Talk:Factorization
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| Mathematics grading: | B Class | Mid Importance | Field: Algebra | ||
| Needs more prose to accompany equations; some sections need expanding Tompw 14:56, 5 October 2006 (UTC) | |||||
Whoa! Language alert!
- 15 factors into primes (verb)
- x2 - 4 factorises (verb)
- In mathematics, factorization (noun)
- The aim of factoring (noun)
Is there a good reason why there are two flavours of each? -- Tarquin 21:25 Sep 22, 2002 (UTC)
Unfortunately, both factor and factorize are used as synonymous verbs, each being more common in a different context, and each having its own noun form. When discussing the problem of breaking down large numbers, "factorize" is almost always used. In all other contexts it's usually "factor". I prefer the latter because it's shorter, but use the former when talking about the problem for large integers.
There's also the difference between using an "s" or a "z". That's purely a British vs. American issue, so it would be fine to standardize one way or the other on a given page.
I think there is a fast method of factoring integer into primes, but it requires a quantum computer.
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[edit] Must be worked on (anyone know what the "special rules" are ?):
[edit] Table method (for quadratics)
A less used (hard to teach, easy to learn) method often involves creating a multiplication table.
For example, let's work with 6x2 - 17x + 12.
Multiply first and last terms. (72x2)
What multiplies into 72 (first term) and adds up to -17(x) (middle term)?
-9 and -8
In the table, place the first term in the first box and the last in the last box. Fit the (in this example) -9 and -8 in the remaining boxes. Find the GCF up and down and side to side for each row for the answer.
| 6x2 | -8x |
| -9x | 12 |
The answer would be (3x-4)(2x-3)
This method is very decisive and much faster than the others. However, there are a few special rules surrounding the method, but when all of the rules are followed, it works every time.
[edit] Sum/difference of two cubes
I'm here to learn, but shouldn't the factorization be (x-10)(x^2+10x+100) to expand to (x^3 - 1000)? Sparky 21:10, 15 April 2006 (UTC)
- You're so right. My apologies! --Mets501talk 21:31, 15 April 2006 (UTC)
- Haha no problem! Keep up the good work. Sparky 22:44, 15 April 2006 (UTC)
Also here is a problem: x^2-y^2+8y+4x-12 should I Factori "x^2-y^2" or "y^2+8y-12"first?
Factor it to
- x2 + 4x + 4 − (y2 − 8y + 16) = 12 + 4 − 16
- (x + 2)2 − (y − 4)2 = 0
—Mets501talk 15:14, 28 May 2006 (UTC)
FYI, I'm finding the description here very confusing. I've reread it several times. Is there a typo in the description of a^n - b^n and a^n + b^n - maybe left out an "even" in the wording? Anyway, the description is confusing.
[edit] Order of sections
I would have expected that the prime factorisation of integers would be before factorisation of polynomials, etc. JPD (talk) 16:56, 7 August 2006 (UTC)
- Yes, you're right; that would be more logical. I've changed it. —Mets501 (talk) 19:00, 7 August 2006 (UTC)
[edit] difference of two cubes,forth powers,fifth powers,etc.
an − bn = c2 − d2 like 73 − 43 = 482 − 452 Bhowmickr 07:15, 28 August 2006 (UTC)
[edit] Ratio Method for factoring quadratics
A teacher at my schoool came up with this method. I like it.
You multiply (A)(C). Use B. Find two numbers that multiply to give AC, and add to give B. Let's say they are Z and Y
Now make two columns on paper...each with a ratio. A will be the first number in both ratios. Z will be the second number in the first ratio, and Y will be the second number in the second ratio.
Reduce if possible. Say you end up with A:Z and A:Y. the answer will be (a+z)(a+y).
This is what they teach in grade 10. and it's easier/more convenient than the methods described here. jmatt1122 CVU (Talk) 20:49, 4 November 2006 (UTC)
