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Topic: Factoring
Replies: 52   Last Post: Jan 16, 2012 8:23 PM

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 Marnie Northington Posts: 586 Registered: 12/13/04
Re: Factoring
Posted: Jan 11, 2012 2:33 PM

"Dave L. Renfro" <renfr1dl@cmich.edu> wrote in message
> John Smith wrote:
>

>> A student asked me how to factor x^3-3x^2-4x+12
>> So I took a guess (correctly it seems) that since
>> 12 factors as 2,2,3 one of the factors was likely
>> to be x-2. Dividing by x-2 gives x^2-x-6 which is
>> easy to factor into (x-3)(x+2) so the answer is
>> (x-2)(x-3)(x+2).
>>
>> Is there a way to do that whithout guesswork?
>>
>> Suppose I had a more complicated example such as
>> (2x+1)(3x-2)(4x+5) I haven't multiplied it out but
>> how would I find the factors if I didn't know

>
> When I read this there were already 35 posts in this
> thread and, unless I overlooked it, none of the replies
> mentioned "factoring by grouping". At least in the U.S.,
> basic algebra classes (algebra 2 in high school; intermediate
> algebra and college algebra in community colleges, 4-year
> colleges, and universities) introduce and use only the
> following methods for factoring: common factor, difference
> of squares, quadratic trinomials, factoring by grouping
> (in many classes/texts this method is used, via the
> "AC Method", to factor quadratic trinomials), difference
> of cubes, and sum of cubes (which can be viewed as a
> difference of cubes via a^3 + b^3 = a^3 - (-b)^3). Pretty
> much anytime in one of these courses you have to factor
> a polynomial with 4 terms, you begin by trying factoring
> by grouping.

Yes it does seem obvious now that for my original example I should have
looked for pairs of terms which have a common factor. So x^3-2x^2+4x-8
factors as (x-2)(x^2+2) which can't be factored further without complex
numbers.
I may have learned this method in the past but forgot it because it was so
long ago. It was likely just part of general algebra then rather than having
a specific name.

I certainly remember having to factor quadratics.

I hadn't used the rational root theorem before but after some of the replies
in this thread I figured out that if Ax^3+Bx^2+Cx+D factors as
(a1x+b1)(a2x+b2)(a3x+b3)
then A=a1a2a3 B=(a1b2a3+a3b1a2+b3a1a2) C=(a3b1b2+b3a1b2+b3a1a2) and D=b1b2b3
I think the rational root theorem follows from the A and D part of that, at
least for cubics.
This led me to find my own method of obtaining the answer to the second
example. The method obviously has many limitations and something tells me
that some people here may have a fit when they see it.

I think I have now got everything I wanted from this thead and I would like
to thank everyone who replied. I'm off to install Python.

5 CLS
10 INPUT "Coefficient of x^3"; C3
15 INPUT "Coefficient of x^2"; C2
20 INPUT "Coefficient of x^1"; C1
25 INPUT "Coefficient of x^0"; C0
30 FOR A1 = -10 TO 10
35 FOR B1 = -10 TO 10
40 FOR A2 = -10 TO 10
45 FOR B2 = -10 TO 10
50 FOR A3 = -10 TO 10
55 FOR B3 = -10 TO 10
60 IF (A1 * A2 * A3) <> C3 THEN 100
65 IF (A1 * B2 * A3 + A3 * B1 * A2 + B3 * A1 * A2) <> C2 THEN 100
70 IF (A3 * B1 * B2 + B3 * A1 * B2 + B3 * B1 * A2) <> C1 THEN 100
75 IF (B1 * B2 * B3) = C0 THEN PRINT A1; B1; A2; B2; A3; B3
100 NEXT B3, A3, B2, A2, B1, A1

> Regarding your "more complicated example",
> the U.S.), usually called precalculus (although some
> honors high school algebra 2 courses and some more
> comprehensive college algebra courses will include it),
> students may get introduced to the rational root theorem,
> which someone did bring up in this thread. I say "may get
> introduced", because over the past 20+ years this has more
> and more been eliminated from the syllabus of such courses.
>