HELP! Factoring!

ninjuhtime · #1 $Old$ November 4th, 2008, 11:55 PM

I know how to do the simply factoring such as for $9x^2 - 25$ , it is $(2x-3)(2x+3)$ but I don't get...

How to factor these:

and do some of these can't be factored?

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Thank you! :D

terr13 · #2 $Old$ November 5th, 2008, 02:08 AM

Some of these cannot be factored in the reals.

An easy trick when the $x^2$ has no numbers in front of it is to look at the factors of the last number, and see if any of their sums is the middle number. If the last number is negative, then you look at the differences of the positive factors. This is a bit confusing...

Some examples
$x^2 +5x -14$
14 = 1*14 or 2*7. Do any of these differences (the sign of the last number is negative) make 5? Yes, $7-2 = 5$
so
$x^2 +5x -14 = (x-2)(x+7)$

Another one
$d^2 -4d +3$
3 is prime and positive, so you are looking at the sum of 1 and 3 to make -4. But -3*-1= 3, and -3-1=4, so
$d^2 -4d +3 = (d-3)(d-1)$

Unless I'm mistaken, only the first one cannot be factored in the reals.

terr13 · #3 $Old$ November 5th, 2008, 02:28 AM

If it is really urgent, here are the answers, although I hope you spend the time to understand it rather than just copying it.

Quote:

Originally Posted by ninjuhtime $View Post$

ninjuhtime · #4 $Old$ November 5th, 2008, 08:36 AM

Thank you terr13!
And I will understand it now that I have the answers.

Plus, I have to understand it anyways xD
...tests :|

tsal15 · #5 $Old$ November 5th, 2008, 08:46 AM

Quote:

Originally Posted by ninjuhtime $View Post$

Thank you terr13!
And I will understand it now that I have the answers.

Plus, I have to understand it anyways xD
...tests :|

hey ninjuhtime, u mite not b up 2 the chapter yet, but the first one, $m^2 + 9$ actually is factorizable only by using complex numbers, so the answer would be $x=3i$ , where $i = \sqrt{-1}$ and the other equals $-7 \pm \sqrt{2}i$

If you're interested

ninjuhtime · #6 $Old$ November 5th, 2008, 09:39 PM

Oh! I found out that one of the problems you said wasn't factorable can actually be factored.

$r^2 - 14r - 51 = (r-17)(r+3)$

...took some time

tsal15 · #7 $Old$ November 5th, 2008, 10:44 PM

Quote:

Originally Posted by ninjuhtime $View Post$

Oh! I found out that one of the problems you said wasn't factorable can actually be factored.

$r^2 - 14r - 51 = (r-17)(r+3)$

...took some time

no. Be very careful.

thats a different question, ninjuhtime. the one u gave is $r^2 -14r +51$ $hence + 51$

ninjuhtime · #8 $Old$ November 6th, 2008, 12:22 AM

Quote:

Originally Posted by tsal15 $View Post$

no. Be very careful.

thats a different question, ninjuhtime. the one u gave is $r^2 -14r +51$ $hence + 51$

Oh, whoops! I must have gave the wrong question or typed it wrong then... sorry, my fault...