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#1 2009-03-25 10:26:07

helppme
Guest

factoring

(x-3)/(x^2-2x+15)

for some reason i cant get the bottom to factor correctly =\

#2 2009-03-25 10:37:17

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: factoring

The bottom of that won't factor. (Not with integers, anyway)
If it was -x²-2x+15, you'd get a much nicer answer.


Why did the vector cross the road?
It wanted to be normal.

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#3 2009-03-25 10:39:08

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

Re: factoring

Are you sure you copied the problem correctly? The denominator doesnt factor. It doesnt have any real roots.

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#4 2009-03-25 10:42:44

helppme
Guest

Re: factoring

yeah, i guess thats my answer then, lmao
nooo real roots(:

#5 2009-03-26 16:25:13

smiyc86
Member
Registered: 2009-03-19
Posts: 78

Re: factoring

helppme wrote:

yeah, i guess thats my answer then, lmao
nooo real roots(:

always check for b^2 - 4ac in the equation ax^2+bx+c = 0

when b^2 - 4ac < 0 u wont get real roots ....
                       = 0 the roots will be real and equal ...
                       > 0 the roots will be real and unequal ...

even for b^2 - 4ac > 0 ,

if it is a square of a rational number, the roots will be rational
if it is not, the roots will be irrational.

cool


I love Maths and Music ... dunno which more wink

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#6 2009-04-12 10:55:05

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: factoring

Hi

These are the factors:

so their are no real factors.

Last edited by bobbym (2009-04-28 07:52:25)


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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