Can't understand a simplification

[RapidAssistant]

I have the following equation (it's the result of a differentiation, but that's not where I am stuck)


This then simplifies to:


However I have no idea of the steps to get from one to the other. I feel stupid because the answer is probably easy, but I can't find any resource online that will explain the steps.

 

[Dr.Peterson]

I have the following equation (it's the result of a differentiation, but that's not where I am stuck)

View attachment 29037
This then simplifies to:

View attachment 29038
However I have no idea of the steps to get from one to the other. I feel stupid because the answer is probably easy, but I can't find any resource online that will explain the steps.

Are you sure they are equivalent? Unless I made a mistake, they have different values for x=0, so they can't be equivalent. For that matter, the first is 0 when x=-2, and the second is not. At the least, surely "x+4" in the second should be "2x+4", or vice versa.

Can you show us the context? Is someone saying one comes from the other, or is one your work and the other the "correct" answer, or what? Did you copy both correctly?

 

[Steven G]

I have the following equation (it's the result of a differentiation, but that's not where I am stuck)

View attachment 29037
This then simplifies to:

View attachment 29038
However I have no idea of the steps to get from one to the other. I feel stupid because the answer is probably easy, but I can't find any resource online that will explain the steps.

Factor.
(2x+4)^6(3x-2)^4[14(3x-2) + 15(2x+4)] = (2x+4)^6(3x-2)^4[42x-28 + 30x + 60] = (2x+4)^6(3x-2)^4[72x+32] =(2x+4)^6(3x-2)^4[8(9x+4)]=
64(x+2)^6(3x-2)^4[8(9x+4)]= 512[(x+2)^6(3x-2)^4(9x+4)

The (x+4) should be a (x+2)

 

[RapidAssistant]

Thank you (apologies in the long time responding I have been very ill of late).

I get that there was a mistake in the original answer - thank you. Still confused though.....where does the bit I have highlighted in red bold below come from?

(2x+4)^6(3x-2)^4[14(3x-2) + 15(2x+4)] = (2x+4)^6(3x-2)^4[42x-28 + 30x + 60] = (2x+4)^6(3x-2)^4[72x+32] =(2x+4)^6(3x-2)^4[8(9x+4)]=
64(x+2)^6(3x-2)^4[8(9x+4)]= 512[(x+2)^6(3x-2)^4(9x+4)

 

[Steven G]

[14(3x-2) + 15(2x+4)] Distribute the 14 and then distribute the 15

 

[HallsofIvy]

You have $\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4$.

Do you see that there is a "(2x+ 4)" in both terms? One is to the 6th power and the other is to the 7th power. $\displaystyle (2x+ 4)^7= (2x+ 4)^6(2x+ 4)$ so there is $\displaystyle (2x+ 4)^6$ in both terms and we can factor that out, leaving 2x+ 4 to the first power in the second term.

Do you see that there is a "(3x- 2)" in both terms? One is to the 5th power and the other is to the 4th power. $\displaystyle (3x- 2)^5= (3x- 2)^4(3x- 3)$ so there is $\displaystyle (3x- 2)^4$ in both terms so we can factor that out, leaving 3x- 2 to the first power in the first term.

$\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4= (2x+ 4)^6(3x- 2)^4[(14(3x- 2)+ 15(2x+ 4)]$.

Now we need to "combine" what was left:
14(3x- 2)+ 15(2x+ 4)= 42x- 28+ 30x+ 60= 72x+ 32 so
$\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4= (2x+ 4)^6(3x- 2)^4(72x+ 32)$.

You could, if you like, also take common factors out of the coefficients. 2x+ 4= 2(x+ 2) so $\displaystyle (2x+ 4)^6= 2^6(x+ 2)^6= 64(x+ 2)^5$. 72= 8(9) and 32= 4(8) so 72x+ 32= 8(9x+ 4). 64(8)= 512.

$\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4= 512(x+ 2)^6(3x- 2)^4(9x+ 4)$.

 

[RapidAssistant]

You have $\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4$.

Do you see that there is a "(2x+ 4)" in both terms? One is to the 6th power and the other is to the 7th power. $\displaystyle (2x+ 4)^7= (2x+ 4)^6(2x+ 4)$ so there is $\displaystyle (2x+ 4)^6$ in both terms and we can factor that out, leaving 2x+ 4 to the first power in the second term.

Do you see that there is a "(3x- 2)" in both terms? One is to the 5th power and the other is to the 4th power. $\displaystyle (3x- 2)^5= (3x- 2)^4(3x- 3)$ so there is $\displaystyle (3x- 2)^4$ in both terms so we can factor that out, leaving 3x- 2 to the first power in the first term.

$\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4= (2x+ 4)^6(3x- 2)^4[(14(3x- 2)+ 15(2x+ 4)]$.

Now we need to "combine" what was left:
14(3x- 2)+ 15(2x+ 4)= 42x- 28+ 30x+ 60= 72x+ 32 so
$\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4= (2x+ 4)^6(3x- 2)^4(72x+ 32)$.

You could, if you like, also take common factors out of the coefficients. 2x+ 4= 2(x+ 2) so $\displaystyle (2x+ 4)^6= 2^6(x+ 2)^6= 64(x+ 2)^5$. 72= 8(9) and 32= 4(8) so 72x+ 32= 8(9x+ 4). 64(8)= 512.

$\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4= 512(x+ 2)^6(3x- 2)^4(9x+ 4)$.

Thank you ever so much - that all makes sense now. I was getting myself into all sorts of trouble, by trying to expand all those brackets using the binomial theorem and then trying to simplify it all back down......