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Getting stuck on simplification
- Thread starter Maxtro
- Start date
D
Deleted member 4993
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The book isn't helping me much.
Problem started as 7(3x+1)4(6x-5)2
I got it down to 7(3x+1)4(2)(6x-5)(6)+7(4)(3x+1)3(6x-5)2
and I don't have a clue what to do next.
Are you trying to find derivative of 7(3x+1)4(6x-5)2 ??
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Hello, Maxtro!
How about some arithmetic?
\(\displaystyle f'(x) \;=\;84(3x+1)^4(6x-5) + 84(3x+1)^3(6x-5)^2\)
Then you can factor:
\(\displaystyle f'(x) \;=\;84(3x+1)^3(6x-5)\bigg[(3x+1)+(6x-5)\bigg]\)
\(\displaystyle f'(x) \;=\;84(3x+1)^3(6x-5)(9x-4)\)
Differentiate: .\(\displaystyle f(x) \:=\:7(3x+1)^4(6x-5)^2\)
I got it down to: .\(\displaystyle f'(x) \:=\:7(3x+1)^4(2)(6x-5)(6)+7(4)(3x+1)^3\color{red}{(3)}(6x-5)^2\)
and I don't have a clue what to do next. . Really?
How about some arithmetic?
\(\displaystyle f'(x) \;=\;84(3x+1)^4(6x-5) + 84(3x+1)^3(6x-5)^2\)
Then you can factor:
\(\displaystyle f'(x) \;=\;84(3x+1)^3(6x-5)\bigg[(3x+1)+(6x-5)\bigg]\)
\(\displaystyle f'(x) \;=\;84(3x+1)^3(6x-5)(9x-4)\)
Hello, Maxtro!
How about some arithmetic?
\(\displaystyle f'(x) \;=\;84(3x+1)^4(6x-5) + 84(3x+1)^3(6x-5)^2\)
Then you can factor:
\(\displaystyle f'(x) \;=\;84(3x+1)^3(6x-5)\bigg[(3x+1)+(6x-5)\bigg]\)
\(\displaystyle f'(x) \;=\;84(3x+1)^3(6x-5)(9x-4)\)
Sorry, I'm an idiot when it comes to math.
Where did the 84 come from?
3 * 7 * 4 = 21 * 4 = 84Sorry, I'm an idiot when it comes to math.
Where did the 84 come from?
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
Actually, there are two "84"s.
\(\displaystyle 7(3x+1)^4(2)(6x- 5)(6)+ 7(4)(3x+1)^3(3)(6x- 5)^2\)
\(\displaystyle 7(2)(6)(3x+1)^4(6x- 5)+ 7(4)(3)(3x+ 1)^3(6x- 5)^2\)
7(2)(6)= 14(6)= 84 and 7(4)(3)= 28(3)= 84.
\(\displaystyle 84(3x+1)^4(6x- 4)+ 84(3x+1)^3(6x-5)^2\)
There are "84", "\(\displaystyle (3x+1)^3\)", and "\(\displaystyle (6x-5)\)" in both terms so we can factor them out leaving
3x+ 1+ 6x- 5= 9x- 4 as Soroban said.
\(\displaystyle 7(3x+1)^4(2)(6x- 5)(6)+ 7(4)(3x+1)^3(3)(6x- 5)^2\)
\(\displaystyle 7(2)(6)(3x+1)^4(6x- 5)+ 7(4)(3)(3x+ 1)^3(6x- 5)^2\)
7(2)(6)= 14(6)= 84 and 7(4)(3)= 28(3)= 84.
\(\displaystyle 84(3x+1)^4(6x- 4)+ 84(3x+1)^3(6x-5)^2\)
There are "84", "\(\displaystyle (3x+1)^3\)", and "\(\displaystyle (6x-5)\)" in both terms so we can factor them out leaving
3x+ 1+ 6x- 5= 9x- 4 as Soroban said.