# How to factor hard derivatives to find critical and inflection points

Derivatives became hard for me now that I need to factor them as much as possible in order to find critical and inflection points, Is there any easy way or steps to follow to factor derivatives ?

When I start trying to factor I just get stuck

hard derivatives examples:

$$f''(x) = \frac{-12(x^2 + 3)^2 - (-48x(x^2 + 3)x)}{((x^2 + 3)^4))}$$

The final result is

$$f''(x) = \frac {36(x+1)(x-1)}{(x^2+3)^3}$$

$$f''(x)= 4(3x(x - 5)^2 + (x - 5)^3) + 4(x - 5)^3$$ the final result is $$f''(x)=20(x-5)^2(x-2)$$

• This is far more a question of basic algebra skills and nothing to do with calculus. Being able to recognize and group common factors in your second problem (i.e., $(x-5)^2$) should be second nature. When I taught students calculus the single biggest source of error was a poor functional grasp of algebra, with the second being that of trigonometry. Brush up on these skills, they will serve you well. Aug 18, 2017 at 0:23
• @adfriedman You are right, I had a bad high school, now I suffering because of it, Could you make a list of important topics of algebra that I should study to make factoring less painful. Aug 18, 2017 at 0:32
• I'd honestly just suggest going through exercises in any decent high school algebra textbook. Start off with a couple of the first few in an exercise set to confirm you understand the topic and then immediately move to solving hardest (which are generally at the end). That way you don't waste your time on trivialities, but maintain a fast pace. I really don't think it would take more than a weekend to be far more confident in these skills. Aug 18, 2017 at 3:48
• As far as specific skills: factoring and distributing, polynomial division (synthetic isn't really a necessary skill), multiplying by conjugate of denominator, rationalizing denominator. There are probably plenty more, but those are the skills that immediately come to mind that I believe to be essential but people are often lacking. Aug 18, 2017 at 3:55

Whenever you use the quotient rule repeatedly, you end up with powers of the denominator all over the place. In your example, we have $(x^2+3)$ occurring in both terms of the numerator (see how the numerator consists of two large terms?), and also in the denominator. Calling that factor $A$, we have:

$$\frac{-12A^2 + 48x^2A}{A^4}$$

See how you can reduce that, by canceling one power of $A$?

$$\frac{-12A + 48x^2}{A^3} = \frac{-12(x^2+3)+48x^2}{(x^2+3)^3} = \frac{36x^2-36}{(x^2+3)^3}$$

Similarly on the second example, there are powers of $(x-5)$ all over the place. Call it $B$:

$$4(3xB^2 + B^3) + 4B^3 = 12xB^2 + 8B^3 = B^2(12x+8B)$$

Now, we can simplify $12x+8B$:

$$12x+8(x-5) = 20x-40 = 20(x-2)$$

• What a wonderful answer, thank you very much. But one question, what happened with the x after(x^2+3) Aug 18, 2017 at 0:35
• It combined with the $48x$ to make it $48x^2$. :) Aug 18, 2017 at 0:36
• Great I didn't realize, thank you man. Aug 18, 2017 at 0:37