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**help****Guest**

Factoring (X^2 + 4)

I know it can be done.

**hey****Guest**

Back in High School in my Pre Cal class, I was shown how to factor something like

(X^2 + 4)

Now I can't think of a way how to factor it.

**nu****Guest**

if you want to factor x^2-4 is easy: (x^2 - y^2) = (x + y)*(x - y), in your case (x+2)*(x-2)

if you want to factor x^2+4 you have to use Ruffini, or interpret this as (x^2 - y^2). The y in this case would be √-4=j√4:

(x^2-(√-4)^2) = (x + √-4 )*(x - √-4 ) = (x + j√4)*(x - j√4), j indicates it is an imaginary number (j = √-1).

nu

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Yeah, always remember the differance of two squares. x[sup]2[/sup] - y[sup]2[/sup] = (x + y)(x - y). Always be on the look out for numbers with integer square roots. For instance x[sup]2[/sup] - 25 = x[sup]2[/sup] - 5[sup]2[/sup] = (x+ 5)(x-5). Also variables with even exponants other (then zero) can always be converted to a square. x[sup]4[/sup] = (x[sup]2[/sup])[sup]2[/sup]

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I guess [sup] 2 [/sup] doesn't work here. :-(

Hmm... x[sup]2[\sup]

A logarithm is just a misspelled algorithm.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,664

LOL, maybe I should put that on my to-do list.

If you don't mind I will experiment with your post using the "math" tag:

mikau wrote:

Yeah, always remember the differance of two squares.

.Always be on the look out for numbers with integer square roots. For instance

.Also variables with even exponants other (then zero) can always be converted to a square.

Example of the first one:

`[math]\Large x^2 - y^2 = (x + y)(x - y)[/math]`

Now, using the math tag makes things a *little* slower, so best to use when other options fail. I often just cut and paste the symbols from the top of the forum, but there is no "^4", so that is where the math tag could help.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Thanks, man. :-)

A logarithm is just a misspelled algorithm.

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