Factoring????????????????????

👨🏻‍🏫 學術台數學

An ESL Learner

2009-07-27 11:31 am

Can x^2 + 6x + 4 be factored?

Some people said "no" since it is prime.
Some people said "yes" since -(-x + 3 - √5)(x + 3 + √5) is a pair of multiples.

So, what is the answer? Yes or no?

THX.

更新1:

Sorry, it should be -(-x + √5 - 3)(x + √5 + 3), not -(-x + 3 - √5)(x + 3 + √5).

回答 (7)

[匿名]

2009-07-27 11:46 am

✔ 最佳答案

It all depends on what you are willing to regard as acceptable for the problem in hand - integer, rational, real or complex roots can all be valid. There is no right or wrong in the general case.

Note added: And factoring merely means writing the equation in terms of the roots in the form

(x - r1).(x - r2).....(x - rn) = 0

it does not place any limitation on the form of the roots, or the means by which they were obtained.

參考: For definition, see Wiki 'Factorization'

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[匿名]

2009-07-27 6:41 pm

No I don't think radicals count as factors in these cases. Factors are rational numbers. 5 is a prime number, it has no factors you can't say sqrt5 is a factor of 5

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[匿名]

2009-07-27 6:54 pm

No it can't be factored as we do not have factors of 4, the sum of which is equal to 6.
Of course, it can be solved by using quadratic formula or by completing square method

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[匿名]

2009-07-28 6:49 am

yes

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[匿名]

2009-07-27 6:44 pm

(x-3+/-sqrt5)

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[匿名]

2009-07-27 6:42 pm

yes....factoring is possible

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[匿名]

2009-07-27 6:41 pm

x^2 + 6x + 4
= x^2 + 6x +9 -9+4
= (x+3)^2 -5
=(x+3)^2-(sqrt5)^2
=(x+3+sqrt5)(x+3-sqrt5)

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