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In my maths class I am doing double derivative to find concavity of the equation if I graph it, and getting these big functions.

Plugging in even on online calculators skips from this big thing to this simplified thing as seen in screenshot below:

I can't figure out for the life of me how to get to that simplification. I missed this question on my last exam, can you please help me understand the magic steps that happen here to get to the "simplified" result.

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    $\begingroup$ You have an $(x-1)^4$ in your original denominator and you want to get an $(x-1)^3$ in your target denominator. That suggests you should factor an $x-1$ out from your numerator. As luck would have it, you have two terms in the numerator - both, with an $x-1$. $\endgroup$
    – Mark McClure
    May 23, 2016 at 3:42
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    $\begingroup$ start with cancelling out$ (x-1)$ from the numerator and the denominator. $\endgroup$
    – mike
    May 23, 2016 at 3:42
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    $\begingroup$ seperate out fractions delete like terms. cross multiply recombine cross out like terms $\endgroup$
    – shai horowitz
    May 23, 2016 at 3:43
  • $\begingroup$ Thanks very much everyone! I have up voted you all! $\endgroup$
    – Noitidart
    May 23, 2016 at 4:04
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    $\begingroup$ @Noitidart Thank you very much! $\endgroup$
    – zz20s
    May 23, 2016 at 4:09

2 Answers 2

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We have

$$ \frac{ (2x-2)(x-1)^2-2(x-1)(x^2-2x) }{((x-1)^2)^2} $$

$$ \Leftrightarrow \frac{ (2x-2)(x-1)^2-2(x-1)(x^2-2x) }{(x-1)^4} $$

$$\Leftrightarrow \frac{ (x-1)((2x-2)(x-1)-2(x^2-2x)) }{(x-1)^4} $$

$$\Leftrightarrow\frac{ ((2x-2)(x-1)-2(x^2-2x)) }{(x-1)^3} $$

$$\Leftrightarrow\frac{ 2x^2-4x+2-2x^2+4x }{(x-1)^3} $$

$$ \therefore \frac{2}{(x-1)^3} $$

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    $\begingroup$ Thank you @dydxx so much for this, looking at this and and zz20s below really helped me understand. I was getting tripped up in trying to open absolutely everything. How did you know what to open and what to keep non opened? Like how did you know to expand (x-1)^2 but leave (x-1)^3? $\endgroup$
    – Noitidart
    May 23, 2016 at 4:03
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    $\begingroup$ Dear dydxx I wish I could accept your solution as well. Seeing your solution and also solution of @zz20s below really made things clear to me. $\endgroup$
    – Noitidart
    May 23, 2016 at 4:06
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    $\begingroup$ I basically saw a common factor on the numerator and the denominator. So I noticed I could factor something out on the numerator so (x-1) both cancel out and you are left with something simple :). You can tick my one if it was more helpful than the other guys but it's your choice! $\endgroup$
    – bigfocalchord
    May 23, 2016 at 4:08
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    $\begingroup$ Knowing when to expand and when to look for factors comes from practicing simplifications like this. In general, First look for common factors in the fractions. Then, when no more factors are apparent, try to expand and see what cancels in the expression. Finally, you may only accept one answer. It is entirely your choice which was more helpful to you. $\endgroup$
    – zz20s
    May 23, 2016 at 4:08
  • $\begingroup$ A sincere thank you to dydxx and @zz20s for those tips! I will practice that for my upcoming exam! :) I couldn't simplify so tried to set that monster = 0 to find concavity and just failed at doing that haha $\endgroup$
    – Noitidart
    May 23, 2016 at 4:09
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$$\frac{(2x-2){(x-1)^2}-2{(x-1)}(x^2-2x)}{{((x-1)^2)^2}}=\frac{(2x-2)\color{red}{(x-1)^2}-2\color{red}{(x-1)}(x^2-2x)}{\color{red}{(x-1)^4}}=\frac{(2x-2){(x-1)}-2(x^2-2x)}{(x-1)^3}=\frac{2x^2-2x-2x+2-2x^2+4x}{(x-1)^3}=\frac{2x^2-4x+2-2x^2+4x}{(x-1)^3}=\frac{\color{red}{2x^2}-\color{blue}{4x}+2-\color{red}{2x^2}+\color{blue}{4x}}{(x-1)^3}=\frac{2}{(x-1)^3}$$

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  • $\begingroup$ Thank you very much sir, the colors really helped me understand! I was opening everything even on the bottom and I was getting majorly tripped up there. $\endgroup$
    – Noitidart
    May 23, 2016 at 4:02
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    $\begingroup$ Glad to help! In general, it is usually a good start to look for common factors in the numerator and denominator. $\endgroup$
    – zz20s
    May 23, 2016 at 4:03
  • $\begingroup$ Thanks sir I will try do that as my first step. I think my mistake was my first step was to open up (distribute everything) absolutely everything. $\endgroup$
    – Noitidart
    May 23, 2016 at 4:05

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