I didn't get this method. \begin{equation*} 3a^2-9a+6=3(a^2-3a+2) \\ =3(a^2-1a*3+3^2-3^2+2)= \\ = 3[(a-3)^2-7], \end{equation*} then?
Perfect square method
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2Yes, then what? What are you trying to accomplish? – 2010-10-24
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1By the way, the last equality is wrong. $(a-3)^2=a^2-6a+9$, not $a^2-3a+9$. – 2010-10-24
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0@Hans to continue... The answer is 3(a-2)(a-1) but how to get it? – 2010-10-24
4 Answers
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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4And 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.) – 2010-10-24
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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3Simple once you know the answer. ;) Not much of a general method, however. – 2010-10-24
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1I disagree,this is how I was taught to factorize a quadratic equation in school :) – 2010-10-24
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1So how would you do $3a^2-9a+5$? – 2010-10-25
You can guess and then confirm that $a=1$ is a zero of $3a^{2}-9a+6$. Thus you can factor it as
$3a^{2}-9a+6=3(a-1)(a-x)=3x-3a-3ax+3a^{2}$
and solve for $x$. Comparing the coefficients of $a^{2},a,a^{0}$, you get
$3=3,3x=6,-3-3x=-9$,
which is equivalent to $x=2$ and $1+x=3$. Hence $x=2$ and
$3a^{2}-9a+6=3(a-1)(a-2)$.
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 $\rm\ \: 4\ =\ 2\cdot 2$