Thread: Help simplifying with binomials
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Old November 18th, 2008, 04:29 AM
HallsofIvy HallsofIvy is offline
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Originally Posted by instinctive1 View Post
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:
[math]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][math].

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: [math](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)][math].
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