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I didn't get this method. $3a^2-9a+6 = 3(a^2-3a+2) = 3(a^2-1a*3+3^2-3^2+2)=$ $= 3[(a-3)^2-7]$, then?

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Yes, then what? What are you trying to accomplish? –  Hans Lundmark Oct 24 '10 at 13:26
By the way, the last equality is wrong. $(a-3)^2=a^2-6a+9$, not $a^2-3a+9$. –  Hans Lundmark Oct 24 '10 at 13:34
@Hans to continue... The answer is 3(a-2)(a-1) but how to get it? –  lam3r4370 Oct 24 '10 at 13:38
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4 Answers

up vote 7 down vote accepted

Ah, so you want to factor the polynomial. Here's how:

$$3(a^2-2 \frac{3}{2} a + 2) = 3\left(\left(a-\frac{3}{2}\right)^2 - \frac{9}{4} + 2\right) = 3\left(z^2 - \frac{1}{4}\right)$$ $$= 3\left(z-\frac{1}{2}\right)\left(z+\frac{1}{2}\right) = 3(a-2)(a-1),$$ where I temporarily used the substitution $z=a-\frac{3}{2}$.

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And of course, when you are done you should multiply out $3(a-2)(a-1)$ to check that you really get back to the polynomial that you started with! (Just to verify that you haven't made any errors during the calculation.) –  Hans Lundmark Oct 24 '10 at 13:55
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There is also a simple way that involves no fraction calculation,it goes like this :

$ 3a^{2}-9a+6 $

$ = 3a^{2}-6a-3a+6 $

$ = a(3a-6)-(3a-6) $

$ = (3a-6)(a-1) $

$ = 3(a-1)(a-2) $

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Simple once you know the answer. ;) Not much of a general method, however. –  Hans Lundmark Oct 24 '10 at 17:13
I disagree,this is how I was taught to factorize a quadratic equation in school :) –  Quixotic Oct 24 '10 at 23:05
So how would you do $3a^2-9a+5$? –  Hans Lundmark Oct 25 '10 at 9:18
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You can guess and then confirm that $a=1$ is a zero of $3a^{2}-9a+6$. Thus you can factor it as


and solve for $x$. Comparing the coefficients of $a^{2},a,a^{0}$, you get


which is equivalent to $x=2$ and $1+x=3$. Hence $x=2$ and


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If you desire to factor it by way of completing the square then it is simpler to first multiply by 4, namely

$$\rm\ \ 4\:(a^2-3\: a+2)\ =\ (2\:a-3)^2 - 1\ =\ (2\:a-2)\:(2\:a-4)$$

Finally, divide both sides through by $\rm\ \: 4\ =\ 2\cdot 2$

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