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#### DeusAbscondus

##### Active member

- Jun 30, 2012

- 176

$$f(x)=(3x+1)^2(2x-3)^3 \text{ by using the product method}$$

Here is my working out so far, using product rule:$u'v+uv'$

$$\frac{d}{dx} (3x+1)^2(2x-3)^3= 2(3x+1)\cdot 3 *(2x-3)^3+(3x+1)^2\cdot3(2x-3)^2\cdot2$$

Simplified: $$f'(x)=6(3x+1)(2x-3)^3+6(3x+1)^2(2x-3)^2$$

At this point, I have a question:

1. is it best practice to factor out $(2x-3)^3$ or $(2x-3)^2$?

Is there a general answer to this question, valid for all such factoring situations?

To continue with my calculations, I will factor out the lower exponential factor:

$$\Rightarrow 6(3x+1)(2x-3)^2[(2x-3)+(3x+1)]$$

$$\Rightarrow (18x+6)(2x-3)^2(5x-2)$$

Final question: is there any obvious problem with (or improvement to be made) in the way I have set this out? something I'm doing which could be avoided/changed so as to avoid careless errors creeping in?

I ask, because this took me an inordinate amount of time to get right.

(It is correct, is it not, by the way?)

thanks,

DeusAbs