What is the Factored Form of the Derivative Using Product Rule?

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Homework Statement


use product rule to find the derivative of g(x)=(2x+1)^2(x-7)^3


Homework Equations


i've applied chain rule and product rule to get...


The Attempt at a Solution


4(2x+1)(x-7)^3+3(2x+1)^2(x-7)^2

i need to factor out to get the following..
5(x-7)^2(2x-5)(2x+1)
That's the final answer from the book. But the steps to get there... I'm lost. Can anyone help me out on getting it to the factored form?
 
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cataschok said:

Homework Statement


use product rule to find the derivative of g(x)=(2x+1)^2(x-7)^3


Homework Equations


i've applied chain rule and product rule to get...


The Attempt at a Solution


4(2x+1)(x-7)^3+3(2x+1)^2(x-7)^2

i need to factor out to get the following..
5(x-7)^2(2x-5)(2x+1)
That's the final answer from the book. But the steps to get there... I'm lost. Can anyone help me out on getting it to the factored form?
Look for factors that are common to both terms.
4(2x+1)(x-7)3+3(2x+1)2(x-7)^2
= (2x+1)(x - 7)2[4(x - 7) + 3(2x + 1)]

Now combine the terms in the brackets and pull out the common factor there, and you'll have what you need.
 
thank you kind sir, you have saved me many hours of frustration!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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