got from the first to the second part...

= 28(3x+1)^3 (7x-4)^3 + 9(7x-4)^4 (3x+1)^2

= (3x+1)^2 (7x-4)^3 [28(3x+1) +9(7x-4)]

Any help would be much appreciated!
Thanks.
The main point is the "distributive" law: ab+ ac= a(b+ c).
Looking at $28(3x+1)^3 (7x-4)^3 + 9(7x-4)^4 (3x+1)^2$
I notice that each term includes "3x+1". One term has it squared, the other cubed. But $(3x+1)^2= (3x+1)^2(3x+1)$ so I can think of $(3x+1)^2$ as the "a" in the distributive law:
[tex]28(3x+1)^3 (7x-4)^3 + 9(7x-4)^4 (3x+1)^2= (3x+1)^2[28(3x+1)(7x-4)^3+ 9(7x-4)^4][tex].

Now I see that each term involves 7x- 4, the first term cubed and the second term to the fourth power. I can think of $(7x-4)^3$ as the a: [tex](3x+1)^2[28(3x+1)(7x-4)^3+ 9(7x-4)^4]= (3x+1)^2(7x-4)^3[28(3x+1)+ 9(7x-4)][tex].